Chemistry notes vtu

Sri Shridevi Charitable Trust( R) SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY, TUMKUR-572106 ENGINEERING CHEMISTRY ENGINEERING CHEMISTRY ENGINEERING CHEMISTRY ENGINEERING CHEMISTRY CLASS NOTES CLASS NOTES CLASS NOTES CLASS NOTES 1. Dr. CHANDRASEKHAR. N H.O.D of chemistry 2. Miss. SUJATHA.K 3. Miss. SHWETHA S. RAO NAME :____________________ USN :____________________ BRANCH : ____________________ SECTION : ____________________ SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 1 CONTENTS UNIT TITLE PAGE NO I CHEMICAL ENERGY SOURCES 04-24 II ELECTROCHEMICAL ENERGY SYSTEMS 25-43 III CONVERSION AND STORAGE OF ELECTROCHEMICAL ENERGY 44-56 IV CORROSION SCIENCE 57-71 V METAL FINISHING 72-84 VI LIQUID CRYSTALS AND THEIR APPLICATIONS 85-98 VII HIGH POLYMERS 99-115 VIII WATER TECHNOLOGY 116-132 IX MODEL QUESTION PAPER 133-140 SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 2 Syllabus Syllabus Syllabus Syllabus PART A Unit - I Electrode Potential and Cells Introduction, Differentiation between galvanic and electrolytic cells, Construction of galvanic cell, EMF of a cell , Origin of single electrode potential, Sign convention and cell notation, Standard electrode potential, Derivation of Nernst equation for single electrode potential. Types of electrodes: Reference electrodes – Primary and secondary , Limitations of standard hydrogen electrode, Construction and working of calomel electrode and Ag – AgCl electrode, Measurement of single electrode potential, Numerical problems on electrode potential and EMF of a cell, Ion selective electrode: Glass electrode – Construction , Determination of pH of a solution using glass electrode. 7 Hours Unit - II Batteries and Fuel Cells Basic concepts, Battery characteristics, primary, secondary, reserve batteries and super capacitors with examples. Classical batteries: Construction, working and applications of Zn – MnO2 battery, Lead acid storage battery and Ni – Cd battery. Modern batteries: Construction and working and applications of Zn – air, Ni – metal hydride and Li – MnO2 batteries Fuel cells – Differences between battery and fuel cell, construction and working of H2 – O2 and CH3OH– O2 fuel cells 6 Hours Unit - III Corrosion and its control Electrochemical theory of corrosion, Galvanic series, Types of corrosion- Differential metal corrosion, Differential aeration corrosion(Pitting and water line corrosion) , Stress corrosion (caustic embrittlement in boilers), Factors affecting the rate of corrosion. Corrosion control: Inorganic coatings – Anodizing and phosphating, Organic coating – painting, Metal coatings – Galvanizing and Tinning, Corrosion inhibitors, Cathodic protection. 7 Hours Unit – IV SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 3 Metal Finishing Technological importance, Theory of electroplating, Significance of Polarization, Decomposition potential and Overvoltage in electroplating. Effect of plating variables on the nature of electrodeposit. Electroplating process, Electroplating of Copper and Chromium. Distinction between electroplating and electrolessplating, Electolessplating of copper and nickel. 6 Hours PART B Unit - V Chemical fuels and Photovoltaic cells Introduction, Classification of chemical fuels, Calorific value – High and Low calorific values, Determination of calorific value –solid or liquid fuel using Bomb calorimeter, Gaseous fuel using Buoy’s calorimeter, Numerical problems. Petroleum – Cracking by fluidized catalytic cracking process, Reformation of petrol, Octane and Cetane numbers. Knocking – mechanism and harmful effects. Antiknocking agents – TEL, Catalytic converters – Principle and working, Unleaded petrol, Power alcohol and Biodiesel. Photovoltaic cells – Production of solar grade silicon, Doping of silicon, Construction and working of photovoltaic cell, Advantages. 7 Hours Unit - VI The Phase rule and Instrumental methods of analysis Statement of Gibb’s phase rule and explanation of the terms involved, Phase diagram of one component system – water system, Condensed phase rule , Phase diagram of two component system- Eutectic Pb – Ag system and Fe – C system. Application – Desilverization of lead. Instrumental methods of analysis- Theory , Instrumentation and applications of Colorimetry, Potentiometry , Conductometry and Flame photometry. 6 Hours Unit - VII Polymers Types of polymerization – Addition and Condensation, Mechanism of polymerization – Free radical mechanism taking ethylene as example. SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 4 Glass transition temperature ( Tg) . Structure – property relationship. Types of plastics – Thermosetting and thermoplastics. Compounding resins to plastics, Manufacture of plastics by compression and injection moulding, extrusion method. Synthesis and applications of Teflon, PMMA, Polyurethane and Phenol – formaldehyde resins. Elastomers: Deficiencies of natural rubber, Vulcanization of rubber. Synthesis and applications of Neoprene and Butyl rubber, Silicone rubbers. Adhesives: Synthesis and applications of epoxy resins. Polymer composites - Synthesis and applications of Kevlar and Carbon fibers. Conducting polymers – Definition, Mechanism of conduction in Polyacetylene and Polyaniline. 7 Hours Unit - VIII Water Chemistry Impurities in water, Water analysis – Determination of different constituents in water – Hardness, alkalinity, chloride , fluoride , nitrate , sulphate and dissolved oxygen by Winkler’s method. Numerical problems on hardness and alkalinity. Sewage – BOD and COD, Numerical problems, Sewage treatment. Desalination of water – Reverse Osmosis and Electrodialysis 6 Hours Text Books 1. A text book of Engineering Chemistry P.C. Jain and Monica Jain Dhanpatrai Publications, New Delhi. 2. Chemistry In Engineering and Technology ( Vol. 1 &2) J.C. Kuriacose and J. Rajaram. Reference Books 1. Principles of Physical Chemistry B.R. Puri , L.R.Sharma & M.S. Pathania, S. Nagin chand and Co. 2. Text Book of Polymer Science F.W. Billmeyer John Wiley & Sons. 3. Corrosion Engineering M.G. Fontana Mc. Graw Hill Publications. 4. Environmental Chemistry Stanley E. Manahan , Lewis Publishers. 5. Polymer Science V.R. Gowariker , Wiley Eastern Ltd. SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 5 UNIT UNIT UNIT UNIT – –– – I II I CHEMICAL ENERGY SOURCES: FUELS 1.1 Define a fuel. Explain the classification of fuels with examples. 5 Marks A fuel is defined as naturally occurring or artificially manufactured combustible carbonaceous material which serves particularly as source of heat and light and also in few cases as a source of raw material. Classification of fuels Fuels are classified into a two types. 1) Based on their origin they are classified into a) Primary fuels b) Secondary fuels. a) Primary Fuels: There are naturally occurring fuels which serves as source of energy without any chemical processing. Ex: Wood, Coal, Crude oil, Natural gas, Peat, Lignite, Anthracite.. b) Secondary Fuels: - They are derived from primary fuels & serves as source of energy only after subjecting to chemical processing. Ex: Charcoal, Coke, produsergas, Petrol, Diesel etc., 2) Bases on their physical state fuel are classified into a) Solid b) Liquid c) Gaseous fuels. SOLID LIQUID GASEOUS Primary Fuels Wood, Coal, Peat, Anthracite Crude oil Natural gas. Secondary Fuels Coke, Charcoal Petrol, Gasoline, Diesel LPG, produsergas, Coal gas. 1.2 Define Calorific Value. Explain the types 04 Marks Calorific value is defined as the amount of heat liberated when a unit mass of fuel is burnt completely in presence of air or oxygen. SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 6 Calorific value is of two types as follows:- 1) Higher calorific value. (HCV) or Gross calorific value. (GCV) 2) Lower calorific value. (LCV) or Net calorific value. (NCV) 1) HCV: - It is the amount of heat liberated when a unit mass of fuels burnt completely in the presence of air or oxygen and the products of combustion are cooled to room temperature. Here it includes the heat liberated during combustion and the latent heat of steam. Hence its value is always higher than lower calorific value. 2) LCV: - It is amount of heat liberated when a unit mass of fuel is burnt completely in the presence of air or oxygen and the product of combustion are let off completely into air. It does not include the latent heat of steam. Therefore it is always lesser than HCV. NCV = HCV – Latent heat of steam. = HCV –0.09X % H2 X 587 cal/g 1.3. Mention the SI units of calorific value. 2 Marks In SI system the units of calorific values for solid fuels are expressed in J/Kg and for gaseous and liquid are expressed in J/m 3 . 1.4. Explain the determination of calorific value of solid fuel using Bomb calorimetric method. 6 Marks SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 7 A small quantity of a fuel is weighed accurately (M Kg) and is placed in the Bomb. The bomb is placed in known amount water taken in a copper calorimeter. The initial temp of water is noted as a t1 0 C with the help of thermometer. Oxygen gas is pumped under pressure 20 to 25 atm through the O2 valve provided. The fuel is ignited by passing electric current through the wires provided. As the fuel undergoes combustion and liberates heat, which is absorbed by surrounding water. The water is stirred continuously to distribute the heat uniformly and the final temp attained by water is noted t2 0 C. & gross calorific value of the fuel is calculated as follows:- Calculation: Mass of the fuel = M Kg. Initial temp of the water = t1 0 C Final temp of the water = t2 0 C Change in temp = t = (t2 – t1) 0 C Specific heat of water = S Water equivalent of calorimeter = W Kg. GCV = W x S x t J/Kg or M GCV = (W+w) x S x t J /Kg M NCV = GCV – 0.09 x %H2 x 587 cal/g PROBLEMS: 1) Calculate calorific value coal samples from the following data. 6 Marks Mass of the coal = 1g. Water equivalent of calorimeter = 2 Kg. Specific heat of water = 4.187 J/Kg/c. Rise of temperature = 4.8 0 C. Solution: GCV = W x S x t J/Kg M = 2 x 4.187 x 4.8 0.001 = 40195.2 KJ/Kg. SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 8 2) A coal sample with 93% carbon, 5% of Hydrogen and 2% Ash is subjected to combustion in a bomb calorimeter. Calculate GCV and NCV Given that. 6 Marks Mass of the coal sample = 0.95g Mass of water in copper calorimeter = 2000g. Water equivalent wt of calorimeter = 700g. Rise in temp = 2.8 0 C Latent heat of = 587 cal/g. Specific heat of water = 1 cal/g/ 0 C GCV = (W+w) x S x t M = (2000+700) x 10 -3 kg x 1 cal/g/ 0 C x 2.8 0 C x 4.184 0.95 x 10 -3 kg = 33295.83 J/kg. NCV = GCV – 0.09 x %H2 x 587x4.184 J/kg. = 33295.83 J/kg – 0.09 x 5 x 587 x4.184 J/kg. = 32190.62 J/kg 3) When 0.84g of coal was burnt completely in Bomb calorimeter the increase in temp of 2655 grams of water was 1.85 0 C if the water equivalent calorimeter is 156g Calculate GCV. GCV = (W+w) x S x t M = (2655+156) x 1.85 x 10-3 x 4.187 0.84 x 10-3 = 25921.26 J/Kg 4) Calculate GCV and NCV of a fuel from the following data. Mass of fuel=0.75g, W=350g. t =3.02 0 C, Mass of water = 1150, % H2=2.8. 6 Marks GCV = (W+w) x t x S M = (1150+350) x10 - x 3 3.02 x 4.184 0.75 x 10 -3 GCV = 25271. 36 KJ/Kg NCV = GCV –0.09 x H x 587 x 4.184 SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 9 = 25271.36 – 0.09 x 2.8 x 587 x 4.184 NCV = 24652.44 KJ/Kg 5) Calculate calorific value of a fuel sample of a coal form the following data. Mass of the coal is 0.6g. Water equivalent wt of calorimeter is 2200g. Specific value 4.187 Kg/KJ/C rise in temperature = 6.52 0 C. 3 Marks GCV = (W1+W2) x S x t M = (2200) x 10-3 x 4.184 x 6.52 0.6 x 10-3 = 100025.49 KJ/Kg. 6) Calculate GCV and NCV of a fuel from the following data. Mass of fuel =0.83g, W=3500g. , W = 385 g, t1 =29.2 0 C, t2 = 26.50C, % H2 = 0.7 and S = 4.2 kj/kg/c 6 Marks GCV = (W+w) x t x S M = (3.5 + 0.385) x (29.2 – 26.5) x 4.2 0.83 x 10 -3 GCV = 53079.39 KJ/Kg NCV = GCV –0.09 x H x 587 x 4.184 = 53079.39 – 0.09 x 0.7 x 587 x 4.2 NCV = 52924.07 KJ/Kg SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 10 1.5 Explain the determination of calorific value of gaseous fuel using Buoy’s calorimetric method. Buoy’s Calorimeter The calorific value of a gaseous fuel is determined by burning known volume of gas in steady flow calorimeter. Water is allowed to pass through the tubes surrounding the combustion chamber till the temperatures measured by the thermometers T1 & T2 are the same. The gaseous fuel under steady is introduced into the burner through the gas inlet at a steady rate recorded by the gas meter. The gas is burnt in the combustion chamber and the combustion products are passed over the tubes through which water flows at a constant rate of 6-10 cm 3 / sec. When a temperature of the apparatus reaches a constant value (i.e when T2 shows a constant reading the heat released by the combustion of the fuel in an certain interval of time will be equal to the heat absorbed by water during the same interval of time. Calorific value Q is given by the equation Q = mass of water x rise in temp. x specific heat Volume of gas burnt The rise in temperature should be of the order of about 20 K. Vapors cooled by circulating water are further cooled in condenser tubes to condense water form during the combustion of hydrogen and hydrocarbons in the fuel. Amount of condense water collected in the SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 11 measuring cylinder for the volume of gas burnt could be made use of for calculating the net calorific value. Observations and calculations: i) Volume of gas burnt at NTP in time t = V m 3 ii) Weight of cooling water circulated in time t= W kg iii) Steady temperature of the inflow water = t1 0 C iv) Steady temperature of the outflow water = t2 0 C v) Rise in temperature = (t2-t1) 0 C vi) Weight of water produced from steam condensation = m kg Heat released by the combustion of V cm3 of gas fuel = Heat absorbed by water HCV x V = W (t2-t1) Higher calorific value of the fuel = W (t2-t1) k cal /m 3 V = W (t2-t1) 4.187 kJ/m 3 V m x 587 Latent heat of steam = k cal m3 V W (t2-t1) 4.187 m x 587 x 4.187 Lower (net) calorific value of fuel = - V V =------------- kJ/m 3 Numericals: 1)Calculate gross and net calorific value of a gaseous fuel from the following data obtained from boys experiment: i) Volume of gaseous fuel burnt at STP ----- = 0.09 m 3 ii) Weight of water used for cooling ----- = 25.0 kg iii) Temperature of inlet water ----- = 25.0 0 C iv) Temperature of outlet water ----- = 40.0 0 C v) Weight of water produced by steam condensation ---- = 0.02 kg vi) Latent heat of steam ----- = 587 k cal kg Solution: W x (t2-t1) Higher calorific value of the given gaseous sample = x 4.187 V SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 12 = 25 x (40-25) x 4.187 0.09 = 17445.8 kJ m3 0.02 x 587 x 4.187 Heat released in condensing steam = 0.09 Net calorific value = 546.17 kJ m3 = HCV – latent heat of steam = 17445.8 – 546.17 = 16899.63 kJ m3 2)Calculate the gross and net calorific value of a gaseous fuel at STP from the following data Volume of gas burnt =0.02m 3 Temperature of the gas =293 K Mass of water passing through calorimeter =4.5 kg Rise in temp =18.5 K Absolute pressure of gas =101990 N/m 2 Specific heat of water =4.18J/g/K Amount of water collected =7.5 cm 3 Latent heat of steam at 288K =2.454 kJm 3 Solution: i)Heat absorbed by cold water =mass of water x specific heat x rise in temp =4.5 kg x 4.18 J/g/K x 18.5 K =4500 g x 4.18 J/g/K x 18.5 K =348000 J or 348 kJ ii)Reduce the volume of the gas to that at STP using combined gas law P1V1 P2V2 = T1 T2 101990 N/m 2 x 0.02 m 3 101325 N/m 2 x V = 293 K 273 K V=0.0188 m 3 iii) Calorific value, Q=Heat absorbed by cold water Volume of gas SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 13 =348/0.0188 = 18510 k J/m 3 iv) Net Q: Water condensed during combustion = 7.5cm3 = 7.5 gram Latent heat of steam at 288 K = 2.454 kJ/g Therefore heat to be deducted from gross calorific value 7.5 g x 2.454 kJ/g = 0188 m3 = 979 kJ/m3 Therfore Net Q = 18510 -979 = 17531 kJ/m3 1.6 What is Petroleum cracking? Explain the Fluidized bed catalytic cracking method. 6 Marks Cracking is a process that involves breaking of C-C and C-H bonds in the chains of high boiling hydrocarbons of high molecular weight, to yield simpler, low boiling hydrocarbons of molecular weight. Fluidized bed catalytic cracking method. The catalyst such as alumina or zeolyte is finely powered and mixed with steam and it is pumped to cracking chamber. The feedstock SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 14 (gas oil, heavy oil) is preheated to 500 0 C and forced into the cracking chamber along with the catalyst by a steam blast where a floating turbulent bed forms. In cracking chamber the higher molecular weight hydrocarbons are broken down into simpler fragments. The products are withdrawn and sent to fractionating column for separation. The catalyst becomes inactive after cracking process. It is sent to catalyst activation chamber and it is activated by burning carbon deposited on the catalyst by air blast at 600 0 C. The catalyst can be used again and the process is continuous. 1.7 What is Reformation of Petroleum? Explain the Reformation of Petrol 6 Marks It is a process of bringing structural modifications in the strait chain hydrocarbons (with lower octane number) to increase the octane number and thereby improving the anti-knocking characteristics of petrol. Reforming is a chemical process, which involves modification of the structure of molecules without much change in the molecular masses. Reformation of petroleum involves the following reforming reactions: 1) Isomerisation. 2) Dehydrogenation. 3) Cyclisation & dehydrogenation and 4) Hydro cracking. 1) Isomerisation: - It is a process of converting the straight chain hydrocarbon compound of lower octane number into a branched chain hydrocarbon compound of higher octane number. 2) Cyclisation: Straight chain hydrocarbons undergo cyclisation producing cyclic hydrocarbons which have higher octane number. Ex. CH3 – CH2 – CH2 – CH2 – CH2-CH3 n- Hexane Cyclo hexane SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 15 3) Cyclisation and dehydrogenation: Straight chain hydrocarbons undergo cyclisation to form cyclic compounds, which further undergoes dehydrogenation to form aromatic compounds. Ex. CH3 – CH2 – CH2 – CH2 – CH2-CH3 n- Hexane Cyclo hexane Benzene 4) Hydro cracking: -Straight chain hydrocarbons undergo hydro cracking in presence of hydrogen and platinum catalyst producing low molecular weight gaseous fractions which are removed to improve the octane number. 1.8 What is Knocking? Explain the Mechanism of Knocking 6 Marks The explosive combustion of petrol and air mixture produces shock waves in I.C. engine, which hit the walls of the cylinder and piston producing a rattling sound is known as knocking. Mechanism of Knocking Beyond a particular compression ratio the petrol mixture suddenly burns into flame. The rate of flame propagation increases from 20 to 25m/s to 2500m/s, which propagates very fast, producing a rattling sound. The activated peroxide molecules decomposes to give number of gases products which produces thermal shock waves which hit the walls of the cylinder and piston causing a rattling sound which is known as knocking. The reactions of normal and explosive combustion of fuel can be given as follows taking ethane as an example SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 16 1.9 Define octane number. 2 Marks Octane number is defined as the percentage of isooctane present in a standard mixture of isooctane and n-heptane, which knocks at the same compression ratio as the petrol being tested. Isooctane is the branched chain hydrocarbon has least knocking rate, hence its octane number is arbitrarily fixed as 100. N-heptane a straight chain hydrocarbon has highest tendency to knock hence its octane number is fixed as zero. Octane number of petrol is 80 means it contains 80% by volume isooctane and 20% by volume n- heptane. 1.10 Define Cetane number. 2 Marks It is defined as the percentage of cetane present in standard mixture of a cetane and Alfa- methylnaphthalene, which knocks at the same compression ratio as the diesel fuel being tested. 1.11 What are Antiknocking agents? Explain Leaded & Unleaded Petrol 5 Marks These are the substances added to petrol in order to prevent knocking in I.C. Engines. Ex: TEL Tetra Ethyl led. TML Tetra Methyl led. MTBE Methyl Tertiary Butyl Ether. Leaded Petrol: The petrol containing TEL or TML as anti knocking agents is called leaded petrol. TEL or TML are the very good anti knocking agents but has some disadvantages as follows. a) After combustion lead is deposited as lead oxide on piston and engine walls it leads to mechanical damage. b) Lead is a poisonous air pollutant. c) It spoils the catalyst used in catalytic converter. Unleaded Petrol: The petrol, which contains antiknocking agent other than lead, is known as unleaded petrol. Ex: MTBE is used, as an antiknocking agent in place of TEL or TML and the petrol is known as unleaded petrol. SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 17 1.12 Explain working and principle of Catalytic converter Principle: Use of catalytic converter in the internal combustion engines of automobiles helps in cleaning up the exhaust emissions. Such converters built into the automobile engines promote oxidation-reduction cycles and ensure complete combustion of CO, NOx and HC. Working: The following fig illustrates the action of catalytic converters: use of catalytic converters in 2 stages helps in the elimination of pollutants from exhaust gases before they are discharged into the atmosphere REDUCTION STAGE I AIR OXIDATION STAGE II HC,CO,NO From exhaust gases Released to atmosphere N2 CO2 H2O Pt-H=Platinum hydrogen compound Pt-O= Platinum Oxygen compound HC=Hydrocarbon, CO=Carbon monoxide NO=Nitrogen oxide, N=Nitrogen NH3=Ammonia Flow chart of Catalytic converters for treating auto missions i)In the first converter, nitrogen oxides are reduced to nitrogen and ammonia in the presence of finely divided catalyst Pt ii)In the second converter, air is introduced to provide an oxidizing atmosphere for complete oxidation of CO & HC into CO2 and H2O in the presence of Pt catalyst 1.13 Explain the manufacture of Synthetic petrol by Bergius Process? 6 Marks Coal is considered as black diamond because of its high utility value. The principal elements present in coal are carbon and hydrogen. It is Pt-H+NO Pt- O+N2+NH3 Pt-O+HC+CO Pt- H+CO2+H2O SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 18 therefore coal can be converted into liquid hydrocarbons from which much demand and needed petrol can be obtained. The synthetic petrol can be manufactured by the following methods. 1) Direct hydrogenation of coal. (Bergius process) 2) Direct hydrogenation of coal. (Fischer-Tropsch process) Bergius process: (Direct conversion of coal) In this method the lignite is grinded into a fine powder. This powder is mixed with heavy oil & made into a paste. Iron oxide or nickel is added as catalyst. The mixture is pumped into a reactor. The temperature of the reactor is maintained about 500-550 0 C & a pressure is maintained about 250 atmospheres. H2 gas is passed through the reactor. Lignite under goes hydrogenation to form a mixture of hydrocarbons. These mixture of hydrocarbons are passed through a fractionating column to for separation to obtain synthetic petrol. 2) Fischer-Tropsch process: (Indirect conversion of coal) This method involves the following steps a) Production of water gas: Water gas (CO+H2) is obtained by passing steam over white hot coal. C + H2O (g) CO + H2 (water gas) b) Production of synthesis gas the water gas obtained above is freed from dust, H2S and organic Sulfur compounds and blended with hydrogen to form synthesis gas (CO + 2H2). c) Hydrogenation of carbon monoxide: the Synthesis gas (CO + 2H2) is compressed to 5-10 atm pressure and admitted into a catalytic reactor. SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 19 containing the catalyst (mixture of cobalt (100 parts), thoria (5 parts) and magnesia (8 parts)). The reactor is heated to about 250 0 C. Hydrogenation, reactions takes place to form saturated and unsaturated. These mixture of saturated and unsaturated hydrocarbons are passed through a fractionating column to for separation to obtain synthetic petrol. 1.14 What is Power alcohol? Explain the advantages 4 Marks A mixture of ethyl alcohol and gasoline blend, which can be used as fuel in internal combustion engine, is known as power alcohol or gasohol. Absolute alcohol is mixed with ether, benzene etc compounds and one volume of this is mixed with four volumes of petrol and is used as a fuel. Advantages: The power out put is good. It has better antiknock property. Ethanol is biodegradable; hence it is environmental friendly fuel. The use of ethanol in alcohol increases the oxygen content of the fuels and promotes more and complete combustion of hydrocarbons in gasoline. It reduces carbon monoxide emission. SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 20 1.15 Bio diesel: Biodiesel refers to a vegetable oil- or animal fat-based diesel fuel consisting of long-chain alkyl (methyl, propyl or ethyl) esters. Biodiesel is typically made by chemically reacting lipids (e.g., vegetable oil, animal fat (tallow)) with an alcohol. Biodiesel is meant to be used in standard diesel engines and is thus distinct from the vegetable and waste oils used to fuel converted diesel engines. Biodiesel can be used alone, or blended with petrodiesel. SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 21 SOLAR ENERGY 1.16 What is solar energy? Mention the advantages & disadvantages of solar energy. 4 Marks The radiations reaching earth from the sun and converting them in to different useful forms of energy is called solar energy. The utilization of solar energy is of two types – Direct solar power and indirect solar power. Advantages: • The Solar power is pollution free. • It can operated with little maintenance or intervention after initial setup. • The Solar power is becoming more and more economical as costs associated with production decreases, and the technology becomes more effective in energy conversion. • The Solar power can be viewed as a local resource because of original climatic variances. Disadvantages: • The Solar power is only practical in certain areas with a favorable climate and latitude. That is, areas near the tropics and which are relatively cloud free. • The Solar power is not available at night. • The Solar power decreases during cloudy. • The Solar power must be converted into some other form of energy to be stored. • Solar cell technologies produce DC power which must be converted to the AC power. 1.17 What are photovoltaic cells? 2 Marks Photovoltaic cells or Solar cells are the semiconductor devices which converts sunlight into direct current electricity on illumination. 1.18 Explain the working of photovoltaic cells? 6 Marks The Solar cells or Photovoltaic cells are made out of semiconductors which have the capacity to absorb light. When n-type and p-type semiconductor are bought together a semiconductor diode is formed. The semiconductor diode separates and collects the carriers and conducts the generated electrical current preferentially in a specific direction. SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 22 A typical silicon photovoltaic cell is composed of a thin wafer consisting of an ultra thin layer of phosphorus doped. (n-type) silicon on top of boron doped (p-type) silicon. Hence a p-n junction is formed. A metallic grid forms one of the electrical current contacts of the diode and allows light to fall on the semiconductor between the grid lines. An antireflective layer between the grid lines increases the amount of light transmitted to the semiconductor. The cell’s other electrical contacts is formed by a metallic layer on the back of the solar cell. When light radiation falls on the p-n junction diode, electron – hole pairs are generated by the absorption of the radiation. The electrons are drifted to and collected at the n-type end and the holes are drifted to p- type end. When these two ends are electrically connected through a conductor, there is a flow of current between the two ends through the external circuit. Thus photoelectric current is produced. 1.19 Explain the Importance of Photovoltaic cells: 6 Marks • The conventional energy exhaustible and depleting. Where as Solar energy being ultimate, inexhaustible and renewable energy. There the photovoltaic cells are important means to utilize this continuous energy source. • The Photovoltaic cells can serve for both off-grid and on-grid applications. It can be used for off-grid professional devices and supply systems such as telecommunication equipment, solar home systems, etc. • The Photovoltaic energy conversion environmental friendly as there is no harmful emission of pollutants. SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 23 • Use of or production of solar energy doesn’t produce noise pollution. • The electricity obtained from solar energy is useful in minimizing global warming due to carbon dioxide. • Photovoltaics can be used as roof integrated systems, providing power and also serving as optical shading elements for the space below and preventing overheating in the summer. • Photovoltaic cells provide power for spacecraft and satellites. • Developments in the field of photovoltaic cells will boost the semiconductor industry and storage battery industries. Silicon: The most common material used for solar cells is crystalline silicon, with multicrystalline silicon is most used. Silicon is the second member in the group IV A in the periodic table. It never occurs free in the nature, but occurs as oxides and silicates. The solar cells are made out of three primary categories of crystalline silicon as follow. i) Single crystalline or mono crystalline wafers. ii) Poly or multi crystalline wafers. iii) Ribbon silicon-drawn from molten silicon, having a multicrystalline structure. 1.20 Explain the Physical properties of Silicon relevant to Photovoltaic 6 Marks i) Silicon is a semiconductor with a band gap of 1.12 eV at 25 0 C. ii) Silicon crystallizes into a diamond cubic structure at atmospheric pressure. iii) Usually metals expand on heating and contract on cooling but silicon Contracts on melting and expands during solidification. iv) Silicon can be made into semiconductor by adding impurities. (Providing either free electrons or holes). Impurities from VA group are n-dopants and impurities from IIIA group are p-dopants. v) Silicon has relatively high refractive index which limits the optical applications of silicon. There fore it is coated with antireflective agent. vi) Silicon is brittle even when alloyed with small quantities of impurities. vii) Shaping for photovoltaic applications require sawing and grinding. SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 24 1.21 Explain the Chemical properties of silicon relevant to Photovoltaic 6 Marks i) Silicon is stable in the tetravalent state and has a strong affinity for oxygen, forming stable oxides and silicates. Elemental silicon readily oxidizes, forming a thin protective film of silica (SiO2). ii) Silicon and carbon from a strong Si-C bond and stable products. Silicon carbide also finds various applications in photovoltaics and electrons. Primary uses exploit the abrasive properties of SiC for wafering silicon crystals. Silicon forms hydrides, and monosilane (SiH4) is a key chemical compound for the production of amorphous silicon and the purification of silicon to semiconductor grade. iii) The chemical reactivity of silicon with chlorine is also very important. Trichlorosilane and tetrachlorosilane are both the intermediates and the by-products of the purification processes in upgrading metallurgical grade silicon to semiconductor grade, as these compounds are volatile at low temperature and can be decomposed to elemental silicon at higher temperature. Other chlorosilanes or halogenosilanes are also used in chemical vapor deposition applications. Silicon and germanium are isomorphism and mutually soluble in all proportions. 1.22 Explain the production of semiconductor grade silicon? 6 Marks Production of semiconductor grade silicon involves the following three stages. i) Production of metallurgical grade silicon (Carbo thermic reduction of silica) Metallurgical grade silicon of purity of 98.5% Si is produced in submerged electrical arc furnace. The furnace consists of a crucible filled with quartz and carbon (metallurgical coke or coal). Silicon is formed as follows. SiO2 + 2C(s) Si (I) +2CO (g) Silicon is formed as molten state and is tapped from the bottom of the furnace. The carbon monoxide further oxidized to carbon dioxide and released into the atmosphere. ii) Refining of silicon: The crude silicon obtained in the above method is taken in a large ladle and treated SiO2 and lime/limestone (CaO/CaCO3). The less noble elements than silica such as Al, Ca and Mg are oxidized as their oxides. SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 25 iii) Production of semiconductor grade silicon The metallurgical grade silicon obtained in the above process is further processed in four stages to get semiconductor grade silicon or polysilicon. 1. Synthesis of silicon hydride. 2. Purification of the hydride. 3. Decomposition to elemental silicon. 4. Removal of byproducts. The metallurgical grade silicon is treated with dry HCl gas at 300 0 C to form trichlorosilane & a small amount of tetrachlorosilane. The mixture is distilled to get pure trichlorosilane. The tetrachloro silane (SiCl4) is reduced with hydrogen at 1000 0 in a reactor to get tri chloro silane (HSiCl3). The tri chloro silane is then passed through fixed bed columns containing quaternary ammonium ion exchange resins catalysts. The products obtained in the above process are separated by distillation. Tetrachlorosilane & trichlorosilane are again recycled to the hydrogenation reactor & the exchange resin respectively. Silicon hydride or silane is further purified by distillation & passed into a reactor containing heated silicon seed rods. Silane gets pyrolysed to form polysilicon (semiconductor grade silicon) 1.23 Explain the Doping of Silicon 6 Marks The process of incorporating the desirable impurities in the crystals of semiconductors to get desirable extrinsic semiconductor properties as known as doping techniques. SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 26 Doping of Silicon can be carried out as follows. i) Doping during crystal formation Doping can be carried out at the crystal formation stage itself, by adding calculated amount of dopants into the melt. When silicon is deposited on to the surface of silicon rod by vapor decomposition, doping can be accomplished by simultaneously depositing a dopant with the semiconductor material. This can be done by mixing the reaction mixture gas feed with vapors of suitable compounds of the dopant. For example, calculated amount of PH3 is mixed with the gas feed to get n-type doping and BH3 for p-type doping. ii) Diffusion technique: In this technique, a region of semiconductor material is incorporated with dopant atoms by the diffusion of impurity atoms into the crystal of the material without actually melting it. By this technique, the extent of impurity penetration can be controlled to a very small thickness of the material. For example, a n-type silicon can be obtained by heating a silicon wafer below its melting point in an atmosphere of n-type impurities such as phosphorus. The impurity atoms considered on the surface of the wafer diffuses into the crystal. Similarly p-type silicon can be obtained by heating in an atmosphere of p-type dopants. The extent of diffusion can be regulated by controlling the temperature and the concentration of the impurity atoms. iii) Ion Implantation technique: In this technique, the semiconductor is impacted with an ion beam of impurity ions. This results in the implantation of some dopants atoms into the semiconductor crystal. Extent of implantation is controlled by the energy of the ion beam. SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 27 Objective type questions: 1) Bomb-calorimeter is used for determining the calorific value of a) Solid fuel b) liquid fuel c) Gaseous fuel d) both solid fuel & liquid fuel 2) Octane number is related to the petroleum product a) Diesel b) Kerosene c) Petrol d) Lubricating oil 3) The process by which the higher hydrocarbons are broken into lower hydrocarbons by the application of heat by a) Combustion b) Cracking c) Sparking d) jetting 4) Quality of Diesel fuel is determined by a) Octane rating b) Percentage of Carbon c) Length of Hydrocarbon chain d) Cetane number 5) The tendency of Knocking is high in a) Aromatics b) Olefins c) Straight chain hydrocarbons d) Cycloparaffins 6) A device in which electricity is produced using Solar energy is called a) Fuel cell b) Voltaic cell c) Photovoltaic cell d) Concentration cell 7) Knocking is due to a) Slow combustion b) Instantaneous explosive combustion c) Incomplete combustion d) All 8) Which of the following is the primary fuel a) Producer gas b) Coal gas c) Petroleum d) Water gas 9) HCV is always ----------- than NCV SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 28 a) Less b) More c) Equal d) Not related 10) Main constituent present in photovoltaic cells are a) Silicon b) Sand c) Carbon d) Iron 11) Unit of Calorific Value is a) g/cc b) Cal/ml c) kJ d) kJ/Kg 12) The Knocking tendency of Hydrocarbon decreases in the following order a) Straight chain> Cyclo alkanes > Aromatic > Branched Chain b) Straight Chain > Branched Chain > Cyclo Alkanes > Aromatic c) Aromatic > Cyclo Alkanes > Branched Chain > Straight Chain d) Cyclo Alkanes > Aromatic > Branched Chain > Straight Chain 13) The catalyst employed in the fluidized catalytic cracking is a) Al2O3 + SiO2 b) Fe2O3+ SiO2 c) ZrO2 + SiO2 d) TiO2 + SiO2 14) In photovoltaic cells solar energy is utilized to transform a) Solar energy into light and heat energy b) Solar energy into electrical energy c) Solar energy into electrical chemical energy d) All the above 15) In Fischer- tropsch process of preparation of synthetic petrol ------- -is used as raw material a) CO+H2 b) CO2+H2 c) CO+H2O d) CO2+H2O SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 29 REVIEW QUESTIONS 1. Define a fuel. Explain the classification of fuels with examples. 2. Define calorific Value. Explain the types of calorific value. 3. Mention the SI units of calorific value. 4. Describe the Bomb calorimetric method of determination of calorific Value of solid fuel. 5. What is petroleum cracking? Explain the process of fluidized bed Catalytic cracking method 6. What is reformation of Petroleum? Give the reactions involved in reforming. 7. What is Knocking? Explain its Mechanism 8. Define octane number. 09. Define cetane number. 10. What are anti knocking agents? Explain leaded and unleaded petrol. 11. What is synthetic petrol? Explain the production of synthetic petrol 12. What is power alcohol? Give its advantages as fuel. 13. What are photovoltaic cells? 14. Explain how a photovoltaic cell works? 15 Explain the Chemical properties of silicon relevant to Photovoltaic 16 Explain the Physical properties of Silicon relevant to Photovoltaic 17. Explain the production of semiconductor grade silicon 18. Explain the Doping of Silicon? SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 30 UNIT UNIT UNIT UNIT – –– – II II II II ELECTRO CHEMICAL ENERGY SYSTEMS ELECTRODE POTENTIAL AND CELLS 2.1 What is an electrochemical cell Explain the classification with examples 5 Marks An electrochemical cell is a device, which is used to convert chemical energy into electrical energy and vise versa. These electrochemical cells are classified into two types as follows. 1) Galvanic or Voltaic cells: These are the electrochemical cells, which converts chemical energy into electrical energy. Ex. Daniel cell, Dry cell, etc. 2) Electrolytic cell: These are the electrochemical cells, which are used to convert electrical energy into chemical energy. Ex: Lead acid battery, Nickel cadmium battery etc., Galvanic or Voltaic cells: Galvanic or Voltaic cells are again classified into three types as follows a) Primary cells: These are the cells which serve as a source of energy only as long as the active chemical species are present in the cell. The cell reactions are irreversible. These are designed for only single discharge and cannot be charged again. Ex: Dry Cell, Zn – Hgo cell, Zn-Ag2o cell etc. b) Secondary cells: These cells are chargeable and can be used again and again. The cell reactions are reversible and are often called as reversible cells. During discharging the cells acts like voltaic cell converting chemical energy into electrical energy. During charging the cell acts like electrolytic cell by converting electric energy into chemical energy, hence these batteries are called as storage battery. Ex: Lead acid Battery, Ni-cd cells. Lithium ion cells etc. SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 31 c) Concentration of cells: These are the electrochemical cells consisting of same metal electrodes dipped in same metal ionic solution in both the half cells but are different in the concentration of the metal ions. Ex: copper concentration cell, Zinc concentration cell 2.2 Explain the construction and working of Daniel cell 6 Marks The Daniel cell consisting of two half cells in which zinc and copper electrodes are immersed in zinc Sulphate and copper Sulphate solution respectively. The two half cells are internally connected by a salt bridge and externally by a metallic wire. The zinc electrode undergoes oxidation and looses electron. The electron liberated migrates to another half cell. Oxidation Zn (s) Zn 2 + + 2e - In the other half-cell the cupric ions accepts the electrons, undergoes reduction, and get deposited on copper electrode as copper atoms. Reduction Cu ++ (aq) + 2e - Cu (s) SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 32 Due to the above simultaneous oxidation and reduction reactions the Daniel cell generates electrical current, which is indicated by voltmeter or ammeter. The Daniel cell can be represented as Zn (s) /Zn 2+ (aq) //Cu 2+ (aq) /Cu. 2.3 Mention the Cell notations and Cell Conventions Cell notations 4 Marks / Single vertical line represents phase boundary between the metal and it’s solution. // Double vertical line represents salt bridge. Arrow mark indicates the direction of flow of electrons. Cell Conventions i) The half cell which undergoes oxidation is always written towards left side of the salt bridge. ii) The half cell which undergoes reduction is always written towards right side of the salt bridge. iii) If the direction of arrow mark indicates from left to right. The electrons flow from anode to cathode and the cell reactions are spontaneous and if the direction of arrow mark indicates from right to left, then electrons flow from cathode to anode. Then cell reactions are non spontaneous. iv) The term electrode potential always refers to the reduction potential. 2.4 Define Single electrode Potential. 2 Marks It is defined as the potential developed at the interphase between the metal and the solution, when a metal is dipped in a solution containing its own ions. It is represented as E 2.5 Explain the origin of electrode Potential: 6 Marks When a metal is dipped in a solution containing its own ions, the metal may undergo oxidation by loosing electrons or the metal ions may undergo reduction and get depositing metal atoms on the metal surface. Consider a metal M is dipped in a solution containing its ions M n+. SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 33 The tendency of metal to pass in to solution (oxidation) can be represented as, M M n+ + ne - Simultaneously the metal ions from the solution tend to deposit on the metal as metal atoms (reduction) M n+ + ne - M The above two opposite tendencies will results in equilibrium as follows M n+ + ne - M When a metal undergoes oxidation it loses positive ions into solution leaving behind a layer of negative charges on its surfaces. This layer attracts positive changes and forms an electric double layer (EDL) because of the formation of EDL electrode potential arises. When metal ions undergo reduction depositing metal atoms on the metallic surfaces the metal surface becomes positively charged. The accumulated positive charge on the metal surface attracts a layer of –ve charges and forms an electrical double layer or Helmotz EDL which causes the origin of electrode potential. 2.6 Explain the measurement of electrode potential 4 Marks The electrode potentials of any metal electrodes can be determined by using reference electrodes like standard hydrogen electrode. (SHE). The SHE is coupled with the electrode whose electrode potential is to be determined and the electrode potential of the electrode is determined by fixing the electrode potential and SHE as zero [at all temperatures] Example: Consider the determination of Single electrode potential of Zinc electrode using Standard Hydrogen electrode. To determine the Single electrode potential of Zinc electrode it is coupled with Standard Hydrogen electrode as follows SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 34 The electrode potential of Zinc electrode can be calculated as E cell = E cathode – E anode E cell = E SHE – E Zn 0.76 = 0 - E Zn E Zn = -0.76 V The electrode potentials can also be determined by using secondary reference electrode such as calomel electrode and Ag / Agcl electrode. 2.7 Define electron motive force. (EMF) 2 Marks It is defined as the potential difference between the two electrodes of a galvanic cell which causes the flow of current from an electrode with higher reduction potential to the electrode with lower reduction potential. It is denoted as E cell. E cell = E right –E left. E cell = E cathode – E anode. Problems: 1) Calculate emf of a cell constructed by combining Cu & Zn electrodes dipping in their respective ionic solutions. The standard electrode potential of Cu and Zn are 0.34 V and 0.76 V respectively at 298K. E cell = E cathode – E anode = Ecu – Ezn = 0.34 – (-0.76) = 1.1volt 2.8 Define Standard electrode potential. 2 Marks SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 35 It is defined as potential developed at the interface between the metal and the solution. When a metal is dipped in a solution containing its own ions of unit ion concentration and at 298K. [If the electrodes involve gases then it is one atmospheric pressure] It is denoted as E 0 2.9 Derive Nernst Equation for single electrode 6 Marks In 1889 Nernst derived a quantitative relationship between the electrode potential and the concentrations of metal ions are involved. The maximum work available from a reversible chemical process is equal to the maximum amount of electrical energy that can be obtained; it shows decrease in free energy. Wmax = W ele Wmax = -– G. The amount of electrical energy available from a cell is equal to the product of no of coulombs and volts (energy). Electrical energy = Volts x coulombs The number of coulombs of current is the product of number of moles of electrons that flow and Faradays constants. Number of volts or energy is the ‘E’ W max = W ele = - n x F x E. (No. of moles of electrons) x(Faradays constant) x (No.of volts or energy). SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 36 Where, E = Electrode potential E 0 = standard electrode potential n = no. of electrons [M n+ ] = Concentration of metal ions R = Universal gas constant T = Temperature(In Kelvin). SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 37 Problems: 1) Calculate the emf of the cell Fe / Fe ++ (0.01) // Ag + (0.1) /Ag at 298K if standard electrode potentials of Fe and Ag electrodes are – 0.42 and 0.8 V respectively. 6 Marks E 0 cell = E 0 cathode – E 0 anode = E0 Ag + /Ag – E 0 Fe ++ / Fe = 0.8 – (– 0.42) = 1.22 V. = E0cell + 0.0591 log10 [Ag + ] 2 n [Fe ++ ] = 1.22 + 0.0591 log10 (0.1) 2 2 (0.01) = 1.22 + 0.02955 log 1 = 1.22 V. 2) A cell is constructed by coupling Zn electrode dipped in 0.5 M ZnSO4 and Ni electrode dipped in 0.05 M NiSO4. Write the cell representation, cell reaction. Calculate the EMF of cell, given that reduction potentials of Zn and Ni as – 0.76 and – 0.25 volt respectively. 6 Marks Cell representation: Zn / ZnSO4(0.5M ) // NiSO4 ( 0.05M ) / Ni. Cell reactions: Zn (S) + Ni ++ (aq) → Zn ++ (aq) + Ni (S) E 0 cell = E 0 cathode – E 0 anode = E 0 Ni ++ /Ni – E 0 Zn ++ /Zn = – 0.25 – (– 0.76) = 0.51 V. SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 38 = E 0 cell + 0.0591 log10 [Ni ++ ] n [Zn ++ ] = 0.51 + 0.0591 log10 (0.05) 2 (0.5) = 0.51 + 0.02955 log 0.1 = 0.4805 V. 3) Calculate the potential of Ag – Zn cell at 298 K if the concentrations of Ag + and Zn ++ are 5.2 x 10 –6 M and 1.3 x 10 –3 M respectively. E 0 of the cell at 298K is 1.5 V. Calculate the change in free energy G for the reduction of 1 mole Ag + . 1 faraday = 96.5 k J/ V / mole. 6 Marks Cell reaction: 2Ag + (aq) + Zn (S) → Zn ++ (aq) + Ag (S) = E 0 cell + 0.0591 log10 [Ag + ] 2 n [Zn ++ ] = 1.5 + 0.0591 log10 (5.2 x 10 –6 ) 2 2 (1.3 x 10 –3 ) = 1.5 + 0.02955 log (20 x 10 –9 ) = 1.2729 V. For the reduction of 2 moles of Ag + ions, 2 electrons are required For the reduction of 1 mole of Ag + ions, 1 electron is required G = – n F E cell. = – 1 x 96.5 x 1.2729 = – 122.83 K J / mol SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 39 4) A cell is constructed by coupling Ni electrode dipped in 0.01 M NiSO4 and Pb electrode dipped in 0.5 M PbSO4. Write the cell representation, cell reactions. Calculate the EMF of cell, given that reduction potentials of Ni and Pb are – 0.24 and – 0.13 volt respectively. 6 Marks Cell representation: Ni / Ni ++ (0.01 M) // Pb ++ (0.5M) / Pb. Cell reactions: At anode Ni → Ni ++ + 2e - At cathode Pb ++ + 2e - → Pb Net cell reaction: Ni (S) + Pb ++ (aq) → Ni ++ (aq) + Pb (S) E 0 cell = E 0 cathode – E 0 anode = E0 Pb ++ /Pb – E 0 Ni ++ /Ni = – 0.13 – (– 0.24) = 0.11 V. = E 0 cell + 0.0591 log10 [Pb ++ ] n [Ni ++ ] = 0.11 + 0.0591 log10 (0.5) 2 (0.01) = 0.11 + 0.02955 log 50 = 0.1602 V. 2.10 What are concentrations cells? Explain with an example 6 Marks These are the galvanic cells consisting of same metal electrodes dipped in same metal ionic solution in both the half cells but are different in the concentration of the metal ions. Ex: Consider the following concentration cell constructed by dipping two copper electrodes in CuSO4 solutions of M2 molar and M1 molar where M2M > M1M. The two half-cell are internally connected by a salt bridge and externally connected by a metallic wire through voltmeter or ammeter. SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 40 The electrode, which is dipped in less ionic concentrations solutions (M1) act as anode and undergoes oxidation. The electrode, which is dipped in more ionic concentration (M2) act as cathode and undergoes reduction. At anode : Cu (S) Cu 2+ (M1) + 2e - At cathode : Cu 2+ (M2) + 2e - Cu (S) _____________________________________________________ NCR Cu 2+ (M2) Cu 2+ (M1) E of cell = E cathode – E anode. E cell = [ E o + 0.0591 log (M2) ] – [E o + 0.0591 log (M1)] n n Where, (M2) > (M1) SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 41 Problems: 1. Calculate emf of the following concentration cell at 25 0 C Ni (S) / Ni ++ ( 0.01M) // Ni ++ ( 0.1M) / Ni (S) 4 Marks Where, (M2) > (M1) E cell = 0.0591 log (0.1) n (0.01) = 0.02955V 2. Calculate emf of the following concentration cell at 25 0 C Cu (S) / Cu ++ (0.05M) // Cu ++ (5M) / Cu (S) 4 Marks Where, (M2) > (M1) E cell = 0.0591 log (5) n (0.05) = 0.0591V 2.11 Mention the different types of single electrodes 4 Marks i) Metal-Metal ion electrode These electrodes consists of a metal dipped in a solution of its own ions. Example: Zn/Zn ++ Cu/Cu ++ Ag/Ag + ii) Metal-Metal salt electrode These electrodes consists of a metal in contact with a it’s salt Example: Calomel electrode (Hg/Hg2Cl2/Cl - ) Silver – Silver Chloride electrode(Ag/AgCl/Cl - ) Lead – Lead sulphate electrode(Pb/PbSO4/SO4 - ) SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 42 iii) Gas electrode Example: H2 electrode (H2/Pt/H + ) Chlorine electrode (Pt/Cl2/Cl - ) iv) Amalgam electrode Example:Lead amalgam electrode(Pb-Hg/Pb + ) v) Oxidation – Reduction electrode Example: Pt/Fe 2+ , Fe 3+ Pt/Ce 3+ , Ce 4+ Pt/Sn 2+ , Sn 4+ vi) Ion selective electrode Example: Glass electrode 2.12 What are Reference Electrodes? Mention the types with Examples 3 Marks These are the standard electrodes with reference to these, the electrode potentials of any other electrode can be determined. The Reference Electrodes can be classified in to two types i) Primary reference electrodes Ex: Standard hydrogen electrode ii) Secondary reference electrodes Ex: Calomel and Ag/Agcl electrodes 2.13 Explain the construction, working and limitations of standard hydrogen electrode 6 Marks The standard hydrogen electrode consists of platinised platinum foil fused to the glass tube. Mercury is placed at the bottom of the tube and SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 43 a copper wire is used for electrical connections. The platinum foil is immersed in a solution containing unit molar hydrogen ions. Pure hydrogen gas is bubbled about the electrode through the H2 gas inlet at 1atm pressure. The electrode is represented as Pt/H2 (g)/H + If the concentration of H + is 1M, H2 gas bubbled at 1atm pressure and at temperature 298k, then the electrode is called standard hydrogen electrode. And the electrode potential is arbitrarily fixed as zero. The electrode reactions is Limitations of SHE i) The construction of SHE is difficult. ii) It is very difficult to maintain the concentration of H + as 1M and pressure H2 gas at 1atm iii) Platinum electrode is poisoned by the impurities of the gas iv) It can not be used in the presence of oxidizing agents. 2.13 Explain the construction and working of Calomel electrode 6 Marks Calomel electrode consisting of a glass container at the bottom of which mercury is placed above which a layer of mercury and mercurous chloride (called calomel) is placed with 3/4 th of bottle is filled with saturated KCl solution. Electrode potential of the cell depends on the concentration of KCl used. The calomel electrode can be represented as. Hg (l) / Hg2 Cl2 (S) / Saturated KCl. SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 44 The calomel electrode acts as both anode and cathode depending upon the other electrode used. The platinum wire is used for electrical connections. Salt bridge is used to couple with other half cell. When it acts as anode the electrode reactions is, Hg2Cl2 + 2e - 2Hg + 2Cl - When it acts as cathode the electrode reaction is Hg2Cl2 + 2e - 2Hg + 2Cl - Advantages of Calomel Electrode: • It is simple to construct. • The electrode potential is reproducible and stable. • It is used as a reference electrode. 2.13. Explain the construction and working of Silver-Silver chloride electrode. 6 Marks Ag/AgCl electrode is a metal metal salt electrode. It consists of narrow glass tube at the bottom of which agar is placed above which saturated solution of KCl is placed. The silver wire is used for electrical connections and it is coated electrolytically with AgCl. The cell can be represented as. Ag(s) / AgCl (s) / Saturated KCl. Electrode acts as both anode and cathode depending on the other electrode used. When it acts as anode the electrode reaction is Ag + Cl - Ag Cl + e - When it acts as cathode the electrode reaction is AgCl + e - Ag + Cl - . SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 45 2.14. What are Ion Selective Electrodes? Explain the Glass electrode 6 Marks These are the electrodes, which responds to specific ions only and develops a potential against that ions while ignoring the other ions present in the solution. Ex: Glass electrode. Glass electrode is a pH sensitive electrode widely used for pH determinations. It is consisting of a long glass tube at the bottom of SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 46 which a thin and delicate glass bulb, which made up of special type of glass (12 % Ba2O, 6% of Cao, 72% of SiO2) with low melting point and high electrical conductance is used. The glass bulb is filled with 0.1.M Hcl and Ag – Agcl is used as a internal reference electrode. A platinum wire is used for electrical contant. The glass electrode can be represented as Ag/AgCl(s) /0.1M (HCl) / Glass. 2.15 Explain the determination of PH using glass electrode. 6 Marks To determine pH of unknown solution the glass electrode is combined with secondary reference electrode such as calomel electrode and the glass - calomel electrode assembly is dipped in the solution whose pH is to be determined. The two electrodes are connected to potentiometer or pH meter. The combined electrodes can be represented as. Hg(l) / Hg2 Cl2 (S) / Saturated KCl //solution of unknown pH /glass/0.1M Hcl/Ag/AgCl(s) The emf of the above cell is given by E cell = E cathode – E anode E cell = E calomel – E glass E glass = E calomel – E cell The potential of E glass is given by E glass = - RT/nf ln [a2/a1] = RT/nf ln [a1/ a2] SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 47 Where a1 = Concentration of [H + ] ions in glass bulb a2 = Concentration of [H + ] ions in unknown solution E glass = RT/nf ln a1 – RT/nf ln [a2] E glass = Constant – RT/nf ln [a2 ] (Because an [H+] ion in glass bulb is constant) E glass = Constant – 0.0591 log [H + ] E glass = Constant + 0.0591pH pH = E glass – Constant 0.0591 pH = E calomel – E cell - Constant 0.0591 pH = 0.242 – E cell - Constant 0.0591 Objective type questions: 1. _____ Converts chemical energy into electrical a) Galvanic cell b) Daniel cell c) Dry cell d) all 2. In the anodic chamber ----------- reaction takes place a) Oxidation b) Reduction c) Addition d) Substitution 3. In the cathodic chamber ----------- reaction takes place a) Oxidation b) Reduction c) Addition d) Substitution 4. Origin of electrode potential is explained in a) Nernst theory b) Helmholtz double layer theory c) galvanic theory d) Electrochemical theory 5. In two half cells, the one which is having high negative value acts as a) anode b) Cathode c) Dry cell d) None 6. EMF of a concentration cell depends on a) [M1] & [M2] b) No. of charges c) temperature d) All of the above 7. Calomel electrode potential is dependent of a) Cl - concentration b) Hg2Cl2 c) Temperature d) None 8. Glass electrode is an example for a) Reference electrode b) Ion selective electrode c) Primary electrode d) None 9. Standard electrode potential is nothing but single electrode potential at a) Unit concentration b) 1 atm. pressure c) 298 K d) All of the above SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 48 10. Potential of standard hydrogen electrode is taken as a) 1V b) 0 V c) 10 V d) None 11) Metal rod dipped in a solution of its ion. Its electrode potential is independent of a) Temperature of the solution b) Conc. Of the solution c) Area of the metal exposed d) Nature of the metal 12. Double vertical line (||) represents a) Solid liquid interphase b) Salt bridge c) solid - solid or Liquid - liquid interphase d) None 13. In the pH determination of a solution---------- electrode is used along with glass electrode a) Ag-Agcl electrode b) Calomel electrode c) Ion selective electrode d) None. 14. A chemical change that occurs when electric current is passed through an electrolyte is a) Conduction b) Dissociation c) Ionization d) Electrolysis 15. Daniel cell is a combination of standard electrodes of a) Cu & Ag b) Zn & Cd c) Zn & Cu d) Cu & Cd SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 49 REVIEW QUESTIONS 1. What is an electrochemical cell? Explain the types with examples. 2. What are galvanic cells? Explain with example 3. Mention the cell notations and conventions 4. Define single electrode Potential. Explain the origin of electrode Potential 5. Explain the measurement of electrode potential 6. Define electron motive force. (EMF) 7. Define standard electrode potential. Derive the Nernst equation for single electrode potential. 8. What are Concentration of cells? Explain with an example. 9. What are Reference Electrodes? Mention the types with examples. 10. Explain Calomel electrode. Mention its advantages. 11. Explain Silver-Silver chloride electrode 12. What are Ion Selective Electrodes? Explain the Glass electrode 13. Explain the determination of PH using glass electrode. SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 50 UNIT UNIT UNIT UNIT – –– – III III III III CONVERTION AND STORAGE OF ELECTROCHEMICAL ENERGY BATTERY TECHNOLOGY 3.1 What is a Battery? Explain the classification with examples. 6 Marks Battery is a collection of cells connected either in series or in parallel to get required amount of energy. Classification of Batteries Batteries are classified into three types as follows. a) Primary b) Secondary c) Reserved. a) Primary Batteries: These are the batteries which serve as a source of energy only as long as the active chemical species are present in the battery or in the cell. The cell reactions are irreversible. These are designed for only single discharge and cannot be charged again. Ex: Dry Cell, Zn – Hgo cell, Zn-Ag2o cell etc., b) Secondary Batteries: These batteries are chargeable and can be used again and again. The cell reactions are reversible and are often called reversible batteries. During discharging the cell acts like voltaic cell converting chemical energy into electrical energy. During charging the cell acts like electrolytic cell by converting electric energy into chemical energy, hence these batteries are called as storage battery. Ex: Lead acid Battery, Ni-cd battery etc. c) Reserved Batteries: The batteries which can be stored in an active state and made ready for use by activating them prior to the applications (usage) are called as reserved batteries. The key components of the batteries such as electrolyte etc., is separated from the battery. And the battery is stored for a longer time. The electrolyte if filled before its usage. The advantages of the reserved batteries are, SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 51 Batteries can be stored for a longer period. To prevent corrosion at contact points during storage. Self-discharging reactions during storage can be eliminated or avoided. They can be used whenever they are required. Ex: Mg – water activated batteries (Mg- Agcl & Mg cucl), Zn-Ag2O Batteries etc. What are Super capacitors ? An Electric double-layer capacitor, also known as supercapacitor, supercondenser, pseudocapacitor, electrochemical double layer capacitor (EDLC), or ultracapacitor, is an electrochemical capacitor that has an unusually high energy density when compared to common capacitors. 3.2 Explain the Characteristics of a battery. 8 Marks A good battery must have the following characteristics. a) Voltage: The voltage of a battery depends on the energy change in the overall cell reaction and the nature of cell reaction. To obtain maximum voltage from a cell. i) The potential difference between the electrodes must be high. ii) The resistance of the cell must be low. iii) Electrode reactions must be fast to reduce the over potential. b) Capacity (C): The capacity of a battery is the charge or the amount of electric current that can be obtained from a battery. It is charge in ampere hours (Ah). The capacity of a battery depends on i) Size of a battery ii) Discharge conditions The capacity of a battery is to be determined by Where, C Capacity of battery (in Ah) W Weight of the active material n number of electrons. F Faradays constant. M Molar mass. SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 52 c) Energy density: Energy density is the ratio of energy available from a battery to its weight or volume Energy density = Energy available from a battery Weight or volume of cell. d) Power density: It is the ratio of power available from a battery to its weight or volume during discharge power density decreases. Power density = Power available from a cell Weight or volume of cell. e) Energy efficiency: The battery should have high-energy efficiency. The percent energy efficiency can be calculated as. Percentage of Energy efficiency = Energy released on discharge X 100 Energy required for charging. f) Cycle Life: Primary batteries are designed for single discharge and secondary batteries can be chargeable again and again. The number of charge and discharge cycles that are possible in secondary batteries, before failure occurs is called cycle life. The cycle life of batteries must be high. The factors which decrease the cycle life are: a) Corrosion at the contact point. b) Leakage of the current. c) Shortening between electrode due to irregular crystal growth and change in morphology d) Shedding of the active material from the battery. g) Shelf life: The duration of storage under specified conditions at the end of which a cell or a battery retains its ability to work or to produce energy is called shelf life. A good battery should have more shelf life. It depends on self-discharge reactions and corrosion at contact points. 3.3 Explain the construction and working of Lead acid battery. Mention the electrode reactions and advantages 6 Marks Lead acid battery consisting of two flat grids of spongy lead filled with lead (Pb) and lead oxide (PbO2), which acts as anode and cathode respectively. Several such anode and cathode pairs are immersed in 5 M or 37% H2SO4 of specific gravity 1.25 g/cm 3 The battery is encased in plastic or glass container. Each pair of anode and cathode produces a voltage of 2v. SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 53 Due to the formation of water in the cell reactions during discharging the acid gets diluted when its specific gravity falls below 1.25 g/cm3 the battery needs charging. During charging lead and lead oxide is again deposited on anode and cathode respectively and charging reaction is. Applications: 1) Lad acid battery is used for starting motors 2) Used in UPS Systems and research centers etc. 3) Used in marketing areas etc. 4) Used in security and alarm systems. 3.4 Explain the construction and working of Nickel-Cadmium Battery. Mention the electrode reactions and advantages 6 Marks Ni-Cd Battery consisting of anode and cathode compartments. The Anode compartments consisting of 78% cadmium, 18% Iron, 1% Nickel and cathode compartment consisting of 80% of Nickel oxy hydroxide, 18% of graphite & traces of Barium and cobalt. 6MKOH is used as electrolyte. SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 54 The graphite increases the conductivity, Barium, and cobalt compound increases the efficiency of the active materials and cycle life. Applications: 1) Used in electronic devices such as earphones, security and alarm systems. 2) Used in pocket calculators, emergency lights. 3.5 Explain the construction and working of Zinc -air battery. 6 Marks Zinc air battery consisting of anode containing granules of zinc mixed with 20%NaOH electrolyte. Cathode can contains a porous carbon plate which provides site for the reduction reaction and do not involves in the reaction. Carbon is catalytically activated to absorb oxygen gas. The anode and cathode compartments are separated by a separator and both are encased in plastic or ebonite insulator. The reactions are as follows. SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 55 Advantages: i) High energy density. ii) Low cost and compact iii) Does not produce harmful products. Applications: Used in Military radio receivers, transmitters, hearing aids. 3.6 Explain the construction and working of Nickel - metal hydride battery. 6 Marks Nickel Metal hydride battery is made up of anode containing metal hydride such as ZrH2, VH2 and TiH2 with hydrogen storage metal alloy such as La Ni5 or TiNi.Cathode consisting of nickel oxy hydroxide both the compartments are separated by polypropylene. KOH used as electrolyte. Cell reactions are as follows. Applications: Used in electric vehicles, laptops, cellular phones etc. Advantages: i) Resistance to chemical oxidation ii) High energy storage capacity. SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 56 3.7 What are Lithium cells? Explain the construction and working of Lithium cells? Mention the advantages Lithium cells 3 Marks Lithium cells are most popularly used cells, which are available in various configurations such as button type, bobbin type, spiral wound prescematic, rectangle etc configuration. The cells have the following advantages. These are light in weight. High energy density More cycle life. Ex: Li – MnO2 cell Li – MnO2 cell 6 Marks Lithium manganese dioxide cell consisting of anode can containing lithium and the cathode can consisting of specially heat-treated manganese dioxide (MnO2). Both anode and cathode are seperated by a separator made up of polypropylene impregnated with the electrolyte.(metal salt such as LiCl, LiBr, LiAlCl4 which are mixed in a organic solvent such as 1,2 dimethoxy ethane, and propylene carbonate.) The cell delivers an emf of 3V. The cell reactions are as follows:- During the all reactions Mn IV stage reduced to Mn III stage. Uses: Li – MnO2 cells are used in Safety and Security Devices, Calculator, watches, automatic camera, memory batteries, cellular phones etc., 3.8 What are Fuel Cells? Explain the Classification of Fuel cells based on temperature 5 Marks SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 57 Fuel cells are the galvanic cells in which chemical energy of fuel is directly converted into electrical energy. Classification of Fuel cells These are classified into three types as follows. 1] Low temp fuel cells: Which operates at the temp range about 75 o C and contains water base electrolytes. 2] Moderate temp fuel cells: Which operates at the temp range about 600 o C and contains salt electrolyte. 3) High temp fuel cells: Which operates at the temp range about 1000 o C and contains ceramics as electrolyte. Difference between battery and fuel cell Battery Fuel cell 1. A battery stores the chemical reactants, usually metal compounds once used up you must recharge or throw away the battery 1.A fuel cell creates electricity through reactants stored externally 3.9 Explain the construction and working of H2 – O2 fuel cell. 6 Marks SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 58 Hydrogen - oxygen fuel cells consisting of two porous graphite electrodes, which is impregnated with an electro catalyst such as finely, divided Pt – Co - Ru or Pt-Ni-Ru. Concentrated KOH is used as an electrolyte. A wick is placed to maintain water balance. Hydrogen gas and oxygen gas are continuously supplied to the anode and cathode respectively. The hydrogen undergoes combustion generating electric current. The cell delivers an emf of 1.23v. The cell reactions are as follows. Uses: The H2-O2 cells are used in Space vehicles, military and mobile power systems. 3.10 Explain the construction and working of Methanol – Oxygen fuel cell. SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 59 6 Marks It consists of two electrodes made up of platinum in between the electrodes H2SO4 is placed as a electrolyte. Methanol and H2SO4 is supplied at the anode and pure oxygen gas is supplied at the cathode. The ethanol is oxidized to CO2 & H2O with the liberation of electrical energy. The cell delivers an emf of 1.20v. The cell reactions are as follows. Uses: 1) Used in Military applications. 2) Used for large scale power production stations. 3.11 Explain the Classification of fuel cells based on the electrolyte 6 Marks Fuel cells are classified into the following types based on the type of electrolyte used. 1) Alkaline fuel cells. 2) Phosphoric acid fuel cell. 3) Molten carbonate fuel cell. 4) Solid oxide fuel cell. 5) Solid polymer electrolyte fuel cell. SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 60 1) Alkaline fuel cells: These fuel cells containing alkali such as KOH or NaOH as electrolyte. Hydrogen is used as fuel and oxygen gas is used as an oxidant. The cell operates at a temp of 80 0 C. Uses: These are used in emergency lights and portable power generations, space applications, military applications etc., 2) Phosphoric acid fuel cells: These fuels cells consisting of 98% phosphoric acid, 2% water as electrolyte, O2 is used as oxidant. Hydrogen LPG, NPG etc are used as fuels. These operate at a temp 190 to 200 O C . Platinum alloys such as platinum- cobalt- chromium, are used as electro catalyst. Uses: These cells are used to provide light and heat in large buildings. 3) Molten carbonate fuel cells: These fuel cells consisting of molten carbonates such as lithium carbonate 26.2% and potassium carbonate (K2CO3) 23% and lithium. Aluminum carbonate as electrolyte. The anode is made up of nickel and cathode made up of nickel oxide. Operating temperature is 650 0 C. Uses: These are used in chemical industries such as aluminum Cloroalkali industries. 4) Solid oxide fuel cells: These contains ZrO2, Y2O3 are solid electrolytes. Cathode is made up of porous strontium doped with LaMnO3 or In2O3 and SnO2. Anode is made up of cobalt, nickel, or ZrO2. Operating temperature is 1000 0 C. Uses: These cells are used in KW power plants, water heating etc, 5) Solid Polymer Electrolyte: These contain ion exchange membrane as solid electrolyte for ionic conduction “nafion–R” membranes which are chemically and electrochemically stable at 200 0 C are used .Operating temperature of the cell is 80 0 C. The electrodes are made up of platinum and noble metals are used as electro catalysts. Uses: Used in the manned Gemini terrestrial orbital missions. Objective questions: 01. Battery is a a) Collection of cells connected in series b) Collection of cells connected in parallel c) Both a & b d) none 02. Which of the following is a primary battery a) Li-MnO2 battery b) Lead acid battery c) Nicd battery d) None 03. Which of the following is a secondary battery SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 61 a) Zn-MnO2 battery b) Li-MnO2 battery c) Lead acid battery d) Zn-HgO battery 04. In which battery a key component is separated from the rest of the battery prior to activation. a) primary battery b) Secondary battery c) Reserve battery d) None 05. The reaction the takes place at anode of the battery is a) Reduction b) Oxidation c) Neutralization d) Addition 06. Voltage of a battery can be expressed as a) Ohms b) Volts c) Both a & b d) None 07. The electrolyte used in the lead acid battery is a) KOH b) NaOH c) HCl d) H2SO4 08.The electrolyte used in the Zinc air battery is a) KOH b) NaOH c) HCl d) H2SO4 09.The electrolyte used in the Ni-MH battery is a) KOH b) NaOH c) HCl d) H2SO4 10.The electrolyte used in the Li-MNO2 battery is a) KOH b) NaOH c) HCl d) LiCl 11. In Ni-MH battery the anode is a) Ni b) Cd c) ZrH2 d) All 12. In Li-ion battery the anode is a) Ni b) Cd c) ZrH2 d) Li 13. The electrolyte used in alkaline fuel cells a) NaOH b) H3PO4 c) K2CO3 d) ZrO2 14.The electrolyte used in Phosphoric acid fuel cells a) NaOH b) H3PO4 c) K2CO3 d) ZrO2 15. The electrolyte used in Molten carbonet fuel cells a) NaOH b) H3PO4 c) K2CO3 d) ZrO2 SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 62 REVIEW QUESTIONS 1 What is a Battery? Explain the classification of Batteries with examples 2. Explain any 4 characteristics of a battery. 3. Explain the construction and working of Lead acid battery. Mention the electrode reactions and advantages 4. Explain the construction and working of Nickel-Cadmium Battery. Mention the electrode reactions and advantages 5. Explain the construction, working and application of Zinc -air battery. 6. Explain the construction, working and application of Nickel – metal hydride battery. 7. What are lithium cells? Explain the construction, working and application of lithium cell. 8. What are Fuel Cells? Explain the classification of fuel cells based on temperature. 9. Describe the construction and working of H2 – O2 fuel cell. 10. Describe the construction and working of methanol – Oxygen fuel cell. 11. Explain the classification of fuel cells based on the electrolyte used. SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 63 UNIT UNIT UNIT UNIT – –– – IV IV IV IV CORROSION SCIENCE 4.1 Define corrosion 2Marks Corrosion is defined as the destruction or deterioration of a metal or its alloy and consequent loss of metal, caused due to direct chemical action or electrochemical reactions with its environment. 4.2 Explain the electrochemical theory of corrosion 6 Marks According to electrochemical theory, corrosion of metals occurs due to the following changes, when they are exposed to the environment. 1) A large number of minute galvanic cells are formed which acts as anodic and cathodic areas. 2) At anode the metal undergoes oxidation and electrons are liberated which migrates towards cathodic region 3) Oxygen of the atmosphere undergoes reduction at cathodic area in the presence of moisture forming hydroxyl ions at the cathode Anodic reactions: At anode the metal undergoes oxidation-liberating electrons M M n+ + ne - Metal Metal ions Ex: when iron is exposed to the environment it undergoes oxidation as Fe Fe 2+ + 2e _ Cathodic reactions: The electrons released at anode migrates to the cathodic area and reduces oxygen to hydroxyl ions. SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 64 The different cathodic reactions are, a) In acidic medium: In acidic medium and in the absence of oxygen, hydrogen ions are reduced to hydrogen gas 2H + + 2e - H2 b) In alkaline and in the absence of O2. If the solution is alkaline and in the absence of oxygen the cathodic reaction is, 2 H2O + 2e - 2OH - + H2 c) In neutral and aerated medium: when the solution is neutral and aerated, hydroxyl ions are formed as follows. 2 H2O + O2 + 4e - 4OH - d) Formation of corrosion product: The hydroxyl ions migrate towards anode and reacts with metal ions (M n+ ions) and forms corrosion product. In the case of iron OH- reacts with Fe 2+ ions and forms an insoluble hydrated ferric oxide known as brown rust. 2Fe ++ + 4OH - 2Fe (OH) 2 2Fe (OH) 2+ O2 + 2H 2O 2 (Fe 2O 3. 3H 2O) rust. 4.3 What is Galvanic series ? The arrangement of elements in the order of their standard reduction potential is reffered to as emf or electrochemical series. Such a arrangement of few elements given in the table. M n+ / M E 0 (v) M n+ / M E 0 (v) Li + /Li -3.05 Cd 2+ /Cd -0.40 K + /K -2.93 Sn 2+ /Sn -0.14 Ba 2+ /Ba -2.90 Pb 2+ /Pb -0.13 Ca 2+ /Ca -2.87 H + /H2 0.00 Na + /Na -2.71 Cu 2+ /Cu 0.34 Mg 2+ /Mg -2.37 Ag + /Ag 0.80 Al 3+ /Al -1.66 Hg 2+ /Hg 0.85 Zn 2+ /Zn -0.76 Pt 2+ /Pt 1.20 Fe 2+ /Fe -0.44 Au 3+ /Au 1.38 1)A negative value indicates oxidation tendency and a positive value indicates reduction tendency with respect to hydrogen. SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 65 2)The metal with lower electrode potential are more reactive and as the electrode otential increases, the reactivity decreases and metals with higher electrode potentials are more noble. 3)Metals with lower electrode potentials have the tendency to replace metals with higher electrode potential from their solutions for example, Zn displaces Cu, Cu displaces Ag 4)Metals with negative electrode potentials can liberate hydrogen from acidic solutions 4.4 Explain the different types of corrosion. 6 Marks The corrosion can be mainly classified as follows 1. Differential metal corrosion. 2. Differential aeration corrosion 3. Stress corrosion and 4. Grain boundary corrosion 1. Differential metal corrosion. This type corrosion occurs when two different metals are in contact with each other due to the formation of galvanic cell. The metal having less standard reduction potential value under goes oxidation and liberates electrons, which migrates to the cathode. The other metal having high SRP value acts as cathode and reduction reaction takes places on its surface forming OH- ions or any one kind of reduction reactions. The rate of corrosion depends on the potential difference between the two metals. If the difference is more corrosion occurs faster and vice versa. The anodic metal undergoes corrosion and cathodic metal is unaffected .The reactions that occurs are At anode M M n+ + ne - At cathode Depending on the nature of corrosion environment the cathodic reaction are as fallows i) 2H + + 2e - H2 (in acidic environment) ii) H2O + O2 + 4e - OH - (in neutral environment) iii) 2H2O + 2e - 2OH - + H2 (in alkaline environment) Example: Iron metal in contact with Copper metal, Brass tap in contact with Iron pipe etc. SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 66 2. Differential aeration corrosion. This type of corrosion occurs when a metal is exposed to different concentrations of Oxygen. The part of metal which is more exposed to air act as cathode and unaffected. The other part of the metal, which is less, exposed to air act as anode and undergoes corrosion. Example i) Water line corrosion: It is differential aeration type of corrosion observed in water storage tanks, ships, etc.During water line corrosion, the part of the Metal below waterline is exposed to less Oxygen concentration act as anode and undergoes corrosion than the other part which is more exposed to atmospheric Oxygen which act as cathode. At anode Fe Fe ++ + 2e _ At cathode 2Fe ++ + 4OH - 2Fe (OH) 2 2Fe (OH) 2+ O2 + 2H 2O 2(Fe 2O 3. 3H 2O) rust. Example ii) Pitting corrosion: Pitting corrosion occurs when small particle like dust, mud etc get deposited on metals surface. The portion of metal covered by the dust or other particles is less aerated and act as anode. The other portion of the metal exposed to more oxygen of the environment act as cathodic region. Corrosion takes place at the portion below dust and a small pit is formed. Then the rate of corrosion increases due to small anodic area and large cathodic area. SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 67 3. Stress corrosion. Stress corrosion is seen in metals suffering from stress which may result from mechanical operations such as design riveting, cold working, welding, bending, pressing, quenching etc. In an corrosive environment the stressed portion act as anode and undergoes corrosion. The other un stressed part of the metal acts as cathode. Example: Caustic embritlement in boilers or intercrystalling corrosion. It is a form of stress corrosion that takes places in boilers operating at high pressure between 10 to 20 atmospheres. The stressed portion contains fine hair like cracks. Boiler water contains of alkali, which enters into the cracks. The water evaporates leaving behind caustic soda. An electrochemical cell is set up between the iron under stress and iron in the main body. The iron surrounded by dilute NaOH act as cathode and iron under stress act as anode and gets corroded resulting in failure of boiler. 4.5 Explain the affect of factors on the rate of corrosion. 6 Marks The main important factors, which affect on the rate of corrosion, are a) Nature of the metal and b) Nature of the environment SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 68 a) Nature of the metal: - i) Nature of corrosion product: If the nature of corrosion product forms, a protective layer on the metal surface, it prevents the further corrosion of metal .If the corrosion product is thin, invisible and doesn’t form a protective layer it leads to further corrosion of metal. Example: In oxidizing environment metals like Al, Cr, Ti etc. forms protective metal oxide films on their surfaces which prevents further corrosion .Metals like Zn,Fe ,Mg ,etc .do not form protective layer and are readily under goes corrosion .Therefore the rate of corrosion depends on the nature of corrosion product. ii) Electrode Potential of anode and cathode. The rate of corrosion increases if the potential difference between anode and cathode is high and vice versa. When two different metal with large electrode potential difference (SRP values) under goes very fast corrosion Example: The potential difference between copper and iron is 0.78 V. The potential difference between iron and tin is 0.3 V. In these two cases iron undergoes fast corrosion when it is in contact with copper than when it is in contact with tin. iii) Anodic to cathodic areas: If a metal has small anodic and large cathodic area the rate of corrosion increases and vice versa. This is because when anode is small the electrons liberated during oxidation are completely consumed on large cathodic surface for the reduction reactions and rate of corrosion increases. b) Nature of the environment: i) Temperature: Increases in temperature results in an increase in the conductance of the aqueous medium and rate of corrosion also increases and vice versa. ii) PH: In general at lower PH value the rate of corrosion is more at higher pH value (more than pH = 10) the rate of corrosion ceases due to the formation of protective coating of hydrous oxides on the metal. Corrosion rate is maximum between PH 3 and 10 in presence of oxygen SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 69 iii) Anodic polarisation and cathodic polarization The process where there is a variation of electrode Potential due to the inadequate supply of species from the bulk of the solution to the electrode is known as polarisation. Anodic polarization: Increase in anodic polarization, decreases the rate of corrosion. Due to the anodic polarization the tendency for oxidation decreases. Cathodic polarization: Increase in Cathodic polarization, decreases the rate of corrosion. Due to the cathodic polarization the tendency for reduction decreases. Due to inadequate Supply of electrons and oxidation cannot continue. Therefore rate corrosion decreases. 4.6 Mention the corrosion control methods. 2 Marks The important methods used to control corrosion are, 1. Design and selection of the materials. 2. Protective coatings. a) Organic coatings – paints and enamels. b) Inorganic coatings i) Metal coatings – anodic and cathodic ii) Surface conversion coatings – anodizing, phosphating. 3) Corrosion inhibitors 4) Cathodic protection 5) Anodic protection. 4.7 What are Inorganic Coatings? 4 Marks i) Metal coatings – anodic and Cathodic ii) Surface conversion coatings – anodizing, phosphating. i) Metal coatings Deposition of protective metal over the surface of base metal (metal to be procted from corrosion) is known as metallic coatings. It is divided into Anodic and Cathodic metal coatings. a) Anodic metal coatings: Anodic metal coatings involve coating the base metal with more active metals, which are anodic to the base metal. Example: Galvanization b) Cathodic coatings: Cathodic coatings involve coating a base metal with more noble metals, which are cathodic to the base metal. Metals SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 70 such as Copper, Nickel, and Tin Silver etc are coated on Iron. One of the disadvantage of Cathodic coatings is if coating ruptures it leads more corrosion because of small anodic area and large cathodic area. Example: Tinning Surface conversion coatings Surface conversion coatings are chemical conversion coatings. The surface layer of the base metal is converted into a compound by chemical or electrochemical reactions, which prevents the base metal form corrosion. The coating can be done by chemical dip, spray or by electrolytic method. The coating helps in the increased electrical insulation, enhanced adherence for paints and prevention of corrosion. Example: Anodizing , Phosphating 4.8 Explain the process of Galvanization 5 Marks It is a process of coating the base metal surface with Zinc, tin, lead, or aluminum metal. Example: coating Zinc on Iron by hot dipping Method. It involves the following steps. The Iron metal surface is washed with organic solvents to remove oil, grease etc content on the metal surface. Then the metal is passed through dilute sulphuric acid to remove rust and other depositions Finally the metal is washed with water and dried. The metal is then dipped in molten Zinc and passed through Ammonium chloride and Zinc chloride flux to prevent oxidation of Zinc. The excess Zinc is removed by passing through the rollers or by wiping. Uses: Galvanisation is used for roofing sheats, buckets, bolts, nuts, nails, pipes etc. 4.9 Explain the process of Tinning. 5 Marks Tinning is a process of coating the base metal with Tin (Sn).It is carried out by hot dipping method as fallows. The base metal surface is washed with organic solvents to remove oil, grease etc content on the metal surface. Then the metal is passed through dilute sulphuric acid to remove rust and other depositions Finally the metal is washed with water and dried. SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 71 The metal is passed through Ammonium chloride and Zinc chloride flux and then dipped in molten Tin Finally it is dipped in palm oil to prevent oxidation of Tin. The excess Tin is removed by passing through the rollers or by wiping. Tinning is used for food storage articles. 4.10 What is Anodizing? Explain the process of Anodizing of aluminum 6 Marks Anodizing is a process of artificially converting protective passive oxide film on the surface of metals such Al, Zn, Mg, Ti, Zr, Ta, Cr, etc by electrochemical oxidation. ANODIZING OF ALUMINUM. Aluminum when it is made as cathode it allows the passage of electrons but it ceases to conduct when made as anode. It exhibits anodic passivity in the medium like chromic and sulphuric acids. This is known as anodic oxidation or anodizing. Process of anodizing aluminum. Pretreatment: The article be anodized is degreased and followed by electro polished Aluminum is connected to positive terminal i.e., made as anode. Steel or copper is made as cathode The anode and cathode are dipped in electrolyte solution containing 5-10% chromic acid. The temperature of the both is maintained at 35 o c Voltage is applied between 0-50V. First ten minutes potential is increased to 0-40V. After 20 minutes voltage is applied from 40-50V The voltage is kept constant at 50 V for five minutes. SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 72 An opaque oxide layer of 2-8 micrometer thick aluminum oxide layer is obtained. For higher thickness 10% H2SO4 is used as electrolyte. Finally the object is treated with Ni or Cobalt acetate fallowed by boiling water treatment to improve corrosion resistance. Uses: Anodized articles are used as Tiffin carriers, soapboxes, household utensils, window frames, etc. 4.11 Explain the process of Phosphating. 6 Marks Phosphate coating is a process of converting the surface atoms of the base metal into their phosphates by chemical or electrochemical reaction between base metal and certain metal phosphates in aqueous solution of phosphoric acid. Phosphate coating is generally obtaining on steel, Al, and Zn. The metal to be coated (base metal) is degreased, polished washed and dried It is dipped in a solution containing mixture of phosphoric acid, metal phosphates such as Fe / Mn / Zn phosphates. Accelerators such as copper salts, H2O2, nitrate etc. pH of the bath is maintained between 3.2 – 7.8 Temperature is maintained about 35 O C. The metal ions dissolve and react with phosphate ions forming metal phosphate. The metal phosphates deposits on the surface of base metal. Uses: - Phosphate coated metal widely used in automobile industry, bolts, nuts, and refrigerators, washing machining, cars bodies etc. 4.12 Write a note on Organic coatings 4 Marks Organic coatings are inert organic barriers applied on metallic surface to protect metals from corrosion. A thin coat of paint or enamels or lacquers is coated on the metal surfaces. PAINTS: Paint is a mixture of one or more pigments in organic medium containing oil. This is mixed with volatile thinners and applied on metal surface. The volatile thinner solvent evaporates leaving behind the paint on metal surface. SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 73 Enamels: These are dispersions of pigment in varnishes these are applied on metal surface to protect them from corrosion. Uses: Organic coating are applied by different methods such as brushing, spraying dipping roller coating etc. these organic coatings protects the metals form corrosion and also increases the decorative ness of the metal. 4.13 Explain corrosion inhibitors with suitable examples. 5 Marks These are the substances added in small concentrations to a corrosive environment to decrease the rate of corrosion. These slow down the anodic or cathodic reactions. Anodic inhibitors. These inhibit the anodic oxidation reactions. The anodic inhibitors include oxidizing agents such as sodium chromate, molybdates, tungstates, chromates, nitrates etc. These prevent anodic reaction by forming a protective layer on metal surface and prevent further anodic reaction. Cathodic inhibitors These are the substance, which slow down the cathodic reaction. The cathodic reactions involve liberation of hydrogen in acidic solution or OH - ions in alkaline and neutral medium. The cathodic organic inhibitors include amines, thiourea, sulphoxides etc. The two types of cathodic inhibition reactions are liberation of hydrogen, absorption of oxygen and formation of hydroxyl ions. i) Inhibition of oxygen absorption Removing oxygen from the corrosion media by using reducing agents or oxygen scavengers such as hydrazine, sodium sulphite, etc can control absorption of oxygen during cathodic reaction. Thus rate of corrosion can be controlled. O2 + NH2 - NH2 N2 + 2H2O. 2 Na2 SO3 + O2 2Na2 SO4 ii) Inhibition of Hydrogen liberation Adding substances like thiourea can prevent the diffusion of H+ ions to the cathode, amine etc which forms protective film on cathode and prevents the diffusion of H + ions. Increasing hydrogen over voltage by adding AS2O3, Antimony or stats like sodium metal arsenate, can prevent the evolution of hydrogen gas at cathode. SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 74 iii) Inhibition of hydroxyl ions (OH - ) Corrosion can be prevented by adding salts such as ZnSO4, MgSO4, NiSO4, etc which migrates to cathode and reacts with hydroxyl ions and reduces all OH - ions. Zn 2 + + 2 OH - Zn (OH)2 Mg 2+ + 2OH - Mg (OH)2 4.14 What is cathodic protection? Explain Sacrificial anodic protection method and Impressed voltage method 6 Marks Cathodic protection is a method in which the base metal to be protected from corrosion is made to act as cathodic by attaching more active anodic metal to it. The active anodic metal undergoes corrosion and base metal is protected from corrosion the following methods are used to protect metal from corrosion. Example: a) Sacrificial anodic protection method b) Impressed voltage method a) Sacrificial anodic protection method. In this method the more active metals like Zn, Mg, and Al etc are attached to base metal. The anodic metals being more reactive undergoes corrosion but base metal remains unaffected .The sacrificial anodes have to be replaced from time to time after complete corrosion. The method is used for protecting buried pipeline, ship hulls, industrial water tank steel rods in RCC columns. Several hundred kilometers long zinc wire is buried along oil pipe line in Alaska is an example for sacrificial anodic protection method b) Impressed voltage method. In this method the base metal is connected to negative terminal of the D.C source and made it cathode an inert anode consists of resin bonded graphite rod, silicon - iron alloy (or) platinized Ti (or) Ta is connected to positive terminal of the D.C source. The protected base metal does not under goes corrosion. SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 75 4.15 Explain the Anodic Protection method. 6 Marks In this method, the metal to be protected from corrosion is made more anodic by passing external impressed current to make it more anodic and it forms a thin oxide film which then protects the metal from further corrosion .The metal part is connected to potentiostat for maintaining constant potential having a reference electrode The auxiliary electrode acts as cathode and base metal acts an anode after passing current the metal from a passive layer which prevents further corrosion. The advantage of the method is that it requires small current. A drawback of this method is that it cannot be applied to the metals which do not passivate. SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 76 Objective type questions 1. Rusting of iron is an example for a) dry corrosion b) electrochemical corrosion c) acid corrosion d) None 2. Differential metal corrosion is a) galvanic corrosion b) Differential aeration corrosion c) stress corrosion d) water line corrosion 3. Pitting corrosion can be explained on the basis of a) relative areas of cathode and anode b) Differential aeration c) centralized corrosion d) all of the above 4. Coating used for the iron container used for food package coated with a) Zn b) Sn c) Pb d) Al 5. Waterline corrosion in steel tank is an example for a) Differential metal corrosion b) Differential aeration corrosion c) pitting corrosion d) stress corrosion 6. Caustic embrittlement in the boiler is an example of a) pitting corrosion b) stress corrosion c) water line corrosion d) Differential metal corrosion 7. Sacrificial anode method of protecting metal is an example of a) anodic protection b) cathodic protection c) metal coating d) organic coating 8. Corrosion can not take place in a) acidic medium b) neutraql medium c) alkaline medium d) vacuum 9. When 2 metals are in contact with each other, metal which undergoes corrosion has a) less standard reduction potential b) high standard reduction potential c) zero standard reduction potential d) high reduction potential 10. In phosphating process, bath contains a) iron phosphate b) zinc phosphate c) manganese phosphate d) any one of the above 11. Galvanising is the process of coating iron with a) Tin b) zinc c) copper d) nickel SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 77 12. Corrosion process is an example of a) oxidation b) reduction c) Electrolysis d) both a & b 13. Rusting of iron takes place in a) acid medium b) alkaline medium c) neutral medium d) all 14. In differential aeration corrosion, part of the metal which is more exposed to air acts as a) cathode b) anode c) insulator d) electrolyte 15. Which of the following is an cathodic inhibitor a) thiourea b) sulphoxides c) hydrazine d) all SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 78 REVIEW QUESTIONS 1. Define corrosion? Explain electrochemical theory of corrosion. 2. Mention the different types of corrosion 3. Explain differential metal corrosion with suitable example. 4. Explain differential aeration corrosion with suitable example. 5. Explain stress corrosion with suitable example. 6. Explain 5 factors that affect on the rate of corrosion 7. Mention the Corrosion control methods: 8. What is galvanization? Explain the process of galvanization. 9. What is Tinning? Explain the process of Tinning 10. What is anodizing? Explain anodizing of Aluminium 11. What is Phosphating? Explain the process of Phosphating. 11. What is Phosphating? Explain the process of Phosphating. 12. What are organic coatings explain their role in preventing corrosion? 13. What are corrosion inhibitors? Explain anodic and catholic inhibitors. 14. What is Cathodic protection explain with examples. 15. Explain the corrosion control by Anodic Protection method. SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 79 UNIT UNIT UNIT UNIT – –– – V VV V METAL FINISHING 5.1 What is metal finishing? Mention the Technological importance of metal finishing. 6 Marks It is a process of modifying surface properties of metals by deposition of a layer of another metal or polymer on its surface, by the formation of an oxide film. Technological importance of metal finishing. The main technological importance of metal finishing include 1. Imparting the metal surface to higher corrosion resistance. 2. Imparting improved wear resistance. 3. Providing electrical and thermal conducting surface. 4. Imparting thermal resistance and hardness. 5. Providing optical and thermal reflectivity. 6. In the manufacture of electrical and electronic components such as PCB’s, capacitors contacts etc 7. In electro framing of articles, electrochemical machining, electro polishing and electro chemical etching. 8. To increase the decorativeness of metal surface. 9. In electrotyping and to build up material or restoration. 10. To improve wear resistance or solder ability. 5.2 Explain the term Polarization. 3 Marks The process where there is a variation of electrode Potential due to the inadequate supply of species from the bulk of the solution to the electrode is known as polarization. When there is a passage of current, the metal ion concentration in the vicinity of the electrode surface decreases due to the reduction of some of the metal ions into metal atoms. M n+ + ne - M . As a result there will be a change in the electrode potential, however equilibrium is reestablished due to the diffusion of metal ions towards the electrode. If the diffusion is slow the electrode potential changes and the electrode is said to be polarized. Polarized electrode uses more negative potential than required in order to maintain given current. Electrode polarization depends on several factors as follows: SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 80 Nature, size, shape, and composition of electrode. Conductivity and concentration of electrolytic solution. Temperature. Products formed at the electrodes. Rate of stirring of the electrolyte. 5.3 Explain the term decomposition potential. 3 Marks It is defined as the minimum external voltage that must be applied in order to bring about continuous electrolysis of an electrolyte. Electrolysis of an electrolyte occurs only when applied voltage is above certain value below which electrolysis do not occur. This can be determined by an electrolytic cell, if dilute acid or bases are used as electrolytes decomposition occurs when voltage is increased with sudden evolution of H2 and O2 at the electrodes. It requires more than 1.7v. The decomposition potential for Zn and iodine cell is experimentally found to be 4.3volts. The decomposition potential is represented as ED. ED = E cathode – E anode 5.4 Explain the term over voltage. 3Marks It is defined as excess voltage that has to be applies above theoretical decomposition potential to bring the continuous electrolysis of an electrolyte is known as over voltage. The theoretical voltage required for the decomposition of aqueous solution of an acid is equal to the emf of the reversible cell with hydrogen and oxygen gases at one atmosphere. This is known to be about 1.23 volts at ordinary temperature with platinised platinum electrodes. It is however found that the observed decomposition potential is always higher than the theoretical value. Thus with platinum and lead electrodes, a voltage of 1.7 and 2.2 is respectively required for the electrolysis of dilute sulphuric acid as against a theoretical value of 1.23 volt. This difference between the observed and theoretical decomposition potential is called over voltage. Over voltage of an electrolyte depends on i) Nature and physical state of the metal employed for the electrodes. ii) Nature of the substance deposited. iii) Current density iv) Temperature. v) Rate of stirring of electrolyte. SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 81 5.5 Explain the process of electroplating. 6Marks It is a process of deposition of a metal by electrolysis, over the surface of substrate. The substrate may be another metal, polymer, ceramic, or a composite. The principal components of electroplating process are: - 1. Electroplating bath. 2. Cathode. 3. Anode. 4. Electroplating tank. 5. D.C. electrical power source. 6. Reactions at anode and cathode. 1) Electroplating bath: The plating bath contains solution for plating process. It is normally a mixtures of metal ion solution, other electrolytes, complexing agents and various organic additives added to improve the nature of deposit. 2) Cathode: The substrate to be plated is made cathode and suspended as separate bus bars. These cathode electrodes are placed in electrolytic bath solution. 3) Anode: The metal which is to be plated on the other metal is made as anode. Anode is used in the form of a rod, a plate or pellets. The anode is enclosed inside an anode bag to retain impurities. 4) Electroplating tank: It is made of wood or steel. If steel tanks are used these are thermally insulated with ceramic or polymeric materials. The volume of tank may vary from 20-100 dm 3 . 5) D.C.Electrical Power Source: A D.C voltage of 8-12v & an operating current density 1-200 mA/cm 2 is used for plating process. Motor generators or d c rectifiers are used to provide current. 6) Reaction at anode and cathode: Electroplating is the process of electrolytically depositing a layer of metal onto a surface. The anode metal undergoes oxidation liberating metal ions & electrons. The electrolytic bath containing metal ion undergoes reduction to metal atoms and get deposited on the metal or substrate to be plated. SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 82 5.6 Explain the factors influencing on the nature of electro deposit. 6Marks The various factors which influences on the nature of electro deposit are as follows:- 1. Metal ion concentration. 2. Electrolytic concentration. 3. Complexing agents. 4. Organic additives. 5. Current density. 6. P H 7. Temperature. 8. Throwing power of plating bath. 1. Metal ion concentration: For a good adherent deposit, the metal ion concentration should be low. The low metal ion concentration can be achieved by addition of compounds having common ion or by the formation of complex compounds and ions. 2. Electrolyte concentration: Electrolytes are added in high concentrations to increase the conductivity of the plating bath. The electrolytes used do not participate in the electrode reactions. 3. Complexing agents: These are added to convert metal ions to complex ions to keep metal ion concentration at optimal range. The complexing agents are added when plating metal ions are known to react with the cathode and to prevent passivation of anode and the anode dissolves slowly. The complexing agents also increase throwing power. Ex. Cyanides, hydroxides, sulphonates. 4. Organic additives: These are added to improve the nature electrodeposit. They modify the structure, morphology, and properties of the electrodeposit. The different organic additives used are as follows: - a) Brighteners, b) Levelers, c) Structure modifiers, d) Wetting agents. a) Brighteners: These are added to obtain a bright and microscopically fine deposit. Ex. Sulphonates, & compounds containing C = O, N = C = S etc b) Levelers: Certain organic compounds are added as levelers in order to get a level deposit. Sodium allyl sulphonates etc c) Structure modifiers or stress relievers: These additives modify the structure of the deposit and orientation in such a way as to alter the deposit properties. Ex. Saccharin. SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 83 d) Wetting agents: Wetting agents are used to release hydrogen gas bubbles from the surface. The wetting agents also improve the adhesion of the deposit. 5) Current density: It is the current per unit area and expressed in mA/cm 2 of the electrode surface. At low current density the discharge of ions occurs slowly and deposits are coarsely crystalline and loosely held. At higher current density hydrogen evolution occurs and deposits are spongy, irregular and loosely held. Deposits may have a burnt appearance. For good deposit the current density should be optimum. 6) P H : If ph of the medium is low H2 has is evolved causing brittle and burnt deposit. At higher P H values, deposits of metallic oxides or hydroxides may form. Hence an optimum P H between 4 to 8 is employed. 7) Temperature: A good deposit formed at slightly higher temperatures. At very high temperatures hydrogen evolution takes place at the cathode forming a burnt deposit. Therefore moderate temperature between 35 to 60 0 C is used to get good deposit. 8) Throwing power of a plating bath: The ability of a plating bath to give a uniform and even deposit on the entire surface of the object, irrespective of its shape is known as throwing power. Throwing power of plating bath can be determined by using “Haring blum cell.” Haring Blum cell consists of plating bath solution who’s throwing power is to be determined. Anode is placed at the center and two cathodes are placed either side of the anode at a distance C1 & C2. Electroplating is carried out and weight of electro deposit on the two cathodes is weighed. The weight of electro deposit (W1) on cathode 1 which is placed far from anode, is less than another cathode 2, which is very nearer to the anode. Then throwing of plating bath is calculated from the equation, SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 84 100X (x - y) % throwing power = (x + y)-2 Where x = C1/C2 (When C1> C2) y = W2/W1 5.7 Explain the process of surface preparation prior to the electroplating process 4Marks The object before subjecting to electroplating process, it is essential to clean the surface of object. The following methods are used to clean the metal surface. i) Solvent cleaning: Organic solvents such as Trichloro ethylene and methylene chloride are used to remove organic matter and grease on the surface of cathode. Trichloro ethylene is used to remove paint varnishes, films, resins etc. Perchloro ethylene is used to remove high melt waxes and cleaning PCB. The above process is called solvent cleaning. ii) Alkali cleaning: After solvent cleaning the metal is cleaned with alkali solution. Alkali cleaners such as soaps, detergents, sodium carbonate, sodiumhydorxide, sodium phosphate etc are commonly used. iii) Mechanical cleaning: The object is subjected for mechanical cleaning to remove oxide scales, rust and other impurities on the metal surface. The mechanical cleaning methods involves, cleaning with Bristle brushes, mechanical polishing, grinding using polishing machines, buffing and sand blasting. iv) Pickling: In this method the oxides scales are removed by dipping the base metal in a dilute acids. Pickling of steel involves dipping in dilute HCl or dilutes H2SO4 to remove rust and other oxide scales. v) Rinsing with water: After surface cleaning process the metallic surface is thoroughly rinsed with water, dried and subjected for electroplating process. SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 85 5.8 Explain the Electroplating process of chromium. 6 Marks Plating bath Chromic acid and H2SO4 in 100:1 proportion. Temperature 45-60 0 C. Current Density 100-200mA/Cm 2 . Anode Insoluble anodes Pb-Sb or Pb-Sn coated with PbO2. Cathode Object to be plated. Chromium anodes are therefore not used in Cr plating for following reason. • Chromium metal passivates strongly in acid sulphate medium & • Chromium anode gives rise to Cr (III) ions on dissolution. In presence of large concentration of Cr (III) ions, a black Cr deposit is obtained. Applications: a) Decorative chromium provides a durable finish on cycles, automobiles, furniture’s, air craft and surgical instruments etc. b) Hard Chromium is used in cutting tools, piston rings, cylinder liners, crankshafts of marine & aero engines, bearings etc. c) Black Chromium is used in optical instruments, machine tools & electronic parts. d) It is also used for non-glase finishes on automobiles, & as an efficient coating for solar energy collectors. 5.9 Explain the Electroplating process of copper 6 Marks Electroplating of copper Sulphate bath Cyanide bath Plating bath solution 200-250 g CuSO4, 50- 75 g H2SO4, per L of bath solution 40-50 g CuCN, 20-30g KCN, 10 g K 2CO3 per L Operating temperature 20-40 0 C 40-70 0 C Current density 20-50 mA/cm 2 10-40 mA/cm 2 Addition agents Gelatin, dextrin, sulphur containing brighteners, sulphonic acids Sodium thiosulphate Current efficiency(%) 95-99 60-90 SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 86 Anode Phosphorous containing rolled Cu O2 free high conductivity Cu Cathode object to be coated pretreated object to be coated pretreated Applications In PCR boards In Cr plating, PCR boards 5.10 What is Electro less plating? 2 Marks Electro less plating is the controlled autocatalytic deposition of a continuous film of a metal from its salt solution on a catalytically active surface by a suitable reducing agent without using electrical energy. Metal ions + reducing agent Metal + oxidized product. 5.11 Distinction between electroplating and Electroless plating 4 Marks Electroplating Electro less plating 1. Require electrical power source and accessories. 1. Does not require electrical power source and accessories. 2. Deposition can’t be made on non-conductors such as plastics, ceramics etc. 2. Deposition can be made on non-conductors such as plastics, ceramics etc. 3. requires Levelers 3. Does not require Levelers 4. Plating baths don’t have excellent throwing power. 4. Plating baths have excellent throwing power. 5.12 Explain the Electro less plating process of Nickel. 6 Marks Plating bath solution: A solution of NiCl2 20g /litre, Reducing agent: Sodium hypophosphate 20g/litre Buffer: Sodium Acetate 10g/lt Complexing agent: Sodium Succinate. P H : 4.5 Temperature: 93 0 C SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 87 Anodic reaction: H2PO2 - + H2O H2PO3 - + 2H + + 2e - Cathodic reactions: Ni 2+ + 2e - Ni. Overall reaction: Ni 2+ + H2PO2 - + H2O Ni + H2PO3 - + 2H + . Uses: i)Used to plate on industrial components such as pumps and valves, shafts, gears, reaction vessels and other tools. ii) To coat on steel, plastic, PCB’s etc. 5.13 Explain the Electro less plating process of copper. 6 Marks Electroless plating of copper on PCB’s is carried out as follows. Plating bath Solution: CuSO4..5 H2O (12 g per dm 3 ), NaOH (15 g per/L Rochelle salt (14 g per dm 3 ), EDTA (15 g per dm 3 ) Reducing Agent: Formaldehyde (8 g per dm 3 ) Complexing agent: EDTA Solution. P H : 11 - 12 Temperature: 25 0 C Electrode Reactions: At Anode: 2HCHO + 4OH - 2HCOO - + 2H2O + H2 + 2e - At Cathode: Cu 2+ + 2e Cu. Overall reactions: 4OH - +Cu 2+ + 2HCHO - Cu + 2HCOO - + 2H2O +H2 APPLICATIONS: Widely used for moralizing (PCB’s) printed circuit boards. For producing through hole connections. For plating on non-conductors. As an undercoat for electroplating. For decorative plating on plastics. SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 88 Objective Type Questions: 1) In the electroplating process the over voltage potential depends on a) Electrolyte b) Temperature c) Current density d) All 2) In chromium plating anode is a) Soluble Chromium anodes b) Insoluble anodes c) Inert anodes d) both b & c 3) Autocatalytic reaction method of plating is also known as a) Electroplating b) Electrolysis plating c) Electro less plating d) electro refining 4) For a electrolytic mixture containing Zn 2 + , Cu 2+ , Ag + the ion which is going to be discharge first is a) Zn 2+ b) Cu 2+ c) Ag + d) None 5) Conductors and insulators can be plated by a) Electroplating b) Electro less plating c) Electro polishing d) None 6) The phenomenon in which the back EMF produced due to the products of Electrolysis is a) Electroplating b) Electro less plating c) Polarization d) None 7) When the metal structure to plated is irregular the process employed is a) Electroplating b) Electro polishing c) Electro less plating d) None 8) Addition of Complexing agent to the plating bath is to a) Increase the rate of electro deposition b) increase the metal ion concentration c) Decrease the metal ion concentration d) None 9) Over voltage depends on a)current density b) Temperature c) Nature of the substance deposited d) All 10) For good deposition throwing power of a plating bath has to be a) Low b) High c) Zero d) No effect SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 89 11) For a good deposition, metal ion concentration should be a) low b) high c) zero d) none 12) Organic additives are added to bath solution to get a) bright deposit b) level deposit c) fine deposit d) all 13)In electroplating of copper, anode is a) insoluble b) coated with PbO2 c) both a & b d) copper 14) In electrolesss plating process, a) levelers are not required b) current desity is not required c) reducing agent is required d) all 15) In electro less plating process of copper, plating bath contains a) CusO4 b) EDTA c) HCHO d) All SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 90 REVIEW QUESTIONS 1. What is metal finishing? Mention the Technological importance of metal finishing 2. Explain the term Polarisation. Mention the factors that affect Polarisation 3. Explain the term decomposition potential. 4. Explain the term over voltage. Mention the factors that affect over voltage 5. Explain the process of electroplating 6. Explain the factors influencing on the nature of electro deposit. 7. Explain the methods of cleaning a metal surface prior to Electroplating. 8. Explain electroplating of chromium. 9. Explain the electroplating of Gold 10. What is Electroless plating? Mention the advantages of Electroless plating over electro plating. 11. Distinction between electroplating and Electroless plating 12. Explain electroless plating of Nickel 13. Explain electroless plating of copper. SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 91 UNIT UNIT UNIT UNIT – –– – V VV VI II I PHASE RULE 6.1 what is phase diagram? 2 mark A single graph depicting the overall relationships among the various phases (solid, liquid and vapour phases) of a substance is known as a phase diagram. A phase diagram summarizes the conditions at which a substance exists as a solid, liquid or gas. Phase diagrams are of considerable commercial and industrial significance, particularly for semiconductors, ceramics, steels, polymers, composites and alloys. They are also the basis of separation procedures in the petroleum industry and cosmetic preparations. 6.2 what is Gibbs Phase Rule? Explain the terms. 4 Marks The number of degrees of freedom (F) of a heterogeneous system at equilibrium at a definite temperature and pressure is related to the number of components (C) and of phases (P) by the phase rule equation, P + F = C + 2 or F = C - P + 2 Where P=No of phases in the system C= No of Components of the system F= No of degrees of freedom or variables of the system(T,P and Composition or concentration) 6.3 Define PHASE . 2 Marks A phase is defined as any homogeneous physically distinct and mechanically separable part of a system which is separated from other parts of the system by definite boundary lines. Eg. 1) A mixture of Ice, liquid water and water vapour is a two-phase system. 2) Two miscible liquids -is one phase system. e.g. Ethanol and water is a one phase system. 3) Two immiscible liquids- is two phase system.e.g. Chloroform and water constitutes a two-phase system. Ether and water constitutes a two-phase system. 4) An aqueous solution of a solid substance such as sodium chloride or sugar is a one-phase system. 5) All gases mixture freely to form a homogeneous mixture. Therefore, any mixtures of gases, say oxygen and nitrogen is a one- phase system 6)Decomposition of calcium carbonate to calcium oxide and CO2 is a three phase system with two solid phases and one gaseous phase. CaCO3(s) ⇐⇒ CaO(s) + CO2(g) SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 92 6.4 Define COMPONENT : (C) 4 Marks It is the smallest (least) number of independent chemical constituents by means of which the composition of each phase present in the particular system can be expressed, either directly by formula or in the form of a chemical equation. i) One component :- 1) Water system has three phases i.e. Ice(s) ↔ Water(l) ↔ Water vapour(g) The composition of all the three phases is expressed in terms of one chemical constituent, H2O. Thus water system is a one component system. 2) Sulphur has four phases : i) rhombic ii) monoclinic iii) liquid and iv) sulphur vapour. The composition of all these phases can be expressed by one chemical individual, S(sulphur). Thus sulphur system has one component only. 3) A mixture of gases Nitrogen and Oxygen is a one phase two component system. 4) Two miscible liquids is one phase two component system. 5) Two immiscible liquids- is two phase two component system. 6) An aqueous solution of a solid substance such as sodium chloride is a one-phase two component system. 7) A saturated solution of a solid substance such as sodium chloride is a two-phase two component system. 8) Dissociation of ammonium chloride in a closed vessel is a two-phase one component system. NH4Cl(s) ↔ NH3(g) + HCl(g) The composition of both NH3 and HCl can be expressed in terms of NH4Cl vapour. ii) Two component :- 8) Decomposition of calcium carbonate. CaCO3(s) ↔ CaO(s) + CO2(g) It has three phases; solid CaCO3, solid CaO and gaseous CO2. The composition of all the phases can be expressed in terms of any two (of three) chemical substances in equilibrium. 9)Dissociation of copper sulphate is a two component system. CuSO4.5H2O(s) ↔ CuSO4.3H2O(s) + 2H2O(g) Components are CuSO4 and H2O 10)Dissociation of ammonium chloride in a open vessel is a two-phase two component system. NH4Cl(s) ↔ x NH3(g) +y HCl(g) The Components are NH4Cl and NH3 in excess or HCl in excess SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 93 6.5 Define DEGREES OF FREEDOM OR VARIANCE (F): 3 Marks The number of degrees of freedom of a system is the minimum number of independently variable factors, such as temperature, pressure and composition which need to be fixed in order that the condition of a system at equilibrium may be completely defined. Eg. 1) Thus, the state of a pure gas can be specified by two variables, pressure and temperature or pressure and volume. This means that a pure gas has two degrees of freedom. 2) A system consisting of boiling water. In this system, water (l) and steam (g) are in equilibrium and does not have a composition variable since both water and steam contain molecules of the same fixed formula, H2O. To define the system, it is necessary to specify only the steam pressure because the temperature of boiling is automatically fixed (or vice versa). Application of the phase rule to the system gives, P + F = C + 2; F = C- P+2: F = 1-2+2 = 1 i.e. either pressure or temperature but not both. A system with F = 1 is referred to as univariant or monovariant. 3) Ice(s) ↔ Water (l) ↔ Water vapour (g) In the above system all the three phases are present in equilibrium. All the three phases co-exist only at one particular temperature and pressure. If variables like T or P is altered one of the phase disappears. A system with F = 0 is known as invariant. A system with F = 1 is referred to as univariant. A system with F = 2 is called bivariant. 6.6 Calculate the number of components and degrees of freedom in the following equilibria. i). Na2SO4.10H2O (s) ↔ Na2SO4 (s) + 10H2O (g) P = 3; C = 2; F = C-P +2 = 2-3 +2 =1 ii).N2 (g) + O2 (g) ↔ 2NO (g) P = 1; C = 2; F = C-P +2 = 2-1 +2 =3 iii) N2O4 (g) ↔ 2NO2 (g) P = 1; C = 1; F = C-P +2 = 1-1 +2 =2 iv) NH4Cl(s) ↔ NH3(g) + HCl(g), when pNH3 = pHCl P =2; C = 2; F = C-P +2 = 1-2 +2 =1 v) Fe(s) + H2O(g) ↔ FeO (s)+ H2(g) P= 3; C=3; F= C-P +2 = 3-3+2 = 2 6.7 Derive phase rule. 6 Marks Gibbs phase rule is derived from the thermodynamic result Consider a system with component number C is existing altogether in P number of phases in equilibrium. Total number of variables may be calculated as, SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 94 One variable Temperature- same for all phases. One variable pressure - same for all phases . Another variable concentration or composition is independent variable. For a system of one phase, the independent concentration variable in respect of C components is C-1. (The concentration variable is left over value). For a system of P phases it is P (C-1) concentration or composition terms must be specified in order to define the composition completely. In addition to composition or concentration, the temperature and pressure of the system which are the same in all phases must be known. Therefore, the total number of independent variables which must be specified is P (C-1)+2. According to thermodynamics for a system in equilibrium at constant temperature and pressure, the chemical potential (µ) of any given component has the same value in every phase. In a system of phases at equilibrium, the total quantity of material in any phase does not matter, it is the composition of the phase that is important.( Eg., In a saturated solution of a pure solid, the composition of the solution is independent of the total quantity of solution and undissolved solid). The chemical potential of any component is the same in all phases at equilibrium according to equations: µ1(a) = µ1(b) = -------- = µ1(P) µ2(a) = µ2(b) = -------- = µ2(P) µC(a) = µC(b) = -------- = µC(P) for C components in P phases. For one component this give rise to (P-1) restrictions for the P number of phases. For C component this give rise to C (P-1) restrictions for the P number of phases. (Chemical potential µ is a function of temperature, pressure and concentration or composition.) The number of variables remaining undetermined for the system at equilibrium is given by F=Total number of independent variables specified-The number of variables defined. F= [P(C-1) + 2] - [C(P-1)] = C-P + 2 In order to define the system completely, the number of variables must be arbitrarily fixed and hence it must be equal to F. Therefore, F = C-P + 2. SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 95 6.8 Explain the application of phase rule to one component system 6 Marks THE WATER SYSTEM: Consider water system Ice (s) ↔ Water (l) ↔ Water vapour(g) Water can exist in three possible phases, namely, solid ice, liquid water and water vapour. Hence there can be three forms of equilibria, 1. Ice(s) ↔ Liquid water(l) 2. Ice(s) ↔ Water vapour(g) 3. Water(l) ↔ Water vapour(g) The phase diagram or P-T graph for water system is obtained by plotting the pressure against temperature. It consists of i) Three areas, namely, COA, COB and AOB. ii) Three curves, OA, OB and OC. iii) One triple point,O. A A' O C B Ice Water Water vapor 0.006 atm 1 atm 0 100 Temperature P r e s s u r e 0.01 ( C) o Curves: In the phase diagram, the curve OA represents the vapour pressure curve of liquid water. Along this curve two phases co-exist. Water (l) ↔ Water vapour (g) The point N on the curve OA indicates the B.P. of water is 100 o C at 1 atms.pressure. Keeping pressure constant at 1 atms. rise in temperature behind 100 o C brings about a change in the phase from water to water vapour along the line NH. And decrease in temperature below 100 o C reverse is the phase change. The curve OA terminates at A which is the critical point (Critical pressure = 218 atms.pr. & Critical temperature = 374 o C). Behind the critical point water phase merge into vapour phase. The curve OB is known as sublimation curve and it shows the variation of vapour pressure with temperature for the solid ice. The vapour SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 96 pressure increases with increase in temperature. Along this curve two phases exist in equilibrium. Ice (s) ↔ Water vapour (g) The point B is a natural limit at -273 o C, behind this point vapour phase merge into ice. The dotted curve OA | which is the continuous extension of OA is obtained for water below 0 o C. Liquid water may be cooled below its freezing point without solidifying. This is called super cooling. The dotted curve OA | represents the super cooled state of water. This curve represents a meta-stable system. The curve OC represents the melting curve of ice or freezing curve of water. Along this curve two phases exist in equilibrium. Ice (s) ↔ Water (l) The slope of OC towards the pressure axis shows that the melting point of ice is decreases by increase in pressure. It may be seen from the phase diagram that the point M on the curve OC represents the freezing point (0 o C temperature & 1 atms.pr). Any increase in pressure on ice keeping temperature constant ice melts into water along MK. Along the curves OA, OB and OC two phases exist in equilibrium and the number of component of the system is one. By applying the phase rule, F = C - P + 2 F = 1 - 2- + 2 F = 1 It means that only one variable, either pressure or temperature is sufficient to define the system. It implies that pressure is not freely variable, if the temperature is set. Thus each system is univariant. Areas: The phase diagram can be divided into three areas, COB, COA and AOB. Each area contains one pure phase only as mentioned here under. Area Phase COB Ice COA WATER AOB Water vapour Application of phase rule gives (P = 1; C = 1), F = C-P+2; F = 1-1+2; F = 2 Thus each area of phase diagram represents a bivariant system. Both the variables (temperature and pressure) can be varied independently without changing the number of phases. Thus each system namely ice, water & water vapour is bivariant system. SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 97 The phase rule predicts that both the variables T & P are necessary to locate any point G in an area. Triple point: The three curves OA, OB & OC meet at point O, the point O is called triple point. The triple point shows the conditions under which all the three phases solid, liquid and vapour can exist in equilibrium. Ice(s) ↔ Water (l) ↔ Water vapour (g) At the triple point, P = 3, C =1 and F= C-P+2 = 1-3+2 = 0. All the variables are fixed for this system at this point. If any of the variables is altered, the equilibrium will be disturbed and one of the phases will disappear. The triple point is fixed for water system. It occurs at 0.01 o C & 0.006 atms.pressure. Thus the system is zero variant or invariant. Application of phase rule to two component system 6.9 Explain Lead – silver system: 6 Marks Lead-silver system is a two component system. It is a solid / liquid system. Lead-silver system is a condensed system. A system in which only solid and liquid phases are present with gaseous phase being absent is called a condensed system. The degrees of freedom in such a case is calculated by a reduced phase rule equation, F = C-P+1. The phase diagram for lead-silver system is obtained by plotting temperature on Y-axis and % composition on X-axis. SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 98 0% Pb 0 % Ag Composition (%) 2.6% Ag Solid Ag + Eutectic Solid Pb + Eutectic 100 % Pb 100%Ag Liquid + Solid Pb Liquid + Solid Ag E B C D 303 961 Liquid Melt 327 A T e m p e r a t u r e ( C ) o The various features of the phase diagram or P-T graph are illustrated here under. The phase diagram consists of curves, areas and eutectic point. Curves (OA & OB) Pure lead melts at 327 o C and the addition of silver lowers its freezing point along AO. This is called freezing point curve for the lead. Along AO the melting point of lead gradually falls on the addition of Ag till lowest point O (303 o C) is reached where liquid melt gets saturated with respect to Ag and the melting point of Pb does not fall further. On cooling further whole mass (Eutectic composition) crystallize out. Pure silver melts at 961 o C and the addition of lead lowers its freezing point along BO and is called freezing point curve of silver. This curve indicates that the melting point of Ag falls gradually on adding Pb, along BO till the lowest point O (303 o C) is reached Where liquid melt gets saturated with respect to pb and m.p. of Ag does not fall further .On cooling further whole mass (eutectic composition) crystalize out. The phase diagram has the following two curves. Curve OA - Along this curve, solid lead and liquid mixture of Pb + Ag co- exist i.e. P = 2. Curve OB- Along this curve solid silver and liquid mixture of Ag +Pb co- exist i.e. P = 2. Applying reduced phase rule, F = C-P+1 = 2-2+1 = 1 SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 99 The systems solid Pb / liquid mixture of Pb + Ag and solid Ag / liquid mixture of Ag +Pb are univariant systems. For a point on the line change in either T or % composition changes the phase. On the other hand along any of the line one variable T or % C is specified and the other is fixed automatically. Areas : There are three areas in the phase diagram, namely, AOB, AOC and BOD. The phases present in each of these are, AOB - 1 (liquid melt) AOC - 2 (solid Pb + liquid mixture of Pb + Ag) BOD - 2 (solid Ag + liquid mixture of Ag +Pb) Area AOB: By applying phase rule, we have F = C-P+1 = 2-1+1 = 2. It is necessary to specify the temperature as well as the % composition in order to define any point in this area. This system (liquid melt) is a bivariant system. Both T & % C are needed to locate any point in this area AOB. Within this area when two variables are altered no change in phase. Area AOC and BOD: In the areas AOC and BOD, P = 2, C = 2 F = C-P+1 = 2-2+1 = 1. The composition of each phase is fixed, each being a pure component and so only one variable is the temperature. These systems have one degree of freedom. These are invariant systems. To locate any point in these areas either T or %C is needed. Ex: If an alloy of Ag & Pb of %composition Z is heated to a point U & cooled and no solid separates till it reaches a point V on the line BO. At V pure Ag separates out which will be in equilibrium with the liquid melt of Ag+Pb. If it is cooled further to a point Z more silver separates out along VX. The %composition of the liquid melt at Z will be same as that at point X on the line BO. If T is known, the %C of the liquid melt in the areas AOC & BOD can be known or vice-versa. 6.10 What is Eutectic point ? 2 Marks At the point O, where the curves AO and BO meet, both the solids Pb and Ag are in equilibrium with the liquid melt and is called the eutectic point. The point ‘O’ is the lowest temperature at which any liquid mixture of silver and lead will freeze and consequently represent the lowest melting point of any mixture of solid lead and solid silver. At eutectic point, F = C-P+1 = 2-3+1 = 0 SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 100 This is a invariant system or Zero variant system The variables, both temperature (303 o C) and composition (97.4% Pb and 2.6% Ag) are fixed. The eutectic mixture, although has a definite melting point, is not to be regarded as a compound. A mixture containing two components which are not miscible in the solid state is called a eutectic mixture. The eutectic composition is 2.6% Ag & 97.4% Pb. The eutectic temperature is 303 o C. 6.11 What is meant by Desilverisation of lead by Pattinson’s process? 4 Marks The phase diagram of Pb-Ag has a special significance in the desilverisation of lead. The argentiferrous lead contains a very small percentage of silver (0.1%). The argentiferrous lead is first heated well above the melting point of lead so that the system consists only of a liquid phase which is represented by the point, p in the figure. The liquid melt is allowed to cool. The temperature of the melt falls along pp′. As the temperature corresponding to p′ is reached on the curve AC, lead will begin to solidify and the liquid will contain relatively higher silver. . Y P' P A 327 Liquid Melt 961 303 D C B E Liquid + Solid Ag Liquid + Solid Pb 100 % Pb 100%Ag Solid Pb + Eutectic Solid Ag + Eutectic 2.6% Ag % Ag 0% Pb T e m p e r a t u r e ( C ) o Lead continues to separate out and is removed constantly by means of ladles. The melt continues to be richer and richer in silver until the point Y is reached when the percentage of silver rises to 2.6. Thus the original SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 101 argentiferrous lead which contains 0.1% silver now contains about 2.6% silver. The process of raising the relative proportion of silver in argentiferrous lead is known as Pattinson’s process Instrumental Methods of Analysis. 6.12 What is Instrumental Methods of Analysis . 3 Marks It is one of the quantitative analysis methods of a compound. An instrument is used for chemical analysis, converts the property of a substance which is under investigation in to form that can be readily measured. The measured quantity is related to the quantity or quality of the sample The analysis may contain different steps such as generation of signals, transformation of signals, amplification and presentation of signals by displaying on a recorder or read out. 6.13 What are the advantages of Instrumental methods? 2 Marks 1. These methods are faster than chemical methods. 2. These methods requires only small quantities of the sample 3. Easy when a large number of samples have to be analyzed 4. These methods are Accurate. 5. The analytical process can be automated 6.14 Explain the instrumentation and theory of Potentiometric Titrations. 8 Marks Instrumentation. Potentiometer consists of a reference electrode, an indicator electrode & a potential measuring device. The indicator electrode responds rapidly to the changes in the concentration of the analyte i.e. the solution under study. A simple arrangement of Potentiometric titration is depicted in diagram. A is a reference electrode, B is the indicator electrode & C is a mechanical stirrer. The solution to be titrated is taken in the beaker. A known volume of analyte is taken & its potential is determined. The titrant is added in increment of 1ml & the emf is measured each time. At the approach of equivalence point, the emf tends to increase rapidly. At this point, small increments, say, 0.1ml of titrant are added. A few readings are taken beyond the end point. Thus the changes in potential at different volumes of titrant are recorded. SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 102 Theory: Redox titrations can be carried out potentiometrically using platinum- calomel electrode combination in a manner similar to acid- base neutralizations. For the reaction. Reduced from oxidized form + n electrons The potential is given by Nernst equation. Where E o is the standard potential of the system. The potential of the system is thus controlled by the ratio of the concentration of the oxidized to that of the reduced species in the vicinity of the end point of the titration. This may be followed potentiometrically and a change of potential against volume (titration curve) is characterized by a sudden change of potential at the equivalence point. The sudden change in the potential at the equivalence point is explained as follows. The reaction that takes place in the determination of Fe 2+ is Fe 2+ Fe 3+ + e Cr2O7 2- + 14H + + 6e 2Cr 3+ + 7H2O Prior to the equivalence point the potential is determined by the Fe 3+ /Fe 2+ system and the potential is given by the equation. The potential of the solution will be around 0.75V (since the contribution to the potential by the second term is negligible). At the equivalence point the potential is determined by both E ” and E o is given by E 0 Fe 2+ + E 0 Cr2O7 2- 0.75V + 1.33V Ecell = = = 1.04 V 2 2 Beyond the equivalence point the potential is determined by Cr2O7 2- / Cr 3+ system given by the equation. 0.0591 [Cr2O7 2- ] Ecell = Ecr2 = E 0 Cr2O7 2- / Cr 3+ + log 6 [Cr 3+ ] SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 103 0.0591 [Cr2O7 2- ] = 1.33V + log 6 [Cr 3+ ] Thus an abrupt increase in the potential of the solution in the vicinity of the equivalence point is observed. This makes the equivalence point in the experiment, the potential of the cell is determined with reference to saturated calomel electrode. PROCEDURE: Pipette out 25 cm 3 of ferrous ammonium sulphate solution into a beaker. Add one test tube full of dilute sulphuric acid. Immerse the electrode assembly into the solution in the beaker and connect the electrodes to a potentiometer. Measure the potential. Fill the burette with potassium dichromate solution. Add 0.5 cm 3 of K2Cr2O7 to the beaker. Blow the solution carefully using a glass tube and measure the potential after 15 seconds. Continue the procedure till potential shows a tendency to increase rapidly. Determine the end point by differential method i.e., by plotting E against volume (V) as shown in the figure. V Calculate the normality of the ferrous ammonium sulphate solution and determine the amount of iron in the given solution. Advantages. 1. Turbid, fluorescent, opaque or colored solutions can be titrated. 2. Mixture of solutions or very dilute solutions can be titrated. 3. The results are more accurate because the actual end point is determined graphically. SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 104 6.15 Explain the instrumentation and theory of Conductometric estimation. 8 Marks Theory and Instrumentation. Measurement of conductance can be employed to determine the end point in acid-base titrations. In conduct metric titrations there is a sudden increase in conductance of the solution at equivalence point. The principle underlying conductometric titrations is the substitutions of ions by other ions of different mobility. Therefore, the conductance of a solution depends on the number and mobility of ions. The equivalence point is determined graphically by plotting conductance (ordinate) against titer values (abscissa). From the equivalence point the concentration of analyte can be determined. Conductometer consists of a conductivity cell& a conductance measuring device. The cell has 2 platinum electrodes placed at unit distance apart. The assembly responds rapidly to the changes in the concentration of the analyte. The simple arrangement of conductometric titration is represented as follows. Advantages of conductometric titrations 1) Mixture of acids can be titrated more accurately. 2) Colored solutions can be titrated. 3) Very weak acids such as H3 PO3, phenol, which cannot be titrated potentiometrically in aqua solutions can be titrated conduct metric. SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 105 Applications a) Titration of strong acid v/s strong base Conduct metric titration may be applied for the determination of acid present in a. In the titration of a mixture of a weak acid (CH3 COOH) and strong acid (HCl) with a strong base (NaOH), the conductance decreases upon adding NaOH to acid mixture owing to the substitution of highly mobile H + ions of HCl replaced i.e., the strong acid is neutralized. Continued the addition of NaOH raises the conductance moderately, as the weak acid (CH3COOH) is converted into its salt (CH3COONa). Further, addition of NaOH raises the conductance steeply due to the presence of OH - ions (mobility: 198 ohm -1 ). The titration curves in the graph given determine the location of the equivalence points. Estimation of HCL Pipette out 25 cm 3 of the given acid into a clean 100 cm 3 beaker. Dip the conductivity cell in the solution and note down the conductance of the solution i.e., when the volume of NaOH added is zero. Now added standard NaOH solution from the burette in increments of 0.5 cm 3 . After each addition, stir the solution gently and note down the conductance. As the titration proceeds, the conductance first gradually decreases and then rises sharply. Continue until the conductance is more or less same as it was in the beginning. Plot a graph of conductance on y- axis versus volume of NaOH on X- axis to get two straight lines. The point of intersection of the two lines gives the volume of NaOH required to neutralize only HCl. (after drawing a perpendicular lines to x- axis). SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 106 CALCULATION: Estimation of HCL Volume of NaOH required to neutralize HCI V1 = …………cm 3 Normality of HCL = Normality of NaOH x volume of NaOH 25 cm 3 = 0.5 x -------- 25 = -------N Therefore, the weight of HCL/dm 3 = Normality x Eq. wt. of HCL (36.5) = ………..x 36.5 = ………..g/ dm 3 6.16 Explain the Colorimetric estimation of copper. 8 Marks PRINCIPLE: When a monochromatic light of intensity I0 is incident on a transparent medium, apart Ia of it is absorbed, apart Ir is reflected and the remaining part It is transmitted. Io = Ia + Ir + It For a glass- air interface Ir is negligible. Therefore, Io = Ia + It It/Io = T called the transmittance log 1/T = log Io/It is called the absorbance or optical density A. The relation between absorbance A, concentration c (Expressed in mol/liter) And path length 1 (expressed in cm) is given by Beer Lambert’s law. A = log Io/It = €Ct Where C is the molar extinction coefficient, t is the path length. € is a constant for a given substance at a given wavelength. If t- the path length is kept constant, then Aœ C. hence a plot of absorbance against concentration gives a straight line. A series of standard solutions of copper is treated with ammonia to get blue cuprammonium complex and is diluted to a definite volume. The absorbance of each of these solutions is measured at 620 nm since the complex shows maximum absorbance at this wavelength. The absorbance values are plotted against volume of copper sulphate (cm3) to get a calibration curve. A known volume of the test solution is treated with strong ammonia and diluted to the same volume as above. The absorbance of this solution at 620 nm is measured and its concentration is determined from the calibration curve. SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 107 PROCEDURE: Transfer the given copper sulphate solution (stock solution) to a burette and draw out 2.5, 5.0, 7.5, 10.0 and 12.5 cm 3 of the solution into 25 cm 3 volumetric flasks. Add 2.5 cm 3 of ammonia solution to each of them and dilute up to the mark with ion exchange water. Stopper the flasks and mix the solutions well. To the test solution given in a 25 cm 3 volumetric flask, add 2.5 cm 3 ammonia solutions and then dilute up to the mark. Mix well, prepare a blank solution by diluting 2.5 cm 3 of ammonia solution in a 25 cm 3 volumetric flask up to the mark with ion exchange water and mix well. Measure the absorbance of the solutions against blank at 620 nm using photoelectric colorimeter. Tabulate the reading. Draw a calibration curve by plotting optical density (OD) against volume of copper sulphate (cm 3 ). Using the calibration curve find out the concentration of copper in the test solution and calculate the amount of copper in 25 cm 3 of the given solution. Applications. A colometric method may frequently be applied where no satisfactory gravimetric or titrimetric procedure exists i.e., for certain biological substances. .. . 6.17Explain the instrumentation and theory of Flame photometric Estimation Principle: Sodium, Potassium, Calcium and Lithium and other common elements impart characteristic colors with the Bunsen flame. The intensity of the colored flame varies with the amount element introduced. This forms the basis of the flame photometry. When a solution containing a compound of the metal to be investigated is aspirated into a flame, The following process occurs. i) Solvent evaporates leaving behind a solid residue. ii) Vaporizations of the solid coupled with dissociation into its constituents atoms which are initially in the ground state. SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 108 iii) Some gaseous atoms get excited by the thermal energy of the flame to higher energy levels. The excited atoms which are unstable quickly emit photons and return to lower energy state i.e, ground state. Flame photometry involves the measurement of emitted radiation. The relation ship between ground state and excited state populations is given by the Boltzmann equation N1/N0 = (g1/g0) e ∆E/KT N1= Number of atoms in the excited state; N0 = Number of atoms present in the ground state. g1/g0 = Ratio of statistical weight weights for ground and excited states; ∆E = Energy of excitation = hv K = The Boltzman constant; T = Absolute Temperature. From the above equation, it is evident that the ratio N1/N0 is dependent upon both the excitation energy E and the temperature T. An increase in temperature and a decrease in ∆E will both result in a higher value for the ratio N1/N0. Instrumentation of flame photometry: Air, at a given pressure is passed into an atomizer and the suction thus produced draws solution of the sample into the atomizer where it joins the air stream in the form of a fine mist and passes into the burner system. Here the air meets the fuel gas supplied under pressure in a small mixing chamber and the mixture is burnt in the burner. The radiations resulting from the flame pass through an optical filter which permits only the radiations characteristic of the element under the investigation to pass through the photocell. The output from the photocell is measured on a digital read-out system. A calibration curve has to be plotted, first using solution of known concentrations. This is done by aspirating into the flame known concentrations of the element to be determined, measuring the respective readings and plotting them against the respective concentrations of the standard solutions used. SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 109 Estimation of sodium: Procedure: 1. Transfer 2, 4, 6, 8, 10 ml of standard NaCl solution in to 50 ml standard volumetric flasks and dilute up to the mark with distilled water. 2. Set the instrument to read Zero by placing te distilled water in the suction capillary of the instrument. (Using sodium filter (598 nm)) 3. Read the flame emission intensity 2, 4, 6, 8, 10 solutions by placing solutions in the suction capillary of the instrument. (Rinse with distilled water between each reading) 4. Dilute the given test solution up to the mark and shake well and read the flame emission intensity solution as above. 5. Draw a calibration curve by plotting the emission intensity (Y-axis) and volume of NaCl solution (X-axis). 6. From the calibration curve, find out the volume of the given test solution and from which calculate the amount of Na in the water sample. SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 110 Objective type questions: 1) Phase rule is the basis of separation procedures in the a)petroleum industry and cosmetic preparations. b) Cotton industry & Rayon industry c) Pharmaceutical industry d) None of the above 2) At the point O, where the two curves meet in the phase diagram. a) Eutectic point b) Triple point c) Double point d) All 3) The argentiferrous lead contains a very small percentage of a) Gold b) Diamond c) copper d) silver 4) Condensed system consists of a) only solid and liquid phases b) only liquid and gaseous phase c) only Solid and gaseous phases d) Solid, liquid & Gaseous phase 5) super cooling is a) Liquid water may be cooled below its freezing point without solidifying. b) solid water may be heated to evaporation. c) Liquid water may be cooled to freezing point for solidification. d) All of the above. 6) The triple point for water system occurs at a)) 0.001 0C & 0.006 atms.pressure b) 0.01 o C & 0.006 atms.pressure c) 1 o C & 0.006 atms.pressure d) b) 0.01 o C & 0.006 atms.pressure 7) In the estimation of FAS by potentiometry the indicator electrode used is a) Silver – Silver Chloride electrode b) Platinum electrode c) Calomel electrode d) Glass electrode 8) Measurement of optical density using monochromatic light in calorimery involves a) IR range b) Visible range c) UV range d) All 9 Ion selective electrode used in the determination of pH is a) Silver – Silver Chloride electrode b) None of the above c) Calomel electrode d) Glass electrode SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 111 10) Instrumental methods of analysis are widely adopted when compare to classical methods of analysis because a) The methods are much faster b) Applicable at concentrations c) the analytical process can be automated d) All 11) Colorimetry involves measurement of absorbance using monochromatic light in the a) UV range b) IR range c) Visible range d) All 12) Flame photometry is suitable for the detection of a) Li b) Cu c) Fe d) Zn 13) For photometric titrations the analyte should a) Have different oxidation state b) Conduct electricity c) Ionize d) All 14) For potentiometric titration, the analyte should a) have different oxidation states b) conduct electricity c) ionize d) all REVIEW QUESTIONS 1. what is phase diagram? 2. what is Gibbs Phase Rule? Explain the terms. 3. Define Phase. 4. Define Component. 5. Define Degrees of freedom OR variance. 6. Derive phase rule. 7. Explain the application of phase rule to one component system 8. Explain Lead – silver system: 9. What is Eutectic point ? 10. What is meant by Desilverisation of lead by Pattinson’s process? 11 What is Instrumental Methods of Analysis ? 12 Explain the instrumentation and theory of Potentiometric Titrations. 13 Explain the instrumentation and theory of Conductometric estimation. 14 Explain the Colorimetric estimation of copper. 15.Explain the instrumentation and theory of Flame photometer SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 112 UNIT UNIT UNIT UNIT – –– – VII VII VII VII HIGH POLYMERS 7.1 What are Polymers? Explain the Classification of polymers with examples 5 Marks Polymers are the high molecular weight compounds obtained by repeated union of simple molecules. (Monomers). Ex: Starch, Polyvinyl chloride, Polyethylene, Nylon 6, 6 and etc. Classification of polymers. Polymers are classified into two types as follows: i) Natural Polymers. ii) Synthetic (artificial) polymers. i) Natural polymers: These are the polymers obtained naturally by plants and animals. Ex: Silk, wool, natural rubber, protein, starch, cellulose, etc. ii) Synthetic Polymers: These are artificially prepared polymers also known as man made polymers. Ex: PVC, Nylon 6.6, Polythene, Phenol, Formaldehyde, Resin etc., 7.2 What are Monomers? Give examples 2 Marks Monomer is a simple repetitive unit which when joined together in large numbers which give rise to a polymer. These are the building blocks of Polymer Ex: Vinyl chloride, ethene, propylene etc. 7.3 What is Polymerization? Explain the types with examples 5 Marks Polymerization is a process of chemical union of large number of monomers to form a polymer. During polymerization the monomers are linked through covalent leakages to give raise to polymer. Based on the type of polymerization reaction, it is classified into two types as follows. i) Addition Polymerization ii) Condensation Polymerization SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 113 i) Addition Polymerization: It is process in which the monomers undergo simple addition reactions to give raise to a polymer without eliminating by products. Alkenes and substituted alkenes readily undergo addition polymerization reactions. Ex: When large number of ethene molecules undergoes addition polymerization reactions, polyethylene polymer is obtained. ii) Condensation Polymerization:- It is a process in which the monomers undergoes intermolecular condensation reactions to form a polymer with the elimination of simple molecules like water, HCl, ammonia, phenol etc., Ex: When adepic acid and hexamithylene diamine undergoes condensation polymerization reaction to form Nylon 6,6 polymer. n.NH2–(CH2)6-NH2+ n.HOOC–(CH2)4–COOH Hexa methylene di amine Adipic acid (NH – (CH2)6- NH -OC – (CH2)4 – CO-)n + n H2O Nylon 6,6 7.4 What is degree of polemerization (DP). 2 Marks Degree of polymerization is a number, which indicates the number of repetitive units (monomers) present in the polymer. By knowing the value of DP, the molecular weight of the polymer can be calculated. [Molecular wt of the polymer] = DP x Molecular wt of each monomer. DP is represented as ‘n’. Ex: (CH2 – CH2) n Polythene Here ‘n’ is the DP. i) Calculate the molecular weight of the polythene polymer given DP is 100. Molecular weight of the polythene = DP X Molecular weight of Polethene = 100 X 28 = 2800. 7.5 Explain the free radical mechanism addition polymerization by SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 114 taking ethylene as example 6 Marks The free radical mechanism polymerization of ethylene involves the following three steps i) Initiation ii) Propagation iii) Termination i) Initiation: When the initiators are heated or exposed to sunlight, they undergo hemolytic decomposition forming highly reactive species known as free radicals. The free radical attacks the double bond of the monomer and initiates the chain reaction. Example: when dibenzoyl peroxide initiators are heated or exposed to sunlight, they produces phenyl free radicals, which attacks the ethylene monomer and converts it into reactive monomer. the reactive monomer again reacts with another monomer and chain growth is initiated. (C6H5COO) 2 2 C6H5* + 2CO2 Dibenzoyl peroxide phenyl free radical CH2=CH2 + C6H5* C6H5 - CH2 - CH2* Ethylene Reactive monomer Monomer ii) Propagation: The reactive monomer again reacts with another unsaturated monomer and converts it into reactive monomer and growth of the polymer chain continues. C6H5 - CH2 - CH2* + CH2=CH2 C6H5 - CH2 - CH2 - CH2 - CH2* Reactive monomer Ethylene growing polymer chain iii) Termination: Termination growing polymer chain occurs by any one of the following reaction. a) Coupling of two growing polymer chain When two growing polymer chains react with each other termination occurs due to the formation of a dead polymer chain. C6H5 - CH2 - CH2 - CH2 - CH2* + *CH2 - CH2 - CH2 - CH2 – C6H5 growing polymer chain growing polymer chain C6H5 - CH2 - CH2 - CH2 - CH2 – CH2 - CH2 - CH2 - CH2 – C6H5 Dead polymer chain SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 115 b) Coupling of one growing polymer chain with free radical When one growing polymer chains reacts with a free radical, termination occurs due to the formation of a dead polymer chain. C6H5 -CH2 - CH2 - CH2 - CH2* + C6H5* C6H5 - CH2 - CH2 - CH2 -CH2- C6H5 Growing polymer chain Dead polymer chain 7.6 Explain the Methods of polymerization. 6 Marks Polymerization is brought about by the following methods. i) Bulk polymerization. ii) Solution polymerization. iii) Suspension polymerization. iv) Emulsion polymerization. i) Bulk polymerization: This method is used for liquid monomers. The initiators are dissolved in the liquid monomers and form a homogeneous phase. Heating or exposing to the sunlight initiates the polymerization reaction. The reaction is exothermic and viscosity of medium increases as the reaction proceeds. The reaction mixture is stirred constantly to dissipate the heat liberated. Advantages: The method is simple. The products don’t require purification or isolation. The products have high optical clarity. The purity of the products is high Disadvantages Agitation becomes difficult due to viscosity build up The heat control is more difficult Applications: The method is used for making polymers such as PVC, PMMA, and PS etc. ii) Solution polymerization: This method is used for monomers & initiators dissolved in a suitable inert solvent. The inert solvent reduces the viscosity build up during the reaction progress & facilitates heat & mass transfer. In this method the obtained polymer can be easily separated from the solvent by evaporating the solution or it can be directly used for coating & adhesive purposes. Advantages: SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 116 Viscosity produced in the reaction is negligible. Better heat transfer can be achieved. In solution face the polymer can be directly used as paintings, coating, & adhesive. Purposes This method is used for preparation of polythene, pvc, polyacrylonitrile, etc., iii) Suspension Polymerization Method (Pearl Polymerization method): In this method water in soluble monomers are suspended in water as tiny droplets by continuously agitating (shaking). Initiators are soluble in monomers. Monomers and initiator forms heterogeneous mixture with the water. Polymerization is initiator by heating or exposing to radiation. The polymerization takes place within the tiny droplets the polymer is obtained as spherical pearls or beads hence it is also called as pearl polymerization methods. Since the product is insoluble in water. It can be easily separated just by filtration & the product needs no further purification. Advantages: Viscosity build up is negligible Better heat transfer can be achieved. High purity product is obtained. Isolation of the product is easy and it does not require any purification. Uses: This method is used for preparing PVC, PVA (Polyvinyl acetate), styrene, divinyl benzene (ion exchanger) iv) Emulsion Polymerization method: In this method the monomers & water soluble initiators forms the emulsion with water. The surfactants such as soap, detergents, etc are add which hold the monomers & initiators (H2O2 or per sulphate) in the form of a micelles. Polymerization takes place inside the micelle on exposing to heat or radiation. Filtration & deemulsifying agents can isolate the obtained polymer. Advantages: The rate of polymerization is high Viscosity build up is negligible Better heat transfer can be achieved. High purity product is obtained Isolation is easy. This method is used to prepare PVC, adhesive etc., SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 117 7.7 What is Glass transition temperature? Mention the factors which affecting the Tg of a polymer 6 Marks When an amorphous polymer is heated, it gets converted from hard brittle state (glassy state) to a soft flexible state (rubbery state). The temperature at which a polymer transforms from a hard glassy state to soft rubbery state is called glass transition temperature. Glass State viscoelastic state Viscofluid state (Hard & brittle) (Rubbery) (Polymer melt) Tg Tm Significance of Tg : It is used as a measure for evaluating the flexibility of a polymer and the type of response the polymeric material would exhibit to Tg is very useful in choosing the right processing temperature for fabrication (moulding, calendaring and extrusion ) Tg is very useful in determining the coefficient of thermal expansion, heat resistant, refractive index, electrical property, etc., Factors affecting glass transition temperature 1. Chain Flexibility: Linear polymer chains have high degree of freedom for rotation ie., more chain flexibility and low Tg. Aromatic or cyclic groups on the back bone of carbon atoms hinder the freedom of rotation ie., chain flexibility decreases and causes an increase in Tg. 2. Cross-linking and branching: A small amount of branching will tend to lower Tg and a high density of branching reduces chain mobility and elevates the Tg. The cross linking brings the polymer chain closer, lowers free volume and restricts molecular motion and hence rises Tg. 3. Intermolecular forces: The presence of large number of polar groups in the molecules lead to strong intermolecular cohesive forces which restrict the segmental motion. As a result Tg increases. 4. Molecular mass: Higher the molecular mass more is the restriction in the molecular freedom. However, Tg is not significantly affected by molecular masses if the degree of polymerization is above 25o; 5. Presence of plasticizers: Addition of plasticizers reduces the Tg vale; for example, addition of di-isooctyl phthalate to pvc reduces its Tg from 80 o C to below room temperature. 5. Stereo regularity of the polymer: A syndiotactic polymer has a higher Tg than atactic polymer, which in turn has higher Tg than its isotactic stereoisomer. 7.8 Explain relationship between structure and property of a polymer SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 118 6 Marks A polymer should have the following property, which depends mainly on the structure of the polymers. 1. Tensile strength : A polymer should have more tensile strength it depends on a. Molecular weight of the polymer: Low molecular weight polymers are soft gummy, & less resistant to heat. But high molecular weights of the polymers are tough & resistant to heat. b. Structure of the polymer: Cross linked & branched polymers are stronger than linear less branched polymers 2. Crystalinity: A polymer would have high degree of crystalinity, which have more melting point more tensile strength & resistance to heat. The degree of crystalinity depends on a. Structure of polymers: Linear polymers without bulky groups & hydrogen bounding are more crystalline than branched & polymers have bulky groups. Ex: HDPE High density polyethylene has high degree crystalinity than polyvinylactetate. b. Configuration of the polymer: Isotactic Polymers are more crystalline than atactic polymers. Ex; Isotactic polyvinyl chloride is more crystalline than Atactic polyvinyl Chloride (random arrangements of groups) Isotactic Polystyrene Atactic Polystyrene c. It also depends on polar groups & hydrogen bonding for those Polymers having polar groups & hydrogen bonding have high degree of crystaslinity. 3. Elasticity: A polymers should have good elastic property. The elasticity is due to uncoiling & recoiling of the molecular chain on the application of force. It should not break on prolonged stretching. The elasticity of a polymer can be improved. a. By introducing cross linkages at suitable positions. b. Avoiding side group such as aromatic and cyclic structures. c. By introducing more non polar groups on the chain so that the chain does not separate on stretching. d. By introducing internal plasticizers during polymerization. SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 119 4. Chemical. Resistance: Polymer should have more resistivity to chemical attack and should not become soft, swelling, or loosing its strength. The chemical resitivity of a polymer depends on mainly. a. Presence of polar & non polar groups in the polymer chain: Presence of polar groups such as –OH, -COOH, are less resistance to chemical attack. But non-polar groups such as –CH3, -C6H5, NHCO, etc., are not usually attacked by chemicals. Ex: PVC, ABS, PS, etc., have high resistance towards chemicals. b. Degree of crystalinity & Molecular mass: Chemical resistance increases with increase in degree of crystalinity because crystalline regions make the penetration of chemicals or solvent more difficult. Higher the degree of crystalinity higher is the chemical resistance. 7.9 What are resins? Give examples 2 Marks Resins are linear, low molecular weight polymeric materials. These further undergo polymerization and cross linking during curing to form hard and rigid three dimensional networks. These resins are used as coatings, adhesives and molding powders. Ex: Epoxy resin Phenol-formaldehyde resin Urea - formaldehyde resin Polyester resin 7.10 What are plastics? Give examples 2 Marks Plastics are linear, high molecular weight polymeric materials, which can be molded by the application of heat and pressure. During the process of molding, these become hard but retain plasticity. Ex: Polyethylene Polystyrene Poly vinyl chloride etc. 7.11 Explain compounding of resin. 6 Marks The process of mechanical mixing of various additives to resins and converting resins in to plastics is called as compounding of resin. The important additives used in compounding are as follows i) Fillers ii) Plasticizers iii) Stabilizers iv) Colorants i) Fillers: These are organic or inorganic compounds added during compounding to increase the bulk of the polymer. The fillers also enhance the tensile strength, impact resistance and abrasion resistance. Ex: Wood, mica, clay, silica, talk, fabric scraps, glass fibers, carbon SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 120 black, polyesters etc, ii) Plasticizers Plasticizers are the substances, which are added during compounding to increase the plasticity of a polymer. These decrease the Tg and helps in converting a hard, brittle polymer into a soft, flexible polymer. Ex: Tricrysyl phosphate, Tri phenyl phosphate, dioctyl phthalate, fatty acids etc, iii) Stabilizers Polymers have tendency to undergo degradation by air and light. The Stabilizers are added to minimize the degradation of polymers. Ex: Phenyl salicylate, aryl and alkyl phosphates, tri phenyl phosphates. iv) Colorants These are added to impart required color to the polymer. These increases the attractive and decorativeness of a polymer. The colorants include organic and inorganic pigments. Ex: Titanium oxide, lead chromate, BaSo4, carbon black, azo dyes etc, 7.12 Plastic Manufacturing Methods Plastics are made into shapes in many ways: 1. EXTRUSION Hot molten plastic is squeezed through a nozzle to make long lengths of special shapes like pipes, spouting and wallboard joining strips. It is also used to make large thick sheets of plastic for fabrication. 2. BLOW EXTRUSION (Fig 1) This is used for making plastic films and bags. While it is still hot, an extruded tube is blown up like a balloon, with compressed air. This stretches the plastic and makes it thin. The balloon is made long enough to allow the plastic to cool. The end of the balloon is pinched together by rollers, to hold the air in and make it flat. The flat tube is then wound on to a big roll. You can see continuous rolls of plastic bags in a fruit shop. SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 121 3. INJECTION MOULDING Hot molten plastic is squeezed into a mould to make lots of objects all the same. They can be very small like a washer or quite large, like a bowl or a clothes basket. Lots of everyday articles are made this way. 4. BLOW MOULDING (Fig 2) A little bit of hot soft plastic is squeezed into the end of a mould. Compressed air is used to blow a big bubble inside the plastic. The plastic swells out like a balloon until it fills up the whole mould. Many bottles, toys and money boxes are made this way. 5. ROTATIONAL MOULDING Plastic powder is scooped into a mould. The mould is rotated over a big gas burner. As the mould gets hot, the plastic melts and sticks to the mould. This method is used for making big hollow things like water tanks and barrels. 6. COMPRESSION MOULDING This is used for thermoset resins. Dry powder is put in a mould which is squeezed and heated until the plastic is cured. This is used for making ashtrays, cups and plates, and some electrical switches. 7. REACTION INJECTION MOULDING SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 122 Two chemicals are mixed together and squirted into a mould. The chemicals react together. This is how they make car bumpers, some disposable cups and plates, and the meat trays you get from supermarkets. 7.13 Give the manufacture and applications of Teflon. 4 Marks Teflon is the trade name of poly tetra fluro ethylene. The monomers for the preparation of Teflon are tetra fluro ethylene Teflon is prepared by emulsion polymerization method using ammonium persulfate as initiators under pressure. persulfate initiators n. CF2=CF2 (-CF2-CF2-) n Tetrafloroethylene Polytetrafloroethylene Properties: a. It has density about 2.3 gram/cm 3 b. It has high melting point 320 0 C c. It has nonstick property & slippery over wide range of temperature. d. It has got good electric insulating property. Uses: It is used for insulating motors, cables, and generators transformers, capacitors. Used for nonstick cooking utensils Used for gaskets, belts pump and valves packing etc. Used as a dry lubricant. 7.14 Give the manufacture and applications of Poly methyl methacrylate PMMA (Plexi Glass) 4 Marks Plexi glass is the trade name of polymethyl methacrylate. The monomers used for the preparation of plexi glass are methyl methacrylate. Plexi glass is prepared by emulsion polymerization method at 60 to 70c in presence of trace of H2O2 or Acetyl peroxide as initiator. SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 123 Properties: i. Plexi glass is a white transparent thermoplastic ii. It has got high optical clarity iii. It is resistant to chemical action Applications: i. It is used in the preparation of aircraft windows. ii. Attractive sign boards. iii. Manufacturing of transparent moulded articles & tubes. iv. Lenses for automobiles, artificial eyes etc., 7.15 Give the manufacture and applications of Polyurethanes. 4 Marks Ployurethanes are produced by the polymerization of disocyanate and diol or triol (or the addition reaction between 2, 4-tolylene diisocyanate with glycol). In the production of polyurethanes foams, glycol, toylene diisocyanate, catalyst tertiary amines), water and surfactants are mixed together and heated. Properties: i. These can be obtained in the form of foams, fibers, Elastomers, Coatings etc. ii. The foams are available in both rigid & flexible forms. Uses: i. Flexible foams are used for cushions in automobiles & furniture. ii. Rigid foams are used to reinforce hallow structural units. iii. Fibers of Polyurethanes are used in lightweight garments and swim suits because of their stretching property. iv. These are used to coat gymnasium floor and dance floor. v. These are used in tyre treads and industrial wheels. SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 124 7.16 Give the manufacture and applications of Phenol Formaldehyde Resins. 6 Marks Phenol formaldehyde resins are prepared from condensation reaction between phenol & formaldehyde. Commercially these are available into two forms nova lacks & resols.. i) Nova lacks. These are phenol formaldehyde resins obtained by condensation of phenol & formaldehyde in presence of acid as catalyst when phenol to formaldehyde ratio is greater than one Properties: i. Nova lacks are linear polymers. ii. These have got good electric insulating property Uses: i. Used for sealing metal holders to the glass bulbs. ii. Bounding sheets of paper, wood, card boards etc., ii) Resol Resins. These are phenol formaldehyde resins obtained by the condensation of phenol & formaldehyde. In presence of alkali as catalyst when phenol to formaldehyde ratio is less than one. Properties. i. These are cross linked polymers ii. These are non conductor of electricity iii. Resols can be converted into bakelite by adding additives such as wood, dies etc., SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 125 Uses. i) Resol in the form of bakelite is used for the preparation of electrical fittings such as switches plugs, sockets, etc. ii) It is also used for preparing telephone out parts etc. 7.17 What are elastomers? Give examples. 2 Marks These are the high polymers which undergoes very long elongation when they are subjected to an external force and readily regains their original shape when stress is released ( when external force is removed). Ex: Rubber. 7.18 Mention the advantages of Synthetic Rubbers over natural rubber. 5 Marks i. These are more resistance to heat & cold ( nitrile rubber) ii. These are not easily attacked by sunlight & air ( neoprene and nitrile rubber) iii. These have high abrasion and high tensile strength. iv. Rubber property is retained at high temperatures (ex: silicon rubbers- 90 0 - 316 0 c) v. Do not age easily (ex; Polyurethane rubber) vi. Do not swell and can hold organic solvent better than natural rubber (ex; polysulphide rubber) vii. Hold more air and water at high pressures (ex; butyl rubber) 7.19 What is Vulcanization of rubber Vulcanization or vulcanisation is a chemical process for converting rubber or related polymers into more durable materials via the addition of sulfur or other equivalent "curatives". These additives modify the polymer by forming crosslinks (bridges) between individual polymer chains. [1] The vulcanized material is less sticky and has superior mechanical properties 7.20 Give the manufacture and applications of Neoprene rubber 4 Marks Poly propylene (Neoprene) is produced by the polymerization of chloroprene units by Emulsion polymerization method. SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 126 Uses: It is used for hoses, tubes for carrying oils and chemicals, gloves, coated fabric, cables, belts, shoe heels, solid tyres, 7.21 Give the manufacture and applications of Butyl rubber. 4 Marks Butyl rubber is prepared by copolymerization of isoprene (2-methyl 1,3 – Butadiene) Isobutylene by emulsion polymerization method at 90 0 C. Uses. i. Butyl rubber is widely used for preparation of inner tubes for tiers. ii.It is also used for insulating high voltage wires & cables. 7.22 Silicone rubber: Synthesis: Silicone rubber is a rubber-like material composed of silicone — itself a polymer — containing silicon together with carbon, hydrogen, and oxygen. During manufacture heat is required to vulcanize (set or cure) the silicone into its rubber-like form. This is normally carried out in a two stage process at the point of manufacture into the desired shape, and then in a prolonged post-cure process. It can also be injection molded. • SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 127 Properties: Silicone rubber is generally non-reactive, stable, and resistant to extreme environments and temperatures from -55°C to +300°C while still maintaining its useful properties. Due to these properties and its ease of manufacturing and shaping, silicone rubber can be found in a wide variety of products, including: automotive applications; cooking, baking, and food storage products 7.23 What are Adhesives? Give examples 2 Marks Adhesives are the polymeric non metallic compounds which can hold firmly two materials together by surface attachment. Ex: Epoxy resin 7.24 Give the manufacture and applications of Epoxy Resin. 4 Marks Araldite or Epon are the trade names of epoxy resin. It is prepared by condensation polymerization of Bis phenol – A & Epichloro hydrin. USES. i. It can be used to bind cardboard, boats, glass for laminating purposes. ii.Used in shrinkage proof garments. 7.25 Polymer composite: Polymer composite is any material made of more than one component. Polymer composites are composites made from polymers, or from polymers along with other kinds of materials. 7.26 Kevlar fibers: Kevlar is synthesized in solution from the monomers 1,4-phenylene- diamine (para-phenylenediamine) and terephthaloyl chloride in a condensation reaction yielding hydrochloric acid as a byproduct. The result has liquid-crystalline behavior, and mechanical drawing orients the polymer chains in the fiber's direction. Hexamethylphosphoramide (HMPA) was the solvent initially used for the polymerization, but for safety reasons, DuPont replaced it by a solution of N-methyl-pyrrolidone and calcium chloride SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 128 7.27 what are Carbon fibers? Carbon fiber (carbon fibre), alternatively graphite fiber, carbon graphite or CF, is a material consisting of extremely thin fibers about 0.005– 0.010 mm in diameter and composed mostly of carbon atoms A common method of manufacture involves heating the spun PAN filaments to approximately 300 °C in air, which breaks many of the hydrogen bonds and oxidizes the material. The oxidized PAN is then placed into a furnace having an inert atmosphere of a gas such as argon, and heated to approximately 2000 °C, which induces graphitization of the material, changing the molecular bond structure. When heated in the correct conditions, these chains bond side-to-side (ladder polymers), forming narrow graphene sheets which eventually merge to form a single, columnar filament. The result is usually 93–95% carbon. 7.28 7.28 What are Conducting polymers? Give examples 2 Marks These are the organic polymers having the conductivity band similar to that of conductors with highly delocalized Pi electron system. Ex: Polyaniline, Polypyrole, Polythiopene, Polyacetylene, etc Conducting polymers are obtained by doping an oxidizing or reducing agent into organic polymers. With the carbon backbone consisting of alternate double bond & single bonds doping results in delocalization of electrons responsible for conduction. 7.29 Explain the mechanism of conduction in Poly acetylene. 6 Marks The electron is removed from the top of the valence band of Polyacetylene creating a vacancy or hole. The Polyacetylene molecule, now positively charged, is called a radical cation, or polaron. The lone electron of the double bond, from which can electron was removed, can move easily. Therefore, the electron successively moves along the polymer chain.This are called delocalization of electrons. The positive charge, on the other hand, is fixed by electrostatic attraction to the iodide ion; when I2 added as dopant which does not move so readily. If SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 129 Polyacetylene is heavily doped (heavily oxidized), polarons form pairs called Solitons. The Solitons are delocalized over 12 carbon atoms. Due to the formation of Solitons, a new localized electronic state appears in the middle of the energy gap. When doping is high, several charged solution band. This band can later merge with edges of valence & conduction bands thus exhibiting conductivity. 7.30 Mention the structure & applications of conducting poly Aniline 4 Marks i) Used as conducting tracks on PCB’s (Printed Circuit Boards) ii) Used as electrode materials for rechargeable batteries iii) Used in humidity sensors. iv) Used in electrochemical transducers. v) Used in biosensors. vi) Used in artificial nerves and optical computers etc. vii) Used in smart windows which absorb sunlight. SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 130 7.31 Explain the mechanism of conduction in Poly aniline 6 Marks Partial oxidation Ammoniumperoxy di sulphate (NH2 S2 O8) Base form of poly- aniline Protonation Aq. HCl ( 1 M) Protonic acid doping: The synthesis of conducting polyaniline is a typical example of this type of doping technique. In this technique current carrying charged species (-/+) are created by the protonation of imine nitrogen. Polyaniline is partially oxidized, first using a suitable oxidizing agent into a base form of aniline which contains alternating reduced and oxidized forms of aniline polymer backbone. This base form of aniline when treated with aqueous HCl (1M) undergoes protonation of imine nitrogen atom, creating current carrying charged sites (+ve) in the polymer backbone. These charges are compensated by anions (Cl - ) of the doping agent, giving the corresponding salt. This type of protonic acid doping of polyaniline results in an increase of conductivity by approximately 09-10 orders of magnitude. Applications: Conducting polymers are highly promising materials to be used in electric and electronic applications, some of the applications are. 1) As electrode material for commercial rechargeable batteries, for higher power to weight ratio (coin type of batteries) 2) As conductive tracks of on printed circuit boards SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 131 3) As sensors: Humidity sensor, Gas sensor, Biosensor for glucose, Galactose etc. 4) In electro chromic display windows. 5) In information storage device. 6) As film membranes for gas separations. 7) In light emitting diodes. 8) In fuel cells as the electro catalytic materials. Objective type questions. 1) Tetrafluro ethylene is the monomer of a) Nylon 66 b) neoprene c) Teflon d) PVC 2) Phenol-formaldehyde resin is commercially a) PVC b) Bakelite c) Elastomer d) Nylon 3) Sulfur is used particularly in a) Manufacture of Buna-S b) Compounding of plastics c) Corrosion control d) Vulcanization of raw rubber 4) Isoprene is a monomer of a) Natural rubber b) Synthetic rubber c) starch d) PVC 5) A polymer with higher Tg value is a) PVC b) polyethylene c) Polypropylene d) polystyrene 6) The polymer which can be used as synthetic adhesive is a) Neoprene b) Buna-S c) Epoxy resin d) Polystyrene 7) Co-polymer of Isoprene and Isobutylene is known as a) Butyl rubber b) Buna-S c) PTFE d) Polyurethane 8) Which of the following polymer is used as a substitute for glass a) Teflon b) polyurethane c) PMMA d) PVC 9) The monomer phenol is a) Mono-functional b) Bi-functional c) Tri-functional d) Poly-functional 10) Neoprene is obtained by the polymerization of a) Styrene b) Propylene c) Chloroprene d) Isosoprene SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 132 11) Suspension polymerization is also known as ____ polymerization a) bulk b) pearl c) addition d) degree 12) Glass transition is dependent on a) chain felxibbility b) molecular mass c) branching d) all 13) Polar groups in a polymer offer _____ resistance to chemical attack. a) More b) less c) maximum d) no 14) Fillers are added to resins to a) enhance tensile strength b) resist abrasion c) increase bulk density d) all 15) An example for conducting polymer is a) poly aniline b) poly thiophene c) poly acetylene d) all SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 133 REVIEW QUESTIONS 1. What are polymers? Give the classification of polymers with examples. 2. What are monomers? Give examples. 3. What is Polymerization? Explain the types with examples. 4. What is degree of polymerization (DP)? Mention its importance. 5. Explain the free radical mechanism for the polymerization of ethylene. 6. Explain the methods of polymerization. 7. What is Glass transition temperature? Mention it’s significant. Discuss factors affecting the glass transition temperature? 8. Explain the relationship between structure and property of a polymer? 9. What are plastics and resins? Give examples 10. Explain the compounding of resin. 11. Explain the manufacture of Teflon & Mention its properties & uses? 12. Explain the manufacture, properties, & uses of Poly methyl methacrylate? 13. Explain the manufacture, properties & uses of Poly urethanes? 14. Explain the manufacture, properties & uses of Phenol Formaldehyde Resin? 15. What are Elastomers? Mention the advantages of synthetic rubbers Over natural rubbers? 16. Explain the manufacture, properties & uses of Neoprene rubber 17. Explain the manufacture, & uses of Butyl rubber? 18. What are adhesives? Explain the manufacture and uses of Epoxy Resin? 19 What are conducting polymers? Give Examples 20 Explain the Mechanism of conduction in polyacetylene 21 Write the Structure of conducting poly aniline and mention it’s applications. SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 134 UNIT UNIT UNIT UNIT – –– – VIII VIII VIII VIII WATER TECHNOLOGY 8.01 Discuss the different types of Impurities present in natural water with examples. 6 Marks Impurities in water may be broadly classified into four categories: i) Dissolved impurities ii) Suspended impurities iii) Dissolved gases & iv) Organic matter. i) Dissolved impurities. The soluble salt impurities present in water include salts of Ca, Mg, Na in various soluble salt forms oxides of Mn, Fe, Pb & Ar may also present in water. ii) Suspended impurities. Suspended impurities are the dispersion of solid particles, which can be removed by filtration or settling. They are of 2 types:- • Inorganic: Includes clay silica, oxides of Fe & Mn etc. • Organic: Includes wood pieces, disintegrated particles of dead animals, leaf, fishes, Bacteria, Algae, and Protozoa etc. iii) Dissolved gases. Most water contain dissolved gases such as O2,CO2,SO2 , NH3 & oxides of N all of which are derived from atmosphere. iv) Organic matter. Organic compounds derived from the decay of vegetables & animal matter including bacteria, water also gets contaminated with sewage & human excreted matter etc. 8.02 What is water analysis 2Marks SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 135 Analysis of water involves determination of various constituents present in water in order to ascertain the quality of water & thereby the utility of water. 8.03. Explain the method of Determining the Total Hardness of Water. 8 Marks Principle: Hardness of water is due to the presence of calcium and magnesium salts in water. Ethylene diamine tetra acetic acid (EDTA) forms complexes with a large number of cat ions including Ca ++ and Mg ++ ions. Accordingly it is possible to determine the total hardness of water using EDTA reagents. The di sodium EDTA molecule (H2Y) has two easily replaceable hydrogen atoms and the resulting ion after ionization may be represented as H2Y 2- . The later forms complexes with metal ions as fallows. M 2+ + H2Y 2- MY 2- + 2H + ………. (1) Where M 2+ is Ca 2+ and Mg 2+ in water. Reaction (1) can be carried out quantitatively at a P H of 10 using Erichrome black-T indicator. Since the reaction involves the liberations of H + ions a buffer mixture has to be used to maintain a P H of 10. The buffer mixture used in the titration is NH3 – NH4 Cl. The hardness of water is usually expressed in terms of ppm (Parts per million) of CaCO3. Since EDTA (free acid) is sparingly soluble, its disodium salt, Na2H2Y, is used for preparing the reagent. Procedure: Part A: Preparation of a Standard EDTA Solution – Weigh accurately the given EDTA crystals using an electronic weighing balance. Note the weight, transfer the crystals carefully into a funnel placed over a 250 cm 3 volumetric flask and note down the empty weight of the weighing bottle. Add few drops of ammonia and pour ion exchange water through the funnel allowing all the crystals to run down into the flask. Wash the funnel with ion exchange water and remove the funnel. Dissolve the crystals by swirling the flask gently. Add some more water if needed. SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 136 Dilute the solution up to the mark with ion exchange water, Stopper the flask and mix the solution thoroughly by inverting the flask several times so that a homogeneous solution results. Calculate the morality of EDTA. Part B: Determination of total hardness of a sample of water Pipette out 25 cm 3 of the given water sample into a clean conical flask. Add 5 cm 3 of NH3 – NH4 Cl buffer and a pinch of Erichrome black – T indicator. Titrate against standard EDTA solution till the color of the solution changes from wine red to clear blue. Repeat the titration for concordant values. Calculations: Volume of EDTA consumed =…………….cm 3 1000 cm 3 1 M EDTA = 100 g CaCO3 (Molecular mass of CaCO3 = 100) ………cm3 of …….M EDTA = ……..cm3 x ………M x 100 g of CaCO3 1000 x 1 25 cm 3 of the water sample contains ……………………… g of CaCO3 Total hardness of the given water sample = …..g x 10 6 ppm of CaCO3 25 = …………pm of CaCO3 Result: The Hardness of the given water sample = ………… ppm of CaCO3 8.04. What is meant by Alkalinity? 4 Marks Alkalinity in water arises due to the substances that can cause the formation of hydroxyl ions & in turn can react with strong acids. Substances that cause the alkalinity in water are of three types. • Hydroxides (NaOH, Ca(OH) etc) • Carbonates (Na2CO3, CaCO3, etc) • Bicarbonates (NaHCO3, Ca (HCO3)2etc.) When a sample of alkaline water is treated with a strong acid such as HCL, the following reactions occur. SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 137 A) (i) For estimation of alkalinity in a water sample phenolphthalein and methyl orange end points were found to be 30ml and 45ml respectively. The readings indicate the presence of: a) OH - and CO3 2- b) CO3 2- and HCO3 - c) OH - and HCO3 - d) Only HCO3 - Answer:- Alkalinity of water is calculated by titration, By reacting water sample with NaOH using phenolphthalein and methyl orange indicator. The reaction water sample may be represented as NaOH + NaHCO3Na2CO3 + 2H2O The alkalinity in the sample can be combination can be represented as Aph = Ah + 0.5 Ac ------- (1) At = Am.o = Ah +Ac + Ab------ (2) Where Aph= Alkalinity of phenolphthalein, Am.o =Alkalinity of methyl orange, Ah= Alkalinity of Hydrodroxide, Ac= Alkalinity of Carbonate, Ab= Alkalinity of Bicarbonate. a) OH - and CO3 2- In this case carbonate is absent i.e, Aph = Ah + 0.5 Ac Am.o = Ah +Ac Ah= Aph - 0.5 Ac Therefore Am.o = Ah +Ac = Aph - 0.5 Ac + Ac Am.o = Aph + 0.5 Ac b) CO3 2- and HCO3 - in this case hydroxide alkalinity is absent I.e., from equation (1) Aph = 0.5 Ac Am.o = Ab +Ac c) OH - and HCO3 - in this case carbonate alkalinity is absent Aph = Ah Am.o = Ah + Ab d) Only HCO3 - in this case both carbonate and carbonate are absent Aph=0 Am.o = Ab SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 138 8.05 Explain the method of Determining the Alkalinity in water 6 Marks Principle: Alkalinity is determined by titrating a known volume of water sample with indicator against a strong acid. Two types of alkalinity can be evaluated based on the indicator used. Alkalinity when methyl orange is used. When methyl orange is used as indicator, only after all the three reactions, the color change observed & that indicate the end point. Total alkalinity = Alkalinity due to hydroxyls + Alkalinity due to Carbonates + Alkalinity due to bicarbonates. when phenolphthalein is used , only after the two reactions (I&II) before the III reaction occur, the color change observed & that indicate the end point. Alkalinity with phenolphthalein = Alkalinity due to hydroxyls + 1/2 Alkalinity due to Carbonates. Procedure. Pipette out 100ml water sample into a clean conical flask. Add two drops of Methyl Orange indicator. Titrate against standard (say 0.02N) HCL till the color of the solution changes sharply from yellow to orange. Let the volume of HCL consumed be Xml. To another 100ml sample of water, add two drops of Phenolphthalein. Titrate against standard (say 0.02N)HCL till the color of the solution changes sharply from pink to colorless. Let the volume of HCL consumed be Yml. Calculations. a) Alkalinity due to methyl orange. 1000ml of 1N HCl is equivalent to 50 g of CaCO3 (50 being the equivalent weight of CaCO3). 1ml of 1 N HCL is equivalent to 50/1000 g of CaCO3 Xml of 0.02 N HCL is equivalent to 50XxX 0.02/1000 g of CaCO3 10 6 ml of water sample contains 50XxX0.02X10 6 /1000X100 g of CaCO3 = Xg of CaCO3 i.e. Alkalinity of methyl orange = 10X PPM of CaCO3 equivalent b) Alkalinity due to Phenolphthalein. SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 139 This is also in similar way that is Alkalinity of Phenolphthalein is 50 x Y x 0.02X10 6 / 1000X100 = 10 Y PPm CaCO3 8.06 Explain the method of determining the Chloride content in water 6 Marks Procedure: Transfer 100cm 3 of water sample into a clean conical flask. Add about 1 cm 3 of K2CrO4 indicator solution & titrate against standard (say 0.02N) AgNO3 solution until a reddish brown color persists in the white precipitate. Record the volume of AgNO3 consumed (Let ‘a’ cm 3 ). Perform a blank titration taking 100 cm 3 of distilled water. The volume of AgNO3 consumed Let ‘b’ cm 3. Calculation: Volume of AgNO3 required for chlorine estimation = (a-b) cm 3 = V 1000ml of 1 N AgNO3 = 35.45g/Cl - 1 ml of 1N AgNO3 = 0.03545g/Cl - V ml of 0.02 N AgNO3 = 0.03545 X V X 0.02 g/Cl - Cl - content in the sample = 0.03545 X V X 0.02 grams /Cl - 100 Cl - content in the sample = 0.03545 X V X 0.02 X 1000 mg/Cl - 100 8.07 Explain the method of determining the Fluoride content in water. 6 Marks SPADNS METHOD.( Sodium 2-(p-sulphophenylazo)-1,8-dihydroxy-3,6 naphthalene disulphonate). Principle: Under acidic conditions, fluorides react with Zirconium SPANDNS solution & the color of SPADNS reagent gets bleached. Bleaching is a function of fluoride ions & is directly proportional to the concentration of fluoride ions. Procedure: 1. Prepare a blank solution by adding 10ml SPANDNS solution in to 100 ml standard volumetric flask and add a solution HCl (7ml conc. HCl SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 140 diluted to 10ml) and make up to the mark. Use this blank solution to set zero in the colorimeter at 570nm. 2. Prepare a series of standard Sodium fluoride (NaF) solutions in the concentration range of 0.0-2.0mg/l. 3. Add 1 drop of NaAsO2 solution (0.5%) to remove any residual chlorine to each of the standard solutions. 4. Add 10ml of Zirconyl- SPANDNS reagent to each flask and dilute to 100ml with distilled water. Mix well & read the optical density of bleached color at 570 nm in the colorimeter. 5. Take suitable aliquot of water sample as test solution, add 1 drop of NaAsO2 solution (0.5%) to remove any residual chlorine. Add 10ml of Zirconyl- SPANDNS reagent and dilute to 100ml with distilled water. 6. Read the optical density of bleached color at 570 nm in the colorimeter. 7. Draw a calibration curve by plotting concentration versus absorbance and calculate the concentration of Fluoride ion. 8.08 Explain the method of Determining the Nitrate in water. 6 Marks Principle: Nitrate reacts with phenol disulphonic acid (PDA) to produce a nitro derivative, which in alkaline solution develops a yellow color. The color produced follows Beer’s law & is proportional of NO3 - present in the sample. The concentration of NO3 - is determined using a colorimeter or spectrophotometer. Procedure. 1. Prepare a standard nitrate (KNO3) solution in the 5-500 mg/L 2. Transfer 2.5, 5.0, 7.5, 10,1 2.5 cm 3 of standard nitrate (KNO3) solutions to separate beakers & evaporate to dryness on a hot plate. SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 141 3. To each beaker add 2 cm 3 of PDA & dissolve the residue. 4. Add 10 cm 3 of conc. NH3 to each beaker & dilute with distilled water in 100 cm 3 standard volumetric flasks. 5. Take 25 cm 3 of the given water sample in a beaker & evaporate to dryness on a hot plate. Add 10 cm 3 of conc. NH3 dilute up to 100 cm 3 in a standard volumetric flask. 6. Prepare a blank solution excluding the water sample. 7. Measure the absorbance at 410 nm wavelength using a Photo colorimeter. 8. Draw a calibration curve by plotting absorbance against the conc. of NO3 - and find out the conc. of NO3 - ions present in the given water sample. 8.09 Explain the method of determining the Sulphate content in Water by colorimetric method. 8Marks 1. Prepare a standard sulphate solution by dissolving 1.479 g of anhydrous sodium sulphate in distilled water & dilute to 1000ml to get 100 mg /ml of SO4 ion. 2. Transfer 2.5, 5.0, 7.5, 10, 12.5 cm 3 of standard sulphate solutions to separate 250 ml conical flasks. 3. Add 5 ml of conditioning reagent. (50 ml of glycerol+30 ml of con. HCl+300 ml of dil H2O+100 ml of C2H5 OH+75 g NaCl)to each conical flasks. SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 142 4. Add BaCl2 crystals with constant stirring till to get a white precipitate. 5. After 20 minutes, measure the absorbance of BaSO4 precipitate using a colorimeter/spectrophotometer at 420nm. 6. Transfer an aliquot of water sample to a 250ml conical flask & dilute to 100ml & repeat the steps through 3 to 5. 7. Prepare a blank solution excluding the sample. 8. Draw a calibration curve of absorption versus conc. of SO4 ions and Calculate mg SO4/L. 8.10 Explain the method of determining sulphate content in water by gravimetric method. 6 Marks Principle: The sulphate ions in the sample are precipitated by the addition of barium chloride solution to water sample acidified with hydrochloric acid & kept near the boiling point. Procedure 1 Transfer 200ml of water sample to a beaker 2 Add conc. Hydrochloric acid drop wise till to become just acidic. & add three drops in excess. 3 Boil the sample to reduce the volume to 50ml. 4 Add hot barium chloride solution (10 %) slowly with constant stirring until all the sulphate is precipitated. SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 143 5 Digest at its boiling temperatures for a few hours. 6 Filter through a gooch crucible & wash the precipitate with hot distilled water until the washings are free from chlorides. 7 Dry the precipitate & weigh barium sulphate. Calculation: Weight of BaSO4 be W g 233.3 g BaSO4 contains 96.0g of SO4 2- W g of BaSO4 contains 96.3 X W g of SO4 2- = mg 233.3 Sulphate = m X 1000 mg/L 200 8.11. Define biological oxygen demand 2Marks BOD is defined as the amount of oxygen in mg required for the complete oxidation of biologically oxidisable matter present in a liter of sewage effluent by a by micro organisms over a period of five days. 8.12. Define COD. 2Marks COD is defined as the amount of oxygen in mg required for the complete chemical oxidation of total oxidisable matter present in a liter of sewage effluent by a suitable oxidizing agent such as potassium di chromate. Problems 1) In a COD experiment 30 cm 3 of an effluent sample required 9.8 cm 3 of 0.001 M potassium di chromate for oxidation. Calculate the COD of the sample. 6 Marks i) 1000 cm 3 of 1M K2Cr2O7 solution contains 294 g of K2Cr2O7 9.8 cm 3 of 0.001 M K2Cr2O7 solution contains 294 X 9.8 X 0.001 1000 X 1 = 2.8812 mg ii) K2Cr2O7 + H2SO4 K2SO4 + Cr2 (SO4)3 + 4H2O + 3[O] SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 144 294 g of K2Cr2O7 = 48 g of oxygen. 2.8812 mg of K2Cr2O7 = 48 X 2.8812 = 0.4704 mg of oxygen 294 iii) COD of the effluent sample = 0.4704 X 1000 30 = 15.68 mg of oxygen / dm 3 2) In a COD experiment, 25 cm 3 of an effluent sample required 8.3 cm 3 of 0.001 M potassium di chromate for oxidation. Calculate the COD of the sample. 6 Marks i) 1000 cm 3 of 1M K2Cr2O7 solution contains 294 g of K2Cr2O7 8.3 cm 3 of 0.001 M K2Cr2O7 solution contains 294 X 8.3 X 0.001 1000 X 1 = 0.3983 mg ii) K2Cr2O7 + H2SO4 H2SO4 + Cr2 (SO4)3 + 4H2O + 3[O] 294 g of K2Cr2O7 = 48 g of oxygen. 0.3983 mg of K2Cr2O7 = 48 X 0.3983 = 0.4704 mg of oxygen 294 iii) COD of the effluent sample = 0.4704 X 1000 25 = 15.93 mg of oxygen /dm 3 8.13 Explain the Sewage Treatment. 6 Marks The treatment of domestic sewage is carried out in 2 or 3 stages. a) Primary treatment. b) Secondary treatment. c) Tertiary treatment. a) Primary Treatment. SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 145 The removal of course solids in the sewage water is effected by means of racks, screens, grid chambers & skimming tanks. Then the water is passed into a sedimentation tank where it is allowed to settle. The non-settle able solids are removed by coagulation by treatment with coagulating agents like alum, ferric chloride or lime. b) Secondary treatment.( Activated sludge process ). The waste water after the primary treatment is allowed to flow into large tanks where biological treatment is carried out. Activated sludge containing microorganisms is sprayed over the water. The microorganisms present in the sludge form a thin layer & thrive on the organic wastes in the sewage. Air is passed vigorously passed from the center of the tank in order to bring good contact between the organic wastes & bacteria in presence of air & sunlight. Under these conditions, aerobic oxidation of organic matter occurs. The sludge formed is removed by settling or filtration. A part of the sludge is reused & the rest is used as fertilizer. The residual water is chlorinated to remove bacteria & finally discharged into running water or used for watering plants. The activated sludge process operates at 90-95percent efficiency of BOD treatment. If the treated water contains a high concentration of phosphates, heavy metal ions, colloidal impurities & non-degradable organic compounds, the water is subjected to tertiary treatment. c) Tertiary treatment. The process includes a) Treatment with lime for the removal of phosphates as insoluble calcium phosphate b) Treatment with S 2- ions for the removal of heavy metal ions as insoluble sulphides c) Treatment with activated charcoal to adsorb remaining organic compounds & SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 146 d) Treatment with alum to remove the colloidal impurities not removed in the previous treatments to further reduce the BOD level. 8.14 Define Potable water 2 Marks Water that fit for human consumption & meets the stringent microbiological & chemical standards of quality to prevent waterborne diseases & health risks from toxic chemicals is called potable water. 8.15 Define Desalination and explain the methods of desalination. 6 Marks The process of partial or complete demineralization of highly saline water such as the sea water is referred to desalination. Methods of desalination. Reverse Osmosis. A reverse osmosis unit consists of a membrane, a vessel & a high pressure pump. The membranes are generally made up of cellulose acetate or nylon and are usually fabricated in a cylindrical shape. Electrodialysis. Principle: Passage of an electric current through a solution of salt results in migration of cations towards the cathode & anions towards the anode. The use of semi permeable cation or anion exchange membrane in an electrolytic vessel permits the passage of only cations or anions respectively in the solution. An electrodialyzer consists of a chamber carrying a series of compartments fitted with closely spaced alternate cation (C) & anion (A) exchange semi permeable membranes between the electrodes. An electrodialyzer unit will have 200 to 1000 compartments. The feed water is taken in the dialyzer & the electrodes are connected to a source of an electric current. SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 147 The anions pass through the anion permeable membrane towards the anode. However, these ions do not pass through the next membrane which is permeable only to cations. Similarly the cations moving in the other direction will pass through the cation exchange membrane but not the next. These anions & cations collect in the alternate chambers; the water in these is enriched with salt while that in the other compartments is desalinated. Micro porous sieves provided near the electrodes prevent the reentry of any deposit, which might have been formed on the electrodes, into the feed water. The enriched & desalinated waters are withdrawn separately. The former is rejected & the desalinated water is recycled to further reduce the salt content. 8.16 What are Hazardous chemicals? Explain the ill effects of Hazardous chemicals 6 Marks Chemicals which are combustible ,Oxidizers, explosives, flammable, pyrophoric, unstable, water reactive, carcinogens, toxicagents, reproductive toxins, irritants, corrosives, hepatotoxins, nephrotoxins, neurotoxins the release of which may substantially endanger the public health, public welfare or the environment are called hazardous chemicals. Hazardous chemical Source Ill effects Cadmium Mining wastes, effluents from plating industries Renal failure,Bone disease called itai-itai,high BP,Kidney damage,destruction of RBC Chromium Plating wastes Carcinogenic Lead Discharges from Severe dysfunction of SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 148 mining,metallurgical operations,plumbing, lead acid batteries Kidneys,reproductive systems & lever,impairement of central & peripheral nervous systems. Mercury Mineral processing operations,organomercury fungicides,discarded batteries,amalgams,toothfi llings. Neurological damage including paralysis,depression & irritability,blindness,insnity,chro mosomes breakage & birth defects. Arsenic Radioactive wastes Erosion of natural deposits,runoff from otchards,runoff from glass & electronics production wastes. Skin damage,prolems with circulatory systems, increased risk of getting cancer. Pesticides Agricultural practices Headache, dizziness. Large quantities may damage central nervous system. Cyanide Exists in water as HCN, metal cleaning & electroplating 60-90 mg is a fatal dose to living beings. Objective Questions: 1) The indicator used for the estimation of total hardness of a given water sample by EDTA method is a) Starch b) EBT c) Ferroin d) Methyl orange 2) Temporary hardness of water is caused due to the presence of a) Calcium carbonate b) Calcium chloride c) Magnesium bi carbonate d) None 3) The method used for desalination of water is a) zeolite process b) Lime soda process c) Ion exchange process d) Distillation 4) The indicator used in the determination of chloride content of water sample by Mohr’s method. a) Phenolphthalein b) K2Cr2O4 c) Starch d) Ferroin 5) Permanent hardness of water is caused due to the presence of a) Calcium carbonate b) Calcium chloride c) Magnesium bi carbonate d) All SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 149 6) Winkler methods is used to determine a) COD b) BOD c) DO d) Both D & C 7) Which of the following alkalinity is not present in water is a) CO3 2- and HCO 3- b) CO3 2- and OH - c) OH - only d) HCO3 - and OH - 8) Secondary treatment of sewage involves a) Biological treatment b) Physical treatment C) Chemical treatment d) All 9) In reverse osmosis the solute particle moves a) From High to low concentration b) low to high concentration c) Does not move d) both a & b 10) The ill effect o chromium is a) Headache b) fatal b) Birth defect c) Cancer 11) In alkaline condition, nitrate ion reacts with phenol disulphoneic acid to give a) Yellow colour b) white colour c) green colour d) brouwn colour 12) In SPADNS method of estimating fluoride ion concentration, wavelength used is a) 410 nm b) 500 nm c) 570 nm c) 570 nm c) 570 nm c) 570 nm d) 620 nm 13) Sulphate ion is precipitated by the addition of a) barium sulphate b) barium nitrate c) barium chloride d) barium phosphate 14) Impurities present in natural water is a) dissolved b) suspended c) organic d) all 15) Indicator used in the determination of alkalinity of water is a) Phenolphthalein b) starch c) EBT d) methyl orange SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 150 REVIEW QUESTIONS 1. Discuss the different types of Impurities present in natural with examples 2. What is water analysis? 3. Explain the method of Determining the Total Hardness of Water 4. What is meant by Alkalinity? 5. Explain the method of Determining the Alkalinity in water 6. Explain the method of determining the Chloride content in water 7. Explain the method of determining the Fluoride content in water. 8. Explain the method of Determining the Nitrate in water. 9. Explain the method of determining the Sulphate content in water by colorimetric method 10.Explain the method of determining sulphate content in water by gravimetric method. 11.Define biological oxygen demand 12.Define COD. 13.Explain the Sewage Treatment 14.Define Potable water 15.Define Desalination and explain the methods of desalination. 16.What are Hazardous chemicals? Explain the ill effects of Hazardous chemicals SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 151 I/II Semester B.E. Degree Examination, Dec.06/Jan.07 Common to all Branches Engineering Chemistry Time: 3 hrs Max. Marks: 100 Note: Answer any five full questions, choosing at Least two questions from each Part A and Part B Part A 1. a. Distinguish between gross and net calorific value of a fuel. (04 Marks) b. What is meant by cracking of petroleum? Explain fluidized bed catalytic cracking. (07 Marks) c. On burning 0.96 grams of a solid fuel in Bomb calorimeter, the temperature of 3,500 grams of water increased by 2.7°C. Water equivalent of calorimeter and latent heat of steam are 385 grams and 587 cals/gram respectively. If the fuel contains 5% H2, calculate its gross and net calorific values. (06 Marks) d. Write a note on power alcohol. (03 Marks) 2 a. Define electrode potential and derive Nernst equation for electrode Potential. (05 Marks) b. What are the advantages of secondary reference electrodes? Explain the construction and working of Ag/AgCl electrode. (06 Marks) c. What are electrochemical cells? Distinguish primary cells from secondary cells with examples. (05 Marks) d. What are concentration cells? Calculate cell potential of the following cell at 298 k. Ag /Ag+ (0.001 M)// Ag+ (0.50 M) / Ag What will be cell potential, when the concentration of silver ions in the above cell is changed from 0.OO1M to 0.0005 M, at same temperature? (04 Marks) 3. a. How does a fuel cell differ from battery? Explain the construction SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 152 and working of Nickel — metal hydride battery. (08 Marks) b. Explain the construction, working and application of H fuel cell, with cell reaction. (06 Marks) c. Give the classification of batteries with examples. (06 Marks) 4. a. Explain stress corrosion with examples. (04 Marks) b. What are corrosion inhibitors? Explain how corrosion is controlled by using anodic and cathodic inhibitors? (07 Marks) c. Write a brief note on the effect of following factors on the rate of corrosion i) Nature of metal ii) Hydrogen — over voltage iii) Relative areas of anode and cathode. (09 Marks) PART B 5. a. What is electroplating? Give the technological importance of metal finishing. (04 Marks) b. Explain the following factors influencing the nature of deposit: i) Complexing agents ii) Brighteners iii) Levellers and iv) wetting agents. (08 Marks) c. Discuss the electroless plating of copper on PCB. (04 Marks) d. Write a note on over voltage governing the metal finishing(04 Marks) 6. a. Explain the following with examples i) Thermotropic liquid crystal and ii) Lyotropic liquid crystal(06 Marks) b. What is homologues series? Explain the liquid crystalline behaviour of homologues of MBBA. (06 Marks) c. Discuss the instrumentation and applications of conductometric estimation. (08 Marks) 7. a. What are adhesives? Explain the synthesis and applications of epoxy resin. (06 Marks) b. What are elastomers? Mention the advantages of synthetic elastomers. (04 Marks) c. Give the synthesis and applications of butyl rubber. (04 Marks) d. Discuss the mechanism of conductance in Polyacetylene. (06 Marks) 8. a. What is potable water? Discuss the purification of water by reverse osmosis process. (05 Marks) b. Explain the method of determining sulphate content in water by gravimetric method. (05 Marks) c. Explain the determination of dissolved oxygen by Winkler method. Give the involved. (06 Marks) d. Describe the secondary treatment of sewage by activated sludge SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 153 process. (04 Marks) First Second Semester B.E. Degree Examination, July 2007 Common to All Branches Engineering Chemistry Time: 3 hrs.] [Max. Marks: l00 Note: Answer any FIVE full questions, choosing at least TWO questions from each part. PART A 1. a. Describe Fischer-Tropsch method of synthesis of petrol. (06 Marks) b. Explain the process of doping of silicon. (05 Marks) c. What is cracking? Explain fluidized catalytic cracking. (05 Marks) d. Discuss the mechanism of knocking. (04 Marks) 2. a. Explain the origin of single electrode potential. Derive Nernst equation for electrode potential. (07 Marks) b. what are reference electrodes? Mention the limitations of primary reference electrode and advantages of secondary reference electrodes. (04 Marks) C. Write a note on calomel electrode. (04 Marks) d. Represent the cell formed by the coupling of two copper electrodes immersed in cupric sulphate solutions. Concentration of cupric ions in one electrode system is 100 times more concentrated than the other. Write the cell reaction and calculate the potential at 300 K. (05 Marks) 3 a. Define fuel cell. How does it differ from a conventional galvanic cell? (04Marks) b. Explain the following fuel cells: i) Molten carbonate ii) Solid polymer electrolyte. (06 Marks) c. Explain the following battery characteristics: I) Cycle life ii) Shelf life iii) Energy efficiency. (06 Marks) d. Describe the construction and working of zinc-air battery(03 Marks) SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 154 4 a. Define corrosion. Explain electrochemical theory of corrosion, taking iron as a example. (08 Marks) b. Explain differential metal corrosion with suitable example (06 Marks) c. What is cathodic protection? Explain corrosion control by sacrificial anode method. (06 Marks) PART B 5. a. What is electro less plating? Mention its advantages. (04 Marks) b. Discuss the following factors influencing the nature of deposit: 1) Throwing power ii) pH of the electrolytic bath and iii)Temperature. (06 Marks) c. Discuss the electroplating of gold and mention its applications. (05 Marks) d. Explain electrolessplating of copper and its applications. (05 Marks) 6 a. Distinguish between Thermotropic and Lyotropic liquid crystals with examples (06 Marks) b. Explain the working of a liquid crystal in display systems.(06 Marks) c. What are the advantages of instrumental methods? (03 Marks) d. Explain the estimation of amount of strong acid in a given solution conductmetrically. (05 Marks) 7 a. Define polymerization. Explain solution and suspension polymerization techniques. (05 Marks) b. Give the synthesis properties and uses of: i) Teflon ii) Butyl rubber. (10 Marks) c. What are conducting polymers? Give the structure of polyaniline and mention its applications. (05 Marks) 8 a. Discuss the types of impurities present in natural water. (04 Marks) b. Explain the determination of hardness of water by complexometeric method. (06 Marks) c. Explain the electodialysis method of desalination of water.(04 Marks) d.Discuss determination of chloride by Argentometric method. (06 Marks) SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 155 First/Second Semester B.E. Degree Examination, Dec. 07 / Jan. 08 Engineering Chemistry Time: 3 hrs Max. Marks: l0O Note: Answer any five full questions, choosing at least two questions from each part. Part A 1. a. Describe the Bomb calorimetric method of determination of calorific value of a solid fuel. (06 Marks) b. What is reforming of petroleum? Give any three reactions involved in reforming. (05 Marks) c. What is power alcohol? Give its advantages as a fuel. (04 Marks) d. What is a photo voltaic cell? Explain its working. (05 Marks) 2. a. What are ion selective electrodes? Explain the measurement of p of a solution using glass electrode. (07 Marks) b. Explain the origin of single electrode potential. (05 Marks) c. Describe the construction and working of a calomel electrode. (04 Marks) d. Calculate the emf of the cell Fe/Fe 2+ (0.01 M)// Ag+ (0.1 M)/ Ag at 298 K, if standard electrode potential of Fe and Ag electrodes are — 0.42 V and 0.8 V respectively. (04 Marks) 3 a. Explain the construction and working of lead acid battery along with the reactions involved during charging and discharging. Mention its applications. (08 Marks) b. Explain the construction, working and applications of Nickel — metalhydride battery. (06 Marks) c. Explain the construction and working of the hydrogen oxygen fuel cell. (06 Marks) SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 156 4 a. Explain the differential aeration corrosion with a suitable example. (O8 Marks) b. How do the following factors affect the rate of corrosion? i) Nature of corrosion product. ii) Temperature. iii) PH. (09 Marks) c. Write notes on: i) Galvanizing ii Tinning (06 Marks) Part B 5. a. What is electroplating? Explain how the following factors influence the nature of electrodeposit: i) Metal ion concentration. ii Wetting agents. (05 Marks) b. Discuss the electro plating of chromium. (05 Marks) c. Mention the technological importance of metal finishing. (05 Marks) d. Explain electroless plating of Nickel and its applications. (O8 Marks) 6 a. Discuss the instrumentation and applications of colorimetric estimation. (08Marks) b. What are liquid crystals? Explain the molecular ordering in the following liquid crystal phases: i) Nematic crystal phase ii) Chiral Nematic phase iii) Smectic phase. (08 Marks) c. Distinguish between thermotropic and lyotrophic liquid crystals with examples. (04 Marks) 7 a. What are polymers? Explain the free radical mechanism of addition polymerization, taking ethylene as an example. (06 Marks) b. What is glass transition temperature? Mention its significance. Discuss any two factors affecting glass transition temperature. (06 Marks) c. Explain the manufacture of the following polymers and mention their uses: i) Phenol-formaldehyde ii) Polymethyl methacrylate. (08 Marks) 8 a. Discuss the different types of impurities present in natural water with examples. (04 Marks) b. Write a note on reverse osmosis. (05 Marks) c. Explain the method of determining sulphate content in water by gravimetric method. (06 Marks) SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 157 d. Calculate the COD of the effluent sample, when 25 cm of the effluent requires 8.3 c of 0.001 M K for complete oxidation. (05 Marks) First and Second Semester B.E. Degree Examination, June/July 08 Engineering Chemistry Time: 3 hrs. Max. Marks: 100 Note: Answer any FIVE full questions, choosing at least two questions from each part. Part-A 1. a. Describe the experimental determination of calorific value of a solid the using Bomb calorimeter. (06 Marks) b. What is synthetic petrol? Describe the Bergius method of synthesis of petrol. (05 Marks) c. Write a note on reforming of petrol. (04 Marks) d. Explain construction and working of silicon photovoltaic cell. (05 Marks) 2. a. What are reference electrodes? Explain the construction and working of alomel electrode. (08 Marks) b. Calculate the standard electrode potential of Cu 2+ /Cu if its electrode potential at 25° C is O.296V when [ is 0.015 M). (03 Marks) c. What are ion-selective electrodes? Explain the measurement of pH of a solution using glass electrode. (07 Marks) d. Derive Nearns’t equation on electrode potential. (08 Marks) 3. a. Explain the following battery characteristics: i) Energy efficiency ii) Current capacity iii) Cycle life. (06 Marks) b. Describe the construction and working of Lead — acid battery. (08 Marks) c. Describe the construction and working of H —02 fuel cell. (06 Marks) 4. a. Discuss the effect of following on the rate of corrosion. (08 Marks) SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 158 I) Nature of the metal; ii) Nature of the corrosion product; iii) pH iv) area effect. b. Explain differential aeration corrosion with suitable example. (06 Marks) c. What is cathodic protection? How a metal is cathodically protected by sacrificial anode method. (06 Marks) Part – B 5 a. Define polarization, decomposition potential and over voltage. Mention their significance with reference to electrode position. (08 Marks) b. How do the following affect the nature of electroplating? i) Current density ii) temperature iii) pH iv) organic additives. (08 Marks) c. What is electro less plating? Mention any two advantages. (04 Marks) 6 a. What are liquid crystals? Distinguish between thermotropic and lyotropic liquid crystals with examples. (06 Marks) b. Explain the working of liquid crystals in display systems. (06 Marks) c. What are the advantages of conductometric titrations over conventional titrations? (03 Marks) d. Explain the determination of concentration of an unknown solution by colorimetric method. (05 Marks) 7. a. Define polymerization. Explain different types of polymerization processes with suitable examples. (05 Marks) b. Give synthesis and uses of the following polymers i) Teflon ii)Neoprene (06 Marks) c. Write a note on compounding of resins. (04 Marks) d. Write preparation, properties and uses of epoxy resins. (05 Marks) 8. a. Explain temporary and permanent hardness of water. (04 Marks) b. Define B.O.D and C.O.D and mention various steps involved in sewage treatment. (04 Marks) c. Describe the argentometric method of estimation of chloride SHRIDEVI INSTITUTE OF ENGINEERING AND TECHNOLOGY. ENGINEERING CHEMISTRY DEPARTMENT, S.I.E.T., TUMKUR. 159 content of a water sample. (04 Marks) d. What is potable water? Describe the electrodialysis process of desalination of water. (08 Marks)