Electrochemistry.pdf

qwertyuiopasdfghjklzxcvbnmqwerty uiopasdfghjklzxcvbnmqwertyuiopasd fghjklzxcvbnmqwertyuiopasdfghjklzx ELECTR C!E"#$TR%& cvbnmqwertyuiopasdfghjklzxcvbnmq 'Type&the&document&subtitle(& wertyuiopasdfghjklzxcvbnmqwertyui opasdfghjklzxcvbnmqwertyuiopasdfg hjklzxcvbnmqwertyuiopasdfghjklzxc vbnmqwertyuiopasdfghjklzxcvbnmq wertyuiopasdfghjklzxcvbnmqwertyui opasdfghjklzxcvbnmqwertyuiopasdfg hjklzxcvbnmqwertyuiopasdfghjklzxc vbnmqwertyuiopasdfghjklzxcvbnmq wertyuiopasdfghjklzxcvbnmqwertyui opasdfghjklzxcvbnmqwertyuiopasdfg hjklzxcvbnmrtyuiopasdfghjklzxcvbn mqwertyuiopasdfghjklzxcvbnmqwert yuiopasdfghjklzxcvbnmqwertyuiopas dfghjklzxcvbnmqwertyuiopasdfghjklz xcvbnmqwertyuiopasdfghjklzxcvbnm qwertyuiopasdfghjklzxcvbnmqwerty ')ick&the&date( Neeraj Kumar ELECTRO CHEMISTRY ELECTROCHEMISTRY 1. INTRODUCTION: Electrochemistry is the branch of chemistry which deals with inter conversion of chemical energies and electrical energies. In electrochemistry, we discuss electrolytic reactions (reactions occurring due to passage of electricity through solutions) as well as electrogenetic reactions (reactions producing electric energy). 2. (i) ELECTROLYTIC CONDUCTION (ELECTROLYSIS): Electrolytic conduction is exhibited principally by molten form and by aqueous solutions of salts or substances. It takes place due to free movement of ions. Consider electrolytic conduction of molten NaCl between inert electrodes in electrolytic cells as shown in figure. DC Cathode – e– + anode (ii) NaCl Molten Na+ Cl– TABLE – 1 (a) (b) (c) (d) Phenomenon Ions attracted Direction of movement electrons Half cell reaction Sign At cathode Cations of Into cell Reduction + At anode Anions Out of cell Oxidation (Na + e → Na) (Cl– → ½ Cl2 + e) Negative, since it is Positive, since it is attached attached to the negative to the positive end of the end of battery. battery. (iii) Within the cell, current is carried by the movements of ions, cations towards the negative electrode (called the cathode) and anions towards the positive electrode (called the anode). This movement of ions is called electrolytic conduction. Factors, that influence the electrolytic conductivity of solutions of electrolytes are inter ionic attraction, solvation of ions, viscosity of solvent, etc. and these factors depends on the attraction of solute – solute, solute – solvent and solvent – solvent respectively. The average kinetic energy of the solute ions increases as the temperature is raised and, therefore, the conductance of electrolytic conductors generally increases. (iv) (v) 2.1 FARADAY’S LAWS OF ELECTROLYSIS: FIRST LAW: For the same electrolyte, the mass of a substance produced or consumed at an electrode is directly proportional to the quantity of electricity passed through the electrolytic cell. WαQ or, W = Z Q = Z .I.t Where Z = Electrochemical equivalent of the substantce. When I = 1 ampere and t = 1 second, w = Z, so, electrochemical equivalent (ECE) of any substance may be defined as the mass of the substance produced or consumed when a current of 1 ampere for 1 second (i.e. 1 coulomb) is passed through the electrolyte. Sri Chaitanya IIT-JEE Academy, 10-A, Talwandi, Jhalawar Road, Kota- 324005, (Raj). 2 ELECTRO CHEMISTRY SECOND LAW: When same quantity of electricity is passed through different electrolytes, the amounts of products obtained are proportional to their equivalent weights. W∝E 3. ELECTROLYTIC CONDUCTION: Conductance or conductivity is the property of the conductor which facilitates the flow of electricity through it. It is equal to the reciprocal of resistance, i.e., Conductance (G) = 1/R Unit: ohm–1 or mho (Ω–1) 3.1 SPECIFIC CONDUCTANCE, λsp: Reciprocal of specific resistance is called Specific Conductance. We Know that, R α ℓ a + – a a (where a : area of the electrode, ℓ : distance between the electrode) ∴R=ρ ℓ where ρ = specific resistance of substance. a 1 1 ℓ = ρ Ra ℓ Now, λsp. = or, λsp = 1 ℓ = G. ρ a For electrolytic cells, ℓ is known as CELL CONSTANT. a Unit of λsp: Ω–1 cm–1 or Ω –1m–1 3.2 EQUIVALENT CONDUCTANCE: It is the conductance of all ions produced by one gm equivalent of an electrolyte. It is measured by λ eq. = λ sp × V Where, λeq. = Equivalent conductance λsp = Specific conductance in Ω–1cm–1 V = Volume of solution of ml containing 1 gram equivalent of electrolyte Unit of λeq.: Ω–1cm2 equiv–1 3.3 MOLAR CONDUCTANCE: It is the conductance of all the ions produced by ionization of one gram mole of an electrolyte. λ m = λ sp × V Unit of λm: Ω–1cm2 mole–1 3.4 3.5 Conductance of the electrolytes is known to depend upon: nature of electrolyte, i.e., strong or weak, temperature and concentration of electrolyte in solution. Variation of molar conductance with concentration by strong electrolyte is given by Huckel Onsagar equation: λ m − λ m = b C , where b is a constant depending on the nature of the solvent and temperature and λ m is molar conductance at infinite dilution. At infinite dilution the concentration, C tends to be zero. At this large dilution λm = λm° as C → 0 Sri Chaitanya IIT-JEE Academy, 10-A, Talwandi, Jhalawar Road, Kota- 324005, (Raj). 3 λeq<λ°eq ∴Degree of dissociation. The drift velocity being a vector quantity.8 THE DRIFT SPEED AND MOBILITY OF IONS: Due to electrical force the ions gets accelerated but due to viscosity of the medium. 4. Therefore it is accelerated only to some limited velocity which depends upon the external field applied.7 DEGREE OF DISSOCIATION: At infinite dilution the ionization becomes 100%.00 volt. i. Electrode potential depends on the nature of conductor as well as the concentration of electrolytic solution. V. λ∞ = λ+ + λ– 3. been assumed to be 0. From above equation. size of the ion. i. etc. Thus.e. By general agreement normal hydrogen reference electrode is takes as the standard reference electrode. α = λ eq λ eq = Equivalent conductance at a given concentration Equivalent conductance at infinite dilution 3.1 STANDARD ELECTRODE POTENTIAL: As we know electrode potential depends on concentration of metal ions in solution in which metal electrode is dipped and the temperature of the half cell.6 KOHLRAUCH’S LAW: At infinite dilution. Sri Chaitanya IIT-JEE Academy. Kota. instead of v. electrode potential means electrode reduction potential. Thus we must choose one electrode arbitrarity as a standard and measure the potential difference between this standard electrode and any other electrode we require..ℓ Where u is called the mobility of the ion and V is the potential difference across two electrodes at the distance l.324005. charge on the ion. Talwandi.ELECTRO CHEMISTRY 3. ELECTRODE POTENTIAL: The tendency to gain or lose electrons by an electronic conductors is called electrode potential. the term drift speed (denoted as s) having no direction but magnitude equal to the |v| is used to that s = v. E and the viscosity of the solvent. The tentency to lose electron is called electrode oxidation potential and the tendency to gain electron is called electrode reduction potential. The potential difference developed between metal electrode and the solution of its ions of unit molarity (activity) at 298K is called standard electrode potential. The drift velocity determines the rate of conduction of current. it is retarded. The drift speed of an ion is proportional to the applied field E. when dissociation is complete. Normally. 10-A. This terminal limiting velocity is called the drift velocity of the ion in solution and is denoted by v.3 ELECTRO-CHEMICAL SERIES: When the elements are arranged in the order of their standard reduction potentials on the hydrogen scale we get electromotive series or electrochemical series of the elements. Conductivity or mobility of ion depends upon: viscosity of the solution (η).2 REFERENCE ELECTRODE: The potential of a single electrode can not be determined but the potential difference between the two electrodes can be accurately measured. Standard reduction potential (SRP) = –Standard Oxidation Potential (SOP) 4.. the mobility of an ion may be defined as the speed of the ion in an electrical field of unit strength (1 V m–1). 4. (Raj). or or s∝E s=uE s = u. each ion makes a definite contribution towards equivalent conductance of the electrolyte irrespective of the nature of the ion with which it is associated and the value of equivalent conductance at infinite dilution for any electrolyte is the sum of contributions of its constituent ions. Jhalawar Road. η.e. 4. At lower dilution the ionization (dissociation into ions) is less than 100% and equivalent conductance become lower. 4 . Its electrode potential has arbitrarily. 771 +0.045 –2. Kota.82 Half – Cell representation Li+ | Li K+ | K Na+ | Na Mg++|Mg OH– | H2 | Pt Al3+ | Al OH– | H2 | Pt Zn2+ | Zn Fe2+ | Fe Cd2+ | Cd Ag(CN)2–. Talwandi.036 0.095 +0.000 +0.ELECTRO CHEMISTRY STANDARD ELECTRODE POTENTIALS OF SOME ELECTRODES Electrode reaction Li+ + e– ⇌ Li K + + e– ⇌ K Na+ + e– ⇌ Na Mg++ + 2e– ⇌ Mg H2 + 2e– ⇌ 2H– Al3+ + 3e– ⇌ Al 2H2O + 2e– ⇌ H2 + 2OH– Zn2+ + 2e– ⇌ Zn Fe2+ + 2e– ⇌ Fe Cd2+ + 2e– ⇌ Cd Ag(CN)2– + e– ⇌ Ag + 2CN– Cu(OH)2 + 2e– ⇌ Cu + 2OH– AgI + e– ⇌ Ag + I– Sn2+ + 2e– ⇌ Sn Pb2+ + 2e– ⇌ Pb Cu(NH3)42+ + 2e– ⇌ Cu + 4NH3 Fe3+ + 3e– ⇌ Fe 2H+ + 2e– ⇌ H2 AgBr + e– ⇌ Ag + Br– Cu2+ + e– ⇌ Cu+ Sn4+ + 2e– ⇌ Sn2+ AgCl + e– ⇌ Ag + Cl– Hg2Cl2 + 2e– ⇌ 2Hg + 2Cl– Cu2+ + 2e– ⇌ Cu Ag(NH3)2+ + e– ⇌ Ag + 2NH3 O2(g) + 2H2O + 4e– ⇌ 4OH– Cu+ + e– ⇌ Cu ½ I2 + e– ⇌ I– Fe3+ + e– ⇌ Fe2+ ½ Hg22+ + e– ⇌ Hg Ag+ + e– ⇌ Ag O2 + 4H+ + 4e– ⇌ 2H2O Br2(ℓ) + 2e– ⇌ 2Br– Cr2O72– + 14H+ + 6e ⇌ 2Cr3+ + 7H2O ½ Cl2(g) + e– ⇌ Cl– Ce4+ + e– ⇌ Ce3+ (1mol dm3 H2SO4) Au3+ + 3e– ⇌ Au MnO4– + 8H+ + 5e ⇌ Mn2+ + 4H2O Co3+ + e– ⇌ Co2+ E°(V) – 3. Fe2+ | Pt Hg22+ | Hg(Pt) Ag+ | Ag H+ | O2 | Pt Br–.25 –1. Br2 | Pt Cr2O72-.222 +0.789 +0.136 –0. Cu(OH)2 | Cu I– | AgI | Ag Sn2+ | Sn Pb2+ | Pb Cu(NH3)42+ . Cr3+. Co2+ | Pt Sri Chaitanya IIT-JEE Academy.126 –0. Cu+ | Pt Sn4+.153 +0.2676 +0.828 –0. H+ | Pt Co3+. 10-A.324005.455 +1.373 +0.224 –0.401 +0.37 –2.44 –0. Ce3+ | Pt Au3+ | Au MnO4. Sn2+ | Pt Cl– | AgCl | Ag Cl– | Hg2Cl2 | Hg(Pt) Cu+ | Cu Ag(NH3)2+. Mn2+. NH3 | Cu Fe3+ | Fe H+ | H2 | Pt Br– | AgBr | Ag Cu2+.151 –0. 5 .33 +1. Jhalawar Road. H+ | Pt Cl– | Cl2 | Pt Ce4+.31 –0. (Raj).337 +0. I– | Pt Fe3+.521 +0.12 –0.925 –2.535 +0.15 +0. NH3 | Ag OH– | O2 | Pt Cu2+ | Cu I2.359 +1.0562 +1. CN– | Ag OH–.799 +1.51 +1.229 +1.66 –0.44 +1.714 –2.40 –0.763 –0. e. Each half cell consist electrode and electrolytic solution. Jhalawar Road. while Cu–rod gain in its mass. Ni. These react with cold water and evolve H2 gas. (A) Strongly electropositive metals (having SRP near about –2.1 APPLICATION OF ELECTROCHEMICAL SERIES: REACTIVITY OF METALS: The activity of the metal depends on its tendency to lose electrons. the electropositive character of metal decreases from top to bottom. Metals like Fe. Ag.4. One the example of Galvanic cell is Daniell Cell. ELECTROCHEMICAL OR GALVANIC CELLS: Electrochemical or Galvanic cell is a device for converting chemical energy into electrical energy. the electricity begins to flow. This tendency depends on the magnitude of SRP (Standard Reduction Potential). These readily dissolve in acids forming corresponding salts and combine with those substances which accept electrons. 6 Sri Chaitanya IIT-JEE Academy. Pb. The chemical reactivity of metals decreases from top to bottom in the series. Like reactivity.0 and –0. which lie below hydrogen are less reactive and do not evolve hydrogen from water. Talwandi.2V) (C) Weakly electropositive metals (lie below hydrogen and having +ve SRP) 4. . (b) The Zn–rod loses its mass. For example: (a) Alkali metals and alkaline earth metals having high –ve values of SRP are chemically active. which lie little down in series do not react with cold water but react with steam to evolve hydrogen. Kota. Such a metal is said to be chemically active.ELECTRO CHEMISTRY 4. 1 DANIELL CELL: Zinc – electrode e Zinc sulphate Solution (ZnSO4) A e– Copper electrode Copper(II) sulphate solution (CuSO4) Salt bridge (i) When zinc and copper electrodes are joined by a wire the following observations are made: (a) There is a flow of current through the external circuit. 10-A. The electromotive force (EMF) of such cell is directly proportional to the intensity of chemical reaction taking place in it. (b) (c) 4. Co etc.4 4. When the two compartments are connected by a “salt bridge” and electrodes are joined by a wire through galvanometer.0V or more negative) (B) Moderately electropositive metals (having SRP between 0.4. The metal which has high –ve value (or smaller +ve value) of SRP. The chemical reaction responsible for production of electricity normally takes place in two separate compartments.324005.3 DISPLACEMENT REACTIONS: (i) (ii) A metal higher in the series (lower SRP) displace the metal from its salt solution which is lower in the series (higher SRP) A non metal having high value of SRP will displace another non metal with lower SRP (i. occupying position above in the series)..2 ELECTROPOSITIVE CHARACTER OF METALS: The electropositive character also depends on the tendency to lose electrons. These compartments are called half cells. readily loses the electrons and converted into cations. An etc. metals are divided into three groups. Cl2 + 2I–  → 2Cl– + I2 Ex: Fe + Cu2+  → Fe2+ + Cu 5. 5. On the basis of SRP values. Metals like Cu.4. Sn. (Raj). electrons flow from the anode to cathode and their polarity is negative and positive. The electrode where oxidation takes place is called Anode and where reduction occurs is termed Cathode. (Raj). (b) The separation of two phases (state of matter) is shown by a vertical line. Jhalawar Road. For example: In Daniell cell. while the concentration of CuSO4 solution decreases. Chemical energy is converted into electrical energy. Gelatin or Agar-Agar (Plant Gel). The net cell reaction is the sum of two cell reactions. The indirect redox reaction take place in the cell will be: Anode or LHS compartment: Zn → Zn2+ (aq) + 2e– Cathode or RHS compartment: Cu2+ (aq) + 2e– → Cu Net cell reaction is: Zn + Cu2+(aq) → Zn2+(aq) + Cu The half cell in which oxidation occurs is called oxidation half cell and the half cell in which reduction occurs is called reduction half cell. (ii) • • • • • 5. fugacity (pressure) of gas. in left.3 REPRESENTATION OF GALVANIC CELL: The two half cells required to form an electrochemical cell can be represented by following rules: (a) Reduction half cell is represented in right and oxidation half cell. KNO3. the potential difference which arises between two solutions when in contact with each other. Kota. The ends of the U-tube are plugged with cotton wool as to minimize diffusion.2 BRIDGE AND ITS SIGNIFICANCE: Salt bridge is usually an inverted U-tube filled with concentrated solution of inert electrolyte (viz. is dissolved in a hot concentrated aqueous solution of inert electrolyte and the solution thus formed is filled in the U-tube. the cell reaction is Zn + Cu2+ → Zn2+ + Cu. The cell diagram is Zn | Zn2+ | | Cu2+ | Cu 5. respectively. During the passage of electric current through external circuit. (e) The significant features of the substances viz. (c) The various materials present in the same phase are shown together with the help of commas.ELECTRO CHEMISTRY (c) The concentration of ZnSO4 solution increases. are indicated in brackets drawn immediately after writing the substance.324005. Inert electrolyte is one whose ions are neither involved in any electrochemical change nor do the react chemically with the electrolytes in the two half cells. • Salt bridge can be replaced by a porous partition which allows the migration of ions without allowing the solutions to intermix. NH4NO3 etc. The following are the functions of the salt bridge: SALT (a) It connects the solutions of two half cells and completes the cell circuit. (d) The two half cells are joined with the help of double vertical lines. It can be represented as Ecell = Higher reduction potential – Lower reduction potential Or Or Or E0 cell = SRP of Cathode – SRP of Anode E0 cell = SOP of Anode + SRP of Cathode E0 cell = SOP of Anode – SOP of Cathode 7 Sri Chaitanya IIT-JEE Academy. Talwandi. activity (concentration) of ion etc. 5.4 ELECTROMOTIVE FORCE OR EMF: The difference in potential which causes a current to flow from the electrode of higher potential to the one of lower potential is called EMF of the cell. 10-A.). (d) The solution in both the compartments remains electrically neutral. (c) It keeps the solutions in two half cells electrically neutral. KCl..e. i. (b) It prevents transference or diffusion of the solutions from one half cell to the other. (d) It prevents liquid-liquid junction potential. . 10-A.further causes the following secondary reactions: NH4Cl + OH– → NH3 + Cl– + H2O Zn2+ + 2NH3 + 2Cl– → Zn(NH3)2Cl2 The EMF of the cell is 1. ECell = E cell – 0. ∆G = ∆G° + RT ln Q or.1 COMMERCIAL CELLS: Some of the electrochemical cells which are commonly employed as source of electrical energy can be classified as follows: THE PRIMARY CELLS: Such cells can be used only so long the reactive materials are present. it is connected with a source of potential higher than that of the cell in such a way. 1. The dependence of electrode and cell potential on activities of the ions in solution can be obtained from thermodynamic. ZnCl2 | MnO2 | Graphite The reactions involved are: Anode: Zn(s) → Zn+ + 2e Cathode: 2MnO2(s) + H2O(l) + 2e → Mn2O3(s) + 2OH– The OH–. that the cell now acts as the electrolytic cell. which convert the energy of fuel by oxidation reaction into electrical energy.5 Volts 6. the cell can not be recharged by passing current through it. as electrical energy flows through external circuit.5 FREE ENERGY CHANGE IN GALVANIC CELLS: When a cell operates. Pb is deposited on the cathode. –nFEcell = –nF E cell + RT ln Q or. The electrolyte is a nearly 37% H2SO4 solution with specific gravity of about 1. This value is equal to the product of number of moles of electrons and emf of the cell. 6.2 THE SECONDARY CELLS: Such type of cells can be used again and again by recharging the cell.6 NERNST EQUATION: The electrode potential and EMF of the cell depend upon the nature of the electrode. 1M is the concentration and 1 atm.3 FUEL CELL: Fuel cells are electrochemical devices. Example: Leclanche or dry cell: represented as: Zn | NH4Cl (20%). The quantity of electrical work done is measured in terms of free energy change. 6. The reactions are: − Anode: Pb(s) + SO 2 4 (aq) → PbSO4(s) + 2e Cathode: PbO2(s) + 4H+(aq. 2. A fundamental example of fuel cell is the Hydrogen – Oxygen cell.4% and 2.2% of H2SO4. temperature and the activities (concentration) of the ions in solution. E cell = E cell − RT ln Q nF Nernst equation At 25°C.ELECTRO CHEMISTRY 5.9V at 7. In lead storage battery Pb acts as anode and lead impregnated with lead oxide (PbO2) acts as cathode. ∆G = –nFEcell If reactants and products are in the standard states. ∆G° = –nFE°cell 5. PbO2 is formed at the anode and H2SO4 is regenerated.324005. Kota.15 at 298K.0V at 21. Sri Chaitanya IIT-JEE Academy. Jhalawar Road. 8 . To recharge the cell. (Raj). Talwandi. work is done by the system.) + 2e → PbSO4(s) + 2H2O(l) The EMF of the cell depends on the concentration of H2SO4 viz.059 log Q n (iii) ∆G° refers to free energy change for the reaction when the various reactants and products are present at standard conditions. 6.14V at 39. ∆G. One of the examples is the Lead storage battery. Once they exhausted.4%. is the pressure). (298K is the temperature. Electrods reactions are: Anode: H2 + 2OH– → 2H2O + 2e Cathode: ½ O2 + H2O + 2e → 2OH– Net reaction: 2H2(g) + O2(g) → 2H2O. The electrodes are placed in aqueous solution of KOH or NaOH. Sri Chaitanya IIT-JEE Academy.23 volts. Kota. Talwandi.324005. (Raj).ELECTRO CHEMISTRY Ammeter Θ Θ H2 O2 NaOH Solution H2 Fuel cell O2 It consists of two electrodes made of porous graphite. 10-A. The EMF of the cell is 1. H2 and O2 are bubbled into the cell under 50 atm pressure. Platinum is coated on the surface of the electrodes. Jhalawar Road. 9 . The calomel electrode is reversible with respect to (A) Hg22+ (B) H+ (C) Hg2+ (D) Cl– 9. (Raj). The following reactions represent the reduction of IO3− ion into I− ion in acidic and basic medium.80 volts. The standard electrode potential values of the elements A. A metal having negative reduction potential when dipped in the solution of its own ions has a tendency to (A) remain neutral (C) go into the solution 2.68. Jhalawar Road.260 V (A) Acid medium (B) Basic medium (C) Equally in both (D) none of these 5. 10-A. The highest chemical reactivity will be shown by (A) A (B) B (C) C (D) D 8. Indicator electrode is (A) SHE (C) Ag/AgCl electrode (B) Calomel electrode (D) Quinhydrone electrode 4. Kota.ELECTRO CHEMISTRY ELECTRODE POTENTIAL HASEEN SHURUAT (DPP – I) 1. Its electrode potential is independent of (A) temperature of the solution (C) area of the metal exposed (B) concentration of the solution (D) nature of the metal 3. What will happen if a copper spoon is used to stir a solution of aluminium nitrate? (A) the spoon will get coated with aluminium (B) an alloy of copper and aluminium is formed (C) the solution become blue (D) there is no reaction 6.50V respectively.324005. -0.50 and –0. Which one of the following does not get oxidised by bromine water? (A) Fe2+ to Fe3+ (B) Cu+ to Cu2+ (C) Mn2+ to MnO4– (D) Sn2+ to Sn4+ 10. B. Talwandi. (B) become electrically positive (D) be deposited from the solution A metal rod is dipped in a solution of its ions. Eº = + 0. -1. B and C are 0. The order of their reducing power is (A) A > B > C (B) A > C > B (C) C > B > A (D) B > C > A Sri Chaitanya IIT-JEE Academy.05. The metal that can not be obtained on reduction of its oxide by Al is (A) K (B) Mn (C) Cr (D) Fe 7. Eº = + 0. The standard electrode potentials of four elements A. 10 .40 and +0.907 V IO3− + 3H2O + 6e → I−+ 6 OH¯ . C and D are -3. Predict in which medium IO3− ion will act as a better oxidising egent? IO3− + 6 H+ + 6 e → I− + 3H2O. –2. The position of some metals in the electrochemical series in decreasing electropositive character is given as: Mg > Al > Zn > Cu > Ag.66. E O 2 / OH − = 0.36V respectively.49 V (D) .056 V 3.357M-CH3COOH.0.23V (O2 + 4 H3O+ + 4e → 6H2O).52 V 6. Estimate the standard reduction potential of CuS/Cu electrode. Ka for CH3COOH = 1.34 volt. – 4 | Mn 2+ = / Mn 2+ = 1. Talwandi.1V.7 (C) 1 : 3. 10-A.28 V? (A) 22. Ksp for CuS = 8. Eº = + 0. MnO4−/MnO2 in acid solution? Eº MnO 1. Fe |Fe 3+ 2+ 8.5 × 10–36 and E 0 Cu + 0. The Eº value of OCl–/Cl2 will be (A) 1. by oxygen.404 V (B) -0. [Co(H2O)6]2+ or [Co(NH3)6]2+ can be oxidised to the corresponding cobalt (III) complex.8V.324005. We have an oxidation – reduction system: [Fe (CN)6]3– + e ⇌ [Fe(CN)6]4– . |Cu = 10. Calculate the reduction potential at 25ºC for Fe3+/Fe2+ electrode if the concentration of Fe2+ ion is five times that of Fe3+ ion. (Raj). What is the standard reduction potential of oxygen in basic solution? (A) +0. E 0 = 0.42 V (C) –2.82 V 4.36 V. At what ratio of the concentration of oxidised and reduced from will the potential of the system be 0.4V.0.04 V (B) – 0.41 V (B) zero (C) . 11 .20 V (D) 0. What is the standard electrode potential for the electrode. Kota. Given: E 0 C H NH HCl|H (1atm) = −0.94V and 1.188 V 6 5 2 2 7.77 volt. (B) [Co(NH3)6]2+ (C) both (D) none of these (A) [Co(H2O)6]2+ 5.0. Which of the complex ion.7 : 1 2.ELECTRO CHEMISTRY PATHRILA RASTA (DPP – II) 1. Calculate electrode potential for the half cell Pt | H2 (1 atm) | 0. The standard potentials of OCl–/Cl– and Cl–/Cl2 are 0. The standard reduction potential for the reaction [Co(NH3)6]3+ + e → [Co(NH3)6]2+ is 0. Calculate the degree of hydrolysis and hydrolysis constant of aniline hydrochloride in M/32 solution at 25ºC. (B) 1 : 22. The standard reduction potential for the reaction [Co(H2O)6]3+ + e → [Co(H2O)6]2+ is about 1. The electrode potential of hydrogen electrode in neutral solution and 298K is (A) .404 V (C) +2.056 V (D) -2.9 (D) 1 : 15 The standard reduction potential of oxygen in acidic solution is 1. Sri Chaitanya IIT-JEE Academy. Jhalawar Road. Eº MnO 2 2+ 9.23V.51V.74 × 10–5. what is the e. Jhalawar Road.2812V and Eº for the quinohydrone electrode is 0.0 (C) nearly zero (D) nearly 14 10.m.6 For a cell reaction involving a two-electron change the standard emf of the cell is found to be 0.01 V 6. the potential of the normal calomel electrode is 0.41 V (B) 0. Eº = 1. the decrease in free energy at a given temperature is a function of (A) ln C1 (B) ln C2 (C) ln C2/C1 (D) ln C1/C2 8. Tell which of the following statements accurately describes the effect of adding CN to the cathode of a cell with a cell reaction: Cd + 2Ag+ → 2Ag + Cd2+. is 0.212V. both at the same temperature. The e.63 volt.75M-Cr3+ and 0. Assuming that hydrogen behaves as an ideal gas.29 (D) 9. The ECell for Ag(s) | AgI (satd) || Ag+ (0.14 (C) 2. Talwandi.6 (D) 1.417 V.ELECTRO CHEMISTRY GALVANIC CELL HASEEN SHURUAT (DPP – III) 1. (A) 4.7 (C) 7.6996V.10M-AgNO3 solution and a cathode of Pt immersed in a solution of 1. (B) 6. Kota.32 × 10-17 5.005 V (D) +0. 12 .5 × 10– 2 (D) 10 - 3.80V and E0Cr O ¯ |Cr3+ = 1. The emf of the cell at 298K Hg(l) | Hg2Cl2 (s). The equilibrium constant of the reaction at 25ºC will be (A) 1 × 1010 (B) 1 × 10–10 (C) 29. The potential of a cell consisting of an anode of silver in 0.f.m.0 atm pressure and 1.005 V (B) -0. 0. What is the PH of the solution in which the anode is immersed if the cathode is in contact with a solution of PH = 3? (A) 3 2.0 M – Concentration for all species except hydrogen ion? Given: Cl2 + 2e → 2 Cl–.7 × 10-8 (B) 7.71 V Sri Chaitanya IIT-JEE Academy. A depolariser used in dry cell batteries is (A) NH4Cl (B) MnO2 (C) KOH (D) Na3PO4 7. At what PH does the potential (emf) for the disproportionation of chlorine change from a negative value to a positive value. For a electrochemical cell Zn | Zn2+ (C1 M) || Cu2+ (C2M) | Cu.295V at 25ºC.32 × 10-13 (C) 8. (1. KCl sol.01 V (C) +0. (A) 3.0 N) | Quinohydrone | Pt.51 V (D) 0.15 9.33 V) 2 2 7 (A) 0.61 V (C) 0.58 (B) 1.f.55 × 10-9 (D) 7. Eº = 1. Eº = 1.324005.36 volt OCl– + 2H+ + e → Cl2 + H2O. (Raj). 10-A.5M-Cr2O72¯ .10M) | Ag (s) is + 0. of a galvanic cell composed of two hydrogen electrodes is 272 mV. assuming 1. What is Ksp of AgI? (A) 2. of the cell at 25ºC if P1 = 640 mm and P2 = 425 mm: Pt | H2 (P1) | HCl | H2 (P2) | Pt? (A) -0.5 (B) 7.25M-H+ is (E0Ag+|Ag = 0.2V (A) Eº increases because Cd(CN)42− forms (C) Eº increases because Ag(CN)2− forms (B) Eº decreases because Cd(CN)42− forms (D) Eº decreases because Ag(CN)2− forms 4. What is the PH of the quinohydrone solution. 44) = 3.88 ×10 ] –5 –5 (A) 3.10M. The emf of the cell obtained by combining Zn and Cu electrodes of a Daniel cell with normal calomel elctrodes are 1. E° = + 0. Talwandi.88 ×10–5 (C) 3. 10-A.443V with respect : = – 0. Given E 0 Pb |Pb 2+ [Antilog(-4. Kota.6885 V at 25º C. (A) 3. find emf of Daniel cell. EºCell = 0.262 V 14. What is the PH of the buffer solution? EºCalomel = 0. 13 .854 V The emf of the cell: H2 (g) | Buffer || Normal calomel Electrode is 0. what is EºV3+|V2+? (A) -0.92 (C) tending to zero (D) tending to 14 13. This couple was observed to have a potential –0. E° = 1. E° = – 0.5 V Which combination of two half cells would result in a cell with largest potential? (A) i and iii 15.439 V If EºAg+|Ag = 0.5 V (iii) A3– → A2– + e–. What is the Ksp of TlBr.803 V (D) 0. Antilog(-4.46 (B) 6. From the following E° values for the half cells (i) D → D2+ + 2e–.6 ×10 .324005.083V and – 0.854 V (D) -1.101 V (C) 0. respectively at 25º C.28V.256 V (C) +1.10M-KBr with TlBr and allowing Tl+ ions form the insoluble bromide to equilibrate.54) = 2. Jhalawar Road.126V to Pb2+/Pb couple in which Pb2+ was 0. EºCell = 0. (B) i and iv (C) iii and iv (D) ii and iv A Tl+/Tl couple was prepared by saturating 0.799 V.ELECTRO CHEMISTRY 11. (Raj).018V.6 ×10–5 (D) None of these Sri Chaitanya IIT-JEE Academy.256 V 12. when barometric pressure is 760 mm. (A) 1.616 V V3+ + Ag+ + H2O → VO2+ + Ag (s) + 2H+. (B) +0. E° = –1.28 V.5 V (ii) B+ + e– → B. If the potential of normal calomel electrode is – 0.6 ×10–3 (B) 2.065 V (B) 1.5 V (iv) C2+ + e– → C+. Two electrochemical cells are assembled in which the following reactions occur: V2+ + VO2+ + 2H+ → 2V3+ + H2O. Pt | H2 | NaHSO3 (0.359 volt Pb2+ + 2e → Pb(s).0M. 4. Given: EºZn2+|Zn = – 0. Calculate Ka2 of H2SO4. Given Ksp of AgCl and AgI are 1. Calculate what will be the cell voltage if this reaction is run at 25ºC and 1 atm in a cell in which H2 activity is unity and H3O+ and Cl– activities are each at 0. Na2SO3 (6. 5.1M–KCl || satd.01M? 7. Kf for Cu (NH3)42+ = 1 × 1012. AgCl in 0. –237. Calculate the minimum mass of NaOH required to be added in RHS to consume all the H+ present in RHS of the cell of emf + 0.4 KJ/mol. what [Cl–] must be present in the cathodic half cell to achieve the desired emf. 9.80V. Eº = 0. Given EºAg+|Ag = 0. Kota. AgCl. Also report the emf of cell after addition of NaOH. before its use.324005. 14 . Estimate the cell potential of a Daniel cell having 1. 0.2 and –368.45 volt at 25ºC.77 volt AgCl (s) + e → Ag (s) + Cl– (0. Calculate the percentange of lead in alloy.0M-Cu2+ after sufficient ammonia has been added to the cathode compartment to make the NH3 concentration 2. The standard reduction potential at 25ºC of the reaction 2H2O + 2e– ⇌ H2 + 2OH– is – 0.5 ×10–17.8 × 10–10 and 8.152 volt Ag+ + e → Ag (s).34 V. (Raj). Talwandi. Eº = 0.086V. E 0 = 0.3 M) | Zn If EºZn2+|Zn = – 0. 10. the standard free energy of formation of AgCl(s). Jhalawar Road. Write the cell diagram for each of the following reactions and calculate Eº for each cell.7.770 V.535 V. Eº = – 0.001 M). For the reaction.ELECTRO CHEMISTRY PATHRILA RASTA (DPP – IV) 1.1M–AgNO3. 8. respectively.4 M). Given E 0 Fe |Fe I |I 3+ 2+ − 2 3. x M.8277 volt. Sri Chaitanya IIT-JEE Academy.503V at 298K. (a) (b) (c) (d) AgBr (s) + 1/2 H2 (g) → Ag(s) + H+ + Br– E0AgBr|Ag|Br = + 0.50 M = 0. NH4NO3 || 0. If it is desired to construct the following voltaic cell to have ECell = 0.1M–AgNO3 is 82% dissociated. An alloy weighing 1.126 volt AgI (s) + e → Ag(s) + I–. Eº = + 0. Ag is 0. EºCu2+|Cu = + 0.Cl–) | Ag (s) The voltage of the cell given below is – 0.44 × 10–3 M) || Zn2+ (0. EºZn2+|Zn = – 0.763V.0 M. 6. Eº = – 0. Calculate [Fe3+] at equilibrium when potassium iodide is added to a solution of Fe3+ initially 0.1 M) + e → Fe2+ (1M). Calculate solubility product of AgCl. Eº = – 0.46V. Calculate equilibrium constant for the reaction 2H2O ⇌ H3O+ + OH– at 25ºC. AgI) || Ag+ (sat.0M-Zn2+ and originally having 1.22 volt - 2. Zn | Zn2+ (0.76 V The EMF of the cell: Ag. Pt. PbSO4 + 2e → Pb(s) + SO42−.10 volt.80 volt Fe3+ (0. H2(g) + 2AgCl(s) + 2H2O(ℓ) → 2Ag(s) + 2H3O+(aq) + 2Cl–(aq) at 25ºC. until [I–] = 1. 10-A. An silver electrode was dipped in solution and Ecell of the cell Pt. Ag(s) | Ag+ (sat.1M) || HCl | H2 (1 atm). H2(1atm) | H+(1M) ||Ag+ |Ag was 0.1M–KCl is 85% dissociated and 0.76 V.701V at 25ºC. H2O(ℓ) and (H3O+ + Cl–) (aq) are –109.05g of Pb-Ag was dissolved in desired amount of HNO3 and volume was made 350 ml. 1M-NH3 | Ag.337V. enthalpy change and entropy change at 25ºC.324005. When metallic copper is shaken with a solution of a copper salt. ∆Gº = – 42927 J at 25ºC. Determine E0 and K for the equation Fe3+ (cyt c1) + Fe2+ (cyt b) ⇌ Fe2+ (cyt c1) + Fe3+ (cyt b) 12. When equilibrium is established at 298K. 1M-NH3 | Ag.ELECTRO CHEMISTRY 11. (B) Ag | 3 x 10-3 M AgCl.m.22V for the Fe3+ (cyb c1) | Fe2+ (cyt c1) couple.5H2O (sat. where E0 represents the standard state reduction potentials at PH = 7.1185V at 298K. Sri Chaitanya IIT-JEE Academy. Given E0 = 0. Ecell = 0. 15 .6753V at 25ºC. 1M-NH3 || 3 x 10-3M AgCl. practically all silver ion can be assumed to exist in form of a single ionic species [Agx(NH3)y]x+. of the cell in which the given reaction takes place.1263V at 298K. 14.6915V at 0ºC and +0. For the reaction: H2 (1atm) + 2AgCl(s) ⇌ 2Ag(s) + 2H+ (0. Ecell = 0. Calculate the free energy change.) + 2Ag(s) is +0. 10-A.5H2O (sat.1M) + 2Cl– (0.0 at 25ºC and cyt is an abbreviation for cytochromes. Talwandi.) || AgCl(s) | Ag(s) for the reaction is Cd(s) + 2AgCl(s) + aq → CdCl2. Calculate the e. (Raj).08V for the Fe3+ (cyt b) | Fe2+ (cyt b) couple and E0 = 0. [Cu2+]/[Cu+]2 = 1. The emf of the cell: Cd(s) | CdCl2. 13.f. Jhalawar Road.66 ×106M–1. 0. 15. Kota.4 x 10-3 M AgCl. Compute the values of x and y using the two following cells: (A) Ag | 0. 1M-NH3 || 40 x 10-3M AgCl.1M). the reaction Cu + Cu2+ ⇌ 2Cu+ proceeds. If the standard potential of the Cu2+ | Cu half cell is + 0. what is the standard potential of Cu+| Cu half cell? When silver chloride is dissolved in a large excess of ammonia. 0 hr 7.79V. Hg.324005.05 % (D) 98 % 11.5 hr (D) 41. The molarity of Ni2+ at the end of electrolysis is (A) 1. Sri Chaitanya IIT-JEE Academy. Jhalawar Road. Mg (B) Mg.34V.586 M (D) 2 M 3. Hg2+ | Hg = 0. Mg2+ | Mg = – 2. A galvanic cell is set up from a zinc bar weighing 100g and 1.10 (B) 0. Ag (C) Ag.525 ampere. Two electrolytic cells. If 0. The colourless solution contains (A) platinum sulphate (B) copper sulphate (C) copper hydroxide (D) sulphuric acid 8. Cu. The ratio of iron deposited at cathodes in the two cells will be (A) 3 : 1 (B) 2 : 3 (C) 1 : 1 (D) 3 : 2 6. (Raj).6 hr (B) 26. The elctrochemical equivalent of two substances are E1 and E2 The current that must pass to deposit the same amount at the cathodes in the same time must be in the ratio of (A) E1 : E2 (B) E2 : E1 (C) E1 − E2 : E2 (D) E1 : E2 − E1 4. The amount Faradays required to liberate 1 mole of an element indicates (A) weight of element (C) charge on the ion of that element (B) conductance of electrolyte (D) chemical equivalent 9.37V. The number of equivalents of the ion is (A) 0. one containing acidified ferrous chloride and another acidified ferric chloride are connected in series. With increasing voltage the sequence of deposit of metals on the cathode will be (A) Ag.ELECTRO CHEMISTRY ELECTROLYSIS HASEEN SHURUAT (DPP – V) 1. it was found that colour of CuSO4 disappeared with the evolution of gas at the electrode.7 ampere is passed for 6 hours between Nickel electrodes in 0.001 (D) 0. Hg.8 hr (C) 20. 10-A.0M copper sulphate solution. Two platinum electrodes were immersed in a solution of CuSO4 and electric current was passed through the solution. (B) Silver (C) Copper (D) Zinc A current of 3. Talwandi. 16 . Electrochemical equivalent is more for (A) Hydrogen 2. Hg.0001 10.01 (C) 0. Mg.50 litre of 2M solution of Ni(NO3)2 . Cu (D) Cu. Cu2+ | Cu = 0.0L of 1. Hg(NO3)2 is being electrolysed using inert electrodes. After some time. Ag An ammeter and a copper voltameter are connected in series in an electric circuit through which a constant direct current flows. The values of standard electrode potential in volts are Ag+ | Ag = 0.80V. How long would the cell run if it is assumed to deliver a steady current of 1. what is the percentage error of ammeter? (A) 2.0 ampere? (A) 53.172 M (C) 0. Mg(NO3)2. The amount of hydrogen evolved from H2SO4 as compared to that from HCl is (A) the same (C) one half as such (B) twice as such (D) dependent on concentration 5. The same quantity of electricity is passed through one molar solution of H2SO4 and one molar solution of HCl. Cu.635 gm of Cu is deposited in one hour. A solution containing one mole per litre each of Cu(NO3)2. An ion is a reduced to the element when it absorbs 6×1020 electrons. AgNO3. Kota. The ammeter shows 0.172 M (B) 0.09 % (B) 51 % (C) 2. 01%.324005.118 (C) -(E1 + E2)/ 0. After passage of 140 ampere for 482. 112) and element B (at.ELECTRO CHEMISTRY 12.36 (C) 14. then the H P of the (A) right electrode will increase (C) both electrode will increase 17.89 gm and 20. = 27) form chlorides. The preparation of LiOH by the electrolysis of a 35% solution of LiCl using a platinum anode led to a current efficiency of 95%. What is the percentage of the theoretical yield of sodium hydroxide obtained? (A) 40 % (B) 60 % (C) 30 % (D) 80 % Sri Chaitanya IIT-JEE Academy. During the electrolysis of an aqueous salt solution.88%. silver and gold as impurities.66 gm (D) 2.28 gm (B) 2. the mass of anode decreased by 22. What weight of LiOH was formed by the passage of 2. During the same period of electrolysis. 98.71 (B) 29.5 seconds.118 (D) (E1 – E2)/ 0.26 gm and the cathode increased by 22. The percentage of iron and copper in the original sample are respectively (A) 0.1 gm Element A (at. (Raj).118 16. + 100 ml of a buffer of 1M. Kota.88% (D) 98. 98. The weight of zinc and nickel deposited in a certain time were found to be 22. 0. 3. In the lead storage battery the anode reaction is Pb(s) + HSO4¯ + H2O → PbSO4(s) + H3O+ + 2e.53 gm Assuming that copper contains only iron.6 gm of A was deposited while only 0.NH3 and 1M. Talwandi.01% 19. respectively. The valency of B is 3.9gm of B was deposited.011gm. wt.85%. How many gm of Pb will be used upto deliver 1 amp for 100 hrs? (Pb = 208) (A) 776 gm 13. 98. what is the chemical equivalent weight of nickel? (A) 58. the PH in the space near one of the electrode was increased and the other one was decreased. The salt solution was (A) NaCl (very dilute) (B) ZnCl2 (C) NaCl (Conc.42 18. The PKa value of acetic acid is (A) (E1 + E2)/ 0. 10-A.55 gm.40 gm (C) 0. After electrolysis of an aqueous sodium chloride solution.118 (B) (E2 – E1)/ 0. (B) left electrode will increase (D) both electrode will decrease The same current was passed successively through solution of zinc-ammonium sulphate and nickelammonium sulphate rendered alkaline with ammonia. it was found that the solution is being neutralised by 60 ml N – HCl solution. If electrolysis of water takes place only and the electrode reactions are: Right: 2H2O + O2 + 4e → 4 OH and Left: 2H2O → 4H+ + O2 + 4e. Jhalawar Road. respectively at 250C. Solutions of these chlorides are electrolysed seperately and it is found that when the same quantity of electricity is passed. wt.68A for 1 hr? (A) 2.1%.88% (B) 0.5 amp is passed through both cells for 20 min.) (D) Cu(NO3)2 15.39 (D) 36. Given that the chemical equivalent weight of zinc is 32.The valency of A is (A) 1 (B) 2 (C) 3 (D) 4 14.88% (C) 0. 5. A hydrogen electrode placed in a buffer solution of CH3COONa and CH3COOH in the ratio’s x : y and y : x has electrode potential values E1 and E2 volts. A current of 1. 20.7.18 gm of copper was deposited in a copper voltameter in series. (B) 388 gm (C) 194 gm (D) 0. 17 .NH4 are placed in two voltaic cells separately. Jhalawar Road.0 gm of this alloy at a current of 1. 10. Tl = 204) Iridium was plated from a solution containing IrCl6y for 2.f.m. 30 ml of an alcohol.ELECTRO CHEMISTRY PATHRILA RASTA (DPP – VI) 1. If the average production of ammonium nitrate is 5000 Kg/day. was operated as an electrolytic cell as Cu as the anode and Zn as the cathode.075 amp. what will be the PH of the final solution? For HSO4 . 7. How long should the current be passed to get the above change? An object whose surface area is 78.50 × 103 ampere? Assume that the density of solution is 1.6N-CuSO4 solution is electrolysed between two platinum electrodes till the concentration in the residual liquid is 0. The density of gold is 19. 18 .0 amp is used. 9.) solution of NaClO if the cell current is 2. assuming that the layer of gold builds up evenly. at a current of 0. (Raj).34V. The object is placed in a solution of Au(NO3)3 and a current of 2. when a steady current of 5.5 amp-hour through the lead cathode electrolytic cell. How long will a cell operate to produce 1000 litre of 5. Ka = 1. Talwandi. 100 ml of 0. Kota. y. on this ion? (Ir = 192) The electrolysis of cold sodium chloride solution produces sodium hypochlorite by reacting NaOH and Cl2 throughly.359 gm (a) what is the oxidation number of Ir in IrCl6y? (b) what is the charge. (Au = 197) The following galvanic cell Zn | Zn (NO3)2 (aq) || Cu (NO3)2 (aq) | Cu (100 ml. What would be the e. If [Cu(NH3)4]2+ is detectable upto its concentration as low as 1×10–5M. The hydrogen is used for the production of ammonia and nitric acid (by the oxidation of ammonia). An alloy of lead (valency = 2)-thallium (valency = 1) containing 70% Pb and 30% Tl by weight can be electroplated onto a cathode from a perchloric acid solution. If Cu is produced at one electrode and oxygen at the other. A test for complete removal of Cu2+ ions from a solution of Cu2+ is to add NH3(aq).76V.75 A is applied.10 M. The Ir deposited on the cathode weighed 0.324005.0 gm/ml.1M-CuSO4 is electrolysed by passing a current of 3.3×10–2. 1M) (100 ml. After passage of this charge. 250 ml of 0.6 cm2 is to be plated with an even layer of gold 8. estimate the daily consumption of electricity per day. 4. EºZn2+|Zn = – 0. 5. EºCu2+|Cu = + 0. sufficient quantity of NH3 is added to electrolysed solution maintaining [NH3] = 0. Sri Chaitanya IIT-JEE Academy.2H2O were placed in the cathode space. 6.0 × 10–4 cm thick.1×1013 and thus confirms the presence of Cu2+ in solution.06 amp. 8. A blue colour signifies the formation of complex [Cu(NH3)4]2+ having Kf = 1. 11 gm of HCl and 1gm of SnCl2. hydrogen is produced by the elctrolysis of water.10 amp? (Pb = 208. After passing current at a rate of 26.512 ampere for 1368 second. would a blue colour be shown by the electrolysed solution on addition of NH3.1N. of the cell at 25ºC? Assume that the only electrode reactions occuring were those involving Cu | Cu2+ and Zn | Zn2+. 1M) 2. How many hours would be required to deposit 5. Find the time required for the electroplating to be completed. 20 gm of C6H5NO2. Determine the current yield. 12. A current of 0.0 hour with a current of 0. 3. 50 ml of 0.48 ampere was passed for 10 hours and then the cell was allowed to function as galvanic cell.1M-CuSO4 solution is electrolysed for 12 min.25% (by wt. To reduce nitrobezene to aniline. 250 ml of water. 10-A. At the Nangal fertilizer plant in Punjab.76 gm of aniline was produced.3 gm/ml. 216 Cal/ºC-gm. The platinum electrodes are inserted in each compartment and 1. 15.ELECTRO CHEMISTRY 11. Jhalawar Road. If the specific heat of metal is 0.324005. 13.0M-Na2HPO4 is placed in two compartments (one litre in each) of an electrolytic cell.7% H2SO4 by wt.72L of H2 and 2. Kota. Before charging.0M-NaH2PO4 and 1.25 ampere current is passed for 212 min. 7.35L of O2 were generated at STP. Two litre solution of a buffer mixture containing 1. the specific gravity of the liquid was found to be 1.15. When the same amount of charge is passed through a nitrate solution of metal M. 12. 10-A.11 (15.). what is the formula of metal nitrate? 14. By passing a certain amount of charge through NaCl solution.467 gm of the metal was deposited. H2S2O8 can be prepared by electrolytic oxidation of H2SO4 as 2 H2SO4 → H2S2O8 + 2H+ + 2e. the specific gravity of the liquid was found to be 1. 0.2 litre of chlorine were liberated at STP. Assuming electrolysis of water only at each compartment. calculate the average current used for charging the battery. What is the weight of H2S2O8 formed? The electrode reactions for charging of a lead storage battery are: PbSO4 + 2e → Pb + SO42– PbSO4 + 2H2O → PbO2 + SO42– + 4H+ + 2e The electrolyte in the battery is an aqueous solution of sulphuric acid. 19 . After charging for 100 hours.9% H2SO4 by wt). H2O2 can be prepared by successive reactions: 2 NH4HSO4 → H2 + (NH4)2 S2O8 (NH4)2 S2O8 + 2H2O → 2 NH4HSO4 + H2O2 The first reaction is an electrolytic reaction and second is steam distillation. What amount of current would have to be used in first reaction to produce enough intermediate to yield 100 gm pure H2O2 per hour. (Raj). Assume current efficiency 50%. Perdisulphuric acid. Talwandi. In such an electrolysis. If the battery contained 2 litres of the liquid.28 (36. 9. what will be PH in each compartment after passage of above charge ? PKa for H2PO4– = 2. Sri Chaitanya IIT-JEE Academy. Oxygen and hydrogen are by products. both solutions being taken in the same conductivity cell.05 N . The degree of dissociation of acetic acid in an aqueous solution of the acid is practically unaffected by (A) adding a piece of sodium chloride (B) adding a drop of concentrated hydrochloric acid (C) diluting with water (D) raising the temperature 9.2 (C) 18.01N solution of NH4OH is 9.8. (A) 833. Equivalence conductance at infinite dilution of NH4Cl. while its specific conductance is 3. then the degree of dissociation of NH4OH at this temperature is (A) 2.023 20 Sri Chaitanya IIT-JEE Academy.5 cm apart.4 × 10–4 cm2volt–1sec–1 at infinite dilution. The cell contained 0. 217.67 ohm 5. If equal volumes of solution A and B are mixed. what is the equivalent conductance of the salt solution in ohm-1cm2eq-1.3 (B) 115. If the equivalent conductance of 0.2 N 8. The equivalent conductivity in ohm–1 cm equivalent–1 of 1N acetic acid is (A) 4.4 (D) 0.9 ohm-1. (A) 150 ohm (B) 75 ohm (C) 33.1 N (D) 1. The resistance of a solution A is 50 ohms and that of solution B is 100 ohms.6 × 10–4 and 5.4 and 108. At 18ºC.324005.24 x 10-8 7. At 25ºC.33ohm-1cm2eq-1. the normality of the solution is (A) 0.9 % (D) 4.8 (C) 8.64 x 10-3 (D) 1.NaCl 3.0 N (C) 1. (B) 0. NaOH and NaCl are 129. The resistance of 1N solution of acetic acid is 250 ohm. The difference between a decinormal solution of HCl and a decinormal solution of acetic acid is that (A) one of them conducts electricity and the other does not (B) one of them is corrosive to the skin and the other is not (C) one of them contains undissociated molecules of the acid and the other does not (D) one of them decompose sodium carbonate and the other does not.33 ohm (D) 66. (A) 120 (B) 60 (C) 240 (D) 6000 6. 10. when measured in a cell of cell constant 1. 10-A. If the two electrodes are just half dipped into the solution which has a resistance of 50 ohms. Which pure substance will not conduct electricity? (A) Molten NaCl (B) Molten KOH (C) Liquefied HCl (D) Liquid Hg 4. Kota.7 % Which of the following solution have highest resistance? (A) 1N – NaCl (B) 0. Talwandi. What is the equivalent conductance of ammonium chromate solution in ohm-1cm2eq-1. If the apparent degree of dissociation is 90%.ELECTRO CHEMISTRY CONDUCTANCE HASEEN SHURUAT (DPP – VII) 1.05N solution of a salt. Jhalawar Road. each of area 10cm2 are fixed 1. (Raj).1N – NaCl (C) 2N – NaCl (D) 0.6 (B) 9.825 ohm-1 cm-1. the equivalent conductance at infinite dilution for HCl solution is 425 ohm-1 cm2 eqv-1. the mobilities of NH4+ and ClO4– ions are 6.90 N (B) 1.0 % 2. respectively. . In a conductivity cell. what will be the resistance of the mixture using the same cell? Assume that there is no increase in the degree of dissociation of A and B on mixing. the two platinum electrodes.15 cm–1.039 % (C) 3. 50 cm2 in surface area and 0. What is the solubility of AgCl in water? (A) 9. All measurements were done at 298K.11.ELECTRO CHEMISTRY PATHRILA RASTA (DPP – VIII) 1. 4.32 x 10−5 g/L (C) 1.1 mho cm2 eq–1 at 25ºC. Find the solubility product of AgCl at 25ºC.7. 7. 500 gm of NaCl was then added to the water and the specific conductance after the addition of NaCl. 21 .10N-MgCl2 solution. is measured to be 6400 ohm. the values of symbol λº are 349. What is the equivalent conductance of the NaCl solution. How far would the K+ ion move in 2 hours at 25ºC? Ionic conductance of K+ ion at infinite dilution at 25ºC is 73.52 mho cm2 mol-1.9. Sri Chaitanya IIT-JEE Academy.10 × 10–5 mho cm–1.826 x 10-6 ohm-1 cm-1. what will be its degree of dissociation? Calculate the ionic product of water at 25ºC from the following data: specific conductance of water = 5.36 mho cm2 eq–1 at 25ºC. Jhalawar Road. Kota. 5. The specific conductance at 298K of AgCl solution in water was determined to be 1. when filled with 0. 6.76 x 10−10 g/L A cell whose resistance. λºH+ = 350.324005. the specific conductance of which was 2.0 and λºOH = 198. respectively.50 cm apart is filled with 0. (A) 12.003M-NaCl solution. The equivalent conductance of an infinitely dilute solution of NH4Cl is 150 and the ionic conductance of OH– and Cl– ions are 198 and 76. 9. How much current will flow when the potential difference between the electrodes is 5 volts? For H+ and Na+.8 and 50.6. Calculate Ka of acetic acid if its 0. A cell with electrodes that are 1.8 × 10–8 mho cm–1. 10-A. The specific conductance of a saturated solution of AgCl at 25ºC after substracting the specific conductance of water is 2.10N solution of MgCl2 is 97. 10.0 mho cm2. What will be the equivalent conductance of the solution of NH4OH at infinite dilution? If the equivalent conductance of a 0.28 × 10–6 mho cm–1. in ohm-1cm2eq-1? Equivalent conductance of 0. Λº CH3COOH = 390. Talwandi. ΛºAgCl = 138. when filled with 0.01N–solution of NH4OH is 9.05N – solution has equivalent conductance of 7.1M-KCl is 200 ohm. (B) 1.9 x10−3 g/L (D) 1. Find the capacity of the vessel if it is fulfilled with water. was found to be 3. A dilute solution of KCl was placed between two platinum electrodes 10cm apart.34 ohm-1 cm2 eq-1 respectively.56 ×10–5 mho cm–1.5 (B) 250 (C) 125 (D) 25 3. Calculate the mobilities of these ions and their velocities if they are in a cell in which the electrodes are 5 cm apart and to which a potential of 2 volt is applied. across which a potential of 6 volts was applied.1M-KCl is 120 ohm-1cm2eq-1. (Raj).3 mho cm2. 8. The equivalent conductance of 0. A big irregular shaped vessel contained water. of Ag+ and Cl¯ are 61. ΛºNaCl = 149. The ionic conductances at infinite dilution.2 x 10−8 g/L 2.92 and 76. Ca and Li metals may arranged in the decreasing order of their standard electrode potential as (A) K. From these values and the Nearnst equation. K. 10-A. Ag (B) Cu. The overall formation constant for the reaction of 6 mole of CN– with cobalt (II) is 1 × 1019. (Raj).23 × 1025 (C) 1. Pb.373V.324005. A student made the following observations in the laboratory: (I) Clean copper metal did not react with 1M-Pb(NO3)2 solution (II) Clean lead metal dissoles in 1M-AgNO3 solution and crystals of Ag metal appeared (III) Clean silver metal did not react with 1M-Cu(NO3)2 solution. Ca (C) Li.02 × 10-8 (B) 1. what should be Kf for the Ag(NH3)2+ ion? (A) 6. K.83 V Co3+ + e → Co2+.030 V (C) Increases by 0.059 V (B) Decreases by 0.37 × 10-20 3. respectively. HAVING ONLY ONE CORRECT OPTION By how much would the oxidising power of MnO4¯ /Mn2+ couple change if the H+ ions concentration is decreased 100 times? (A) increases by 189 mV (C) will increase by 19 mV (B) decreases by 189 mV (D) will decrease by 19 mV 2.059 V 9. Kota. Ca. Ag (D) Pb. Four colourless salt solutions are placed in separate test tubes and a strip of copper is placed in each. Ag. K (D) Ca. Cu. Pb (C) Pb.030 V (D) Decreases by 0. If hydrochloric acid present in a hydrogen electrode coupled with a saturated calomel electrode is titrated against ten times stronger solution of NaOH. Jhalawar Road. Which solution finally turns blue? (A) AgNO3 (B) Pb(NO3) 2 (C) Zn(NO3) 2 (D) Cd(NO3) 2 5. 22 .23 × 1063 (B) 8. What is the formation constant for the reaction of 6 moles of CN– with cobalt (III)? Given that Co(CN)63. The order of decreasing reducing character of the three metals is (A) Cu.66 × 107 (C) 7. Eº = + 1. Talwandi. the reduction electrode potential (A) Increases by 0.82 V (A) 8.7991V and +0.22 × 10 -26 (D) 1. Li (B) Li.3 × 1019 (D) 1.22 × 10-64 4. The solution of CuSO4 in which copper rod is immersed is diluted to 10 times.+ e → Co (CN)64-. Ca. Ag. K. Li Sri Chaitanya IIT-JEE Academy. Eº = – 0. The standard reduction potential for the reaction Ag+ + e → Ag and Ag(NH3)2+ + e → Ag + 2NH3 are + 0. the variance in electrode potential is (A) electrode potential changes slowly first and more and more rapidly as the end point approaches (B) electrode potential changes fast initially and very little change is observed as the end point approaches (C) electrode potential does not change at all during titration (D) none of the above 8. Electrode potential will be more for hydrogen electrode with PH (at the same temperature) (A) 4 (B) 3 (C) 2 (D) 5 7.ELECTRO CHEMISTRY CHEMISTRY ARENA (LEVEL # I) OBJECTIVE QUESTIONS 1. Cu 6. For a reaction A(s) + 2B+ → A2+ + B(s). changed to E2.453 V 17.5 × 10-11 13.8 × 10−5. Talwandi. when the concentrations of copper ions were found to be 0.1M) || (0.01M.8 × 10−5 and Kb for NH4OH = 1. when 1. Given EºAu+|Au = 1. F2 gas can not be obtained by the electrolysis of any aqueous fluoride salt because (A) F2 is the strongest oxidizing agent (C) F2 readily combines with the electrodes (B) F2 eaisly combines with water (D) All of these 19.534 V (D) 0.01M) NH4OH | H2 (1atm).f. Sri Chaitanya IIT-JEE Academy.m. When a lead storage battery is discharged (A) SO2 is evolved (C) lead is formed (B) lead sulphate is consumed (D) sulphuric acid is consumed 16.f. Pt. of a Daniel cell at 298K is E1. (A) 0.m. What is the equilibrium constant of the reaction: 2Fe2+ + Au+ ⇌ 2 Fe+ + Au3+. Copper can be deposited from acidified copper sulphate and alkaline cuprous cyanide. What is the relationship between E1 and E2? (A) E1 = E2 (B) E2 = 0 (C) E1 > E2 (D) E1 < E2 15. The e. Kota. If the same current is passed for a definite time (A) the amount of copper deposited from acidic copper sulphate will be higher (B) the amount of copper deposited from alkaline cuprous cyanide will be higher (C) the same amount of copper will be deposited (D) copper will not deposite in either case 20.458 V (C) -0.1M solution of copper sulphate was electrolysed using copper electrodes. Cl2 (P1 atm) | Cl | Cl2 (P2 atm). What volume of dry detonating gas at NTP is evolved from water.2 mV (C) 8. (A) 5 × 1021 (B) 2 × 10-22 (C) 3 × 10-18 (D) 1. EºFe3+|Fe2+= 0. A 0. What is the emf of the cell Pt. 10-A.089 V (D) +0.77 V.324005. Pt (A) P1 > P2 (B) P1 < P2 (C) P1 = P2 (D) P1 = 1 atm 12.m.06M respectively. when the concentration of ZnSO4 is 1. The back emf developed at 27ºC is (A) 20 mV 11. (Raj). Ka for CH3COOH = 1. (B) 5.8 mV - (D) 10 mV The cell reaction for the given cell is spontaneous if Pt. The same current is passed through acidulated water and stannous chloride solution. Jhalawar Road.458 V (B) -0.0M and that of CuSO4 is 0.12M and 0.50 V.f. the e. (1atm) | CH3COOH (0.ELECTRO CHEMISTRY 10.68 V. EºAu3+|Au = 1.89 V 14. The EMF of the cell is (A) 0.20 gm of tin is deposited from the other solution? (Sn = 120) (A) 448 ml (B) 336 ml (C) 224 ml (D) 672 ml When molten ICI3 is electrolysed using platinum electrodes (A) I2 is evolved at cathode and Cl2 at anode (B) Cl2 is evolved at cathode and I2 at anode (C) I2 is evolved at cathode and both I2 and Cl2 at anode (D) electrolysis does not take place 21.354 V (B) 0. Which of the following metals when coupled will give maximum e.708 V (C) 0. 23 . for a voltaic cell? (A) Fe and Cu (B) Pb and Au (C) Cu and Au (D) Ca and Cu 18. Kc has been found to be 1012. 5 min? Assume current efficiency 80%. If the cathode is pulled out of the solution. The amounts of Al. Electrolysis of an acetate solution produces ethane according to the reaction: 2CH3COO– → C2H6(g) + 2CO2 (g) + 2e What total volume of ethane and CO2 would be produced at STP if a current of 0. When 10-6M-HCl is electrolysed (A) O2 is produced at the anode (C) Cl2 is produced at the anode 24. C14H10. what current should be employed to achieve a production rate of 1 mole of S2O82− per hour? (A) 71.5 : 2 : 3 (C) 1 : 1.55 (C) Rs. Most of the copper used to make wire has been electrically refined by depositing it from copper salts solution (divalent) on to a cathode. Assuming 75% current efficiency. the positive ions will start moving randomly (D) positive and negative ions will start moving randomly 23.00033 gm/coul. Electrolysis of a solution of HSO4− ions produces S2O82−.5 amp is passed through the solution for 482. C14H8O2.5 amp (B) 35. When electric current is passed through a cell having an electrolyte.99 gm 26. can be oxidised anodically to anthraquinone.98 gm (B) 1. Anthracene. (A) Rs. 5. Jhalawar Road. the negative ions will stop moving (C) negative ions will continue to move towards the anode.354 V (C) 0.344 L (B) 2. Three faradays of electricity are passed through molten Al2O3.792 L 28. To perform an analysis of a mixture of metal ions by electrodeposition.324005.005 V 29. aqueous solutions of CuSO4 and molten NaCl solutions.7 amp (C) 53.0 amp (D) 44. (Raj).11 27.68 amp (C) 0. What was the average rate of current flow in ampere? (A) 0. What weight of anthraquinone can be produced by the passage of a current of 1 amp of 60 minutes if the current efficiency is 90 %? (A) 6.90 gm (D) 0.536 amp (D) 0. the second metal to be deposited must not being plating out until the concentration ratio of the second to the first is about 106.08M – Na2S2O3 solution. (B) H2 is produced at the anode (D) Cl2 and O2 are produecd at the anode A constant current flowed for 2 hours through a potassuim iodide solution.268 amp (B) 2. Cu and Na deposited at the cathodes will be in the ratio of moles indicated (A) 1 : 2 : 3 (B) 1.16 gm (C) 41.177 V (B) 0.ELECTRO CHEMISTRY 22.00 per KWh and the cell operates at 0. then the (A) positive and negative ions will move towards the anode (B) positive ions will start moving towards the anode. the iodine was titrated with 25 ml of 0. Kota. At the end of the experiment. oxidising the iodide ion to iodine. 11. 2.708 V (D) 0.3 amp Sri Chaitanya IIT-JEE Academy.134 amp 25.016 L (C) 4.5 : 3 (D) 1 : 3 : 2 30.33 volt? The electrochemical equivalent of copper is 0.11 (B) Rs. What must be the minimum difference in standard potential of the two metals which form dipositive ions in order for such an analysis to be feasible? (A) 0. positive ions move towards the cathode and negative ions towards the anode. Talwandi. 10-A. 1. What is the cost of electrical energy required per kg of copper if the cost of electricity is Rs 4. 24 .22 (D) Rs.032 L (D) 1. (A) 1. 60 (C) 0. 25 . 10-A. cobalt electrode acts as anode (C) Standard potential increases with increasing concentration of the electrolyte (D) Calomel electrode is a reference electrode having 0. (Raj). 33. 36. Kota.20 38. 31.97 (B) 4.30 mole of HNO2 are produced in the cell that contains 0.50 mole of Hg22+ and 0. Hg(l). AgCl(s) | KCl (0. (B) Decreases on decreasing concentration of Cl– ions. How long will it take to produce 0. COMPREHENSION . 37. Talwandi. 35.ELECTRO CHEMISTRY MULTIPLE CHOICE QUESTIONS MAY HAVE ONE OR MORE THAN ONE CORRECT OPTION.60 mole of Hg2+ and 0. In which dish.76 (C) Both (C) 2.5 sec Sri Chaitanya IIT-JEE Academy. Jhalawar Road.I is measured under standard conditions in the electrochemical cell shown in the accompanying diagram. The cell potential (A) Increases on increasing concentration of Cl– ions.40 How many moles of electrons pass through the circuit when 0. Which one of the following statement is/are incorrect regarding an electrochemical cell? (A) the electrode on which oxidation takes place is called anode (B) anode is the negative pole (C) the direction of the current is same as that of the direction of flow of electrons (D) the flow of current is partly due to flow of electrons and partly due to flow of ions Which of the following statement(s) differentiate between electrochemical cell and electrolytic cell? (A) spontaneous or non-spontaneous nature of the chemical process (B) chemical reactions occuring at the electrodes (C) positive and negative nature of anode (D) dependence on Faraday’s law Pick up the false statement(s): (A) Galvanic cell reactions are always redox reactions (B) In a galvanic cell made of cobalt and cadmium electrodes.18 (D) None (D) 1. the solution is acidic? (A) Dish A (B) Dish B What is the equilibrium constant for the reaction? (A) 1.30 (B) 0.00 volt potential Consider the cell: Ag (s). 34.40 mole of NO3 at the beginning of the reaction? (A) 0.15 (D) 1.324005. (C) is independent of concentration of Cl– ions (D) is independent of amounts of AgCl and Hg2Cl2 COMPREHENSION BASED QUESTIONS HAVING ONLY ONE CORRECT OPTION The cell potential for the unbalanced chemical reaction: Hg22+ + NO3 + H3O+ → Hg2+ + HNO2 + H2O 32.1M) | Hg2Cl2 (s).10 mole of HNO2 by this reaction if a current of 10 A passes through the cell? (A) 965 sec (B) 96500 sec (C) 1930 sec (D) 482. 46 V (D) 0. Talwandi.76 V (D) -0.615 V 40.8 kJ mol–1 and ∆G° = –237.34 V 47.76 V (C) +1. Cu – H2 Cell (A) -0. The cell voltage will be (A) reduced by half (C) increased by a factor of 4 (B) increased by a factor of 2 (D) unchanged 43.53 % (B) 100 % (D) 97.10 V (C) 0. are needed to produce 23.739 KJ of useful work under ideal conditions? (A) 2. EºCu2+|Cu = + 0.314 × 298 × ln 286 J How many litres of gaseous H2. Kota. Such conversions are possible because the combustion reactions are essentially redox reactions and highly exothermic as well as highly exergonic. when combined with excess O2 in the fuel cell at 25°C and 1.0 V (D) Indeterminate (B) +1.88L (C) 1. (B) 1.34 volt) 45. Suppose the concentration of hydroxide ion in the cell is doubled. the following reaction take place: Anode reaction: 2H2 (g) + 4OH–(aq) → 4H2O(l) + 4e– Cathode reaction: O2(g) + 2H2O(l) + 4e– → 4OH–(aq) Overall reaction: 2H2(g) + O2(g) → 2H2O(l) The overall all reaction has a value of ∆H° = –285.00 atm.24L 42. Zn – H2 cell (A) +0.H. (Raj).34 V 46.324005. Jhalawar Road.10 V (B) -1. What would this do to the observed voltage under standard condition for each of the following: (EºZn2+|Zn = – 0.39 kJ mol–1 at 25°C.ELECTRO CHEMISTRY COMPREHENSION . where the fuel is burnt directly. What is the approximate value of ∆S° for the fuel cell reaction at 25°C? (A) –0.23 J 41. Electrical energy can be obtained indefinitely from a fuel cell as along as the outside supply of fuel is maintained.II A fuel cell is the device to convert the energy of a fuel into electrical energy without the use of heat engine. Zn – Cu Cell (A) +1.164 JK–1 44. was arbitrarily assigned a value of 1.164 JK–1 (B) –164 JK–1 (C) 164 JK–1 (D) 0.23 J (D) 8. 26 . What is the standard EMF of the cell? (A) 0.74 V Which of the following expressions gives ∆G° for the reaction in the fuel cell? (A) 4 × 96500 × 1.34 V (B) -0.00 volt for 2H+ + 2e → H2(g). The theoretical efficiency of the fuel cell is given by (A) 83.III Suppose that the S.76 V (C) -1.34 V Sri Chaitanya IIT-JEE Academy. In hydrogen – oxygen fuel cell.06 % (C) 67.88 % COMPREHENSION .23 J (C) –8. (B) – 4 × 96500 × 1.314 × 298 ln 1.76 volt. 10-A.22L (D) 2.23 V (C) 2. 39.44L (B) 4.E.76 V (D) +0. 6 mm (C) 6.IV A current of 15. (Raj).324005. Kota. The density of nickel = 8. (B) 32.0 A is employed to plate nickel in a NiSO4 solution.85 gm What is the thickness of the plating if the cathode consists of a sheet of metal 4.0 cm2 which is to be coated on both sides? (A) 13.76 L (C) 2.0 mm 50.27 L (B) 3. Talwandi. The current efficiency with respect to formation of Ni is 60%. 27 .43 gm 49. Both Ni and H2 are formed at the cathode. Jhalawar Road. (Ni = 58. What volume of H2 at STP is formed per hour? (A) 6.9 gm/ml. How much of nickel is plated on the cathode per hour? (A) 16.85 gm (C) 19.01 L Sri Chaitanya IIT-JEE Academy.9 mm (D) 23.7 gm (D) 9. 10-A.5 L (D) 5.7) 48.ELECTRO CHEMISTRY COMPREHENSION .8 mm (B) 27. 29 × 10–7 at 25ºC and E 0 Cl − 2. 10. The concentration of CuSO4 in the cell with higher emf value is 0.1M-H2S. Eº = 1.ELECTRO CHEMISTRY AGNI PARICHHA (LEVEL # II) 1.5 × 10–7 t2 where t is in a degree celcius. Eº = –1. Eº = 0.001158V. Calculate ∆Gº. Carbon. The solubility product of CuCl is 2. ∆Hf H2O(ℓ) = −56. Kota. the anode is made up of Zn and cathode of carbon rod surrounded by a mixture of MnO2. 8. Calculate the concentration of hydrogen ions at the positive electrode. Two students use same stock solution of ZnSO4 and a solution of CuSO4. (Raj). 10-A.35 V.229V for the O2 | H2O. As soon as OH– ions are produced. If electrolysis was made for 115 sec using 15. 6.01M) | O2(0. Lactic acid.05 kcal/mol. Talwandi. Ksp of Ag2S = 2 × 10–49 and E0Ag+|Ag = 0. what was the percentage of lactic acid in muscle tissue. The cathodic reaction is 2MnO2(s) + Zn2+ + 2e → Zn Mn2O4(s) If 8 gm MnO2 is present is cathodic compartment.5M.324005.74 kcal/mol and energy of ionisation of H2O(ℓ) = 19. Calculate Eº for the half reaction: Ag(NH3)2+ + e → Ag + 2NH3. Determine potential of the cell Pt | Fe2+. Given.185V for the Mn2+ | Mn couple and 1.62 × 10–4 t + 8. 3. The temperature dependence of the standard state cell potential for the lead storage cell is Eº (volt) = 2. HCl (0.001M).129V. 11. The emf of the cell is 0. 9. Dissociation constant for Ag(NH3)2+ into Ag+ and NH3 is 6 × 10–14.75M.03 V higher than the other. they react with lactic acid and at complete neutralisation. 5. [Cr2O72-] = 2M. 4. | CuCl | Cu is 0.1 × 10–13. Ka1 = 10–8 and Ka2 = 1. Jhalawar Road. HC3H5O3. 25ºC) and are interconnected through a salt bridge.118 volt at 25ºC. In a Zn–MnO2 Cell. Find out the concentration of CuSO4 in the other cell (2.80V. [Fe3+] = 0. For H2S. produced in 1 gm sample of muscle tissue was titrated using phenolphthalein as indicator against OH− ions which were obtained by the electrolysis of water. Fe3+ || Cr2O72-. ∆Sº and ∆Hº at 25ºC.5M. [Cr3+] = 4M and [H+] = 1M. What will be the reversible emf at 25ºC of the cell Pt | H2(g) (1atm) | H+ || OH¯ | O2(g) (1atm) | Pt if at 26ºC the emf increasing by 0.77 V 14 H+ + 6e + Cr2O72− → 2Cr3+ + 7H2O. Calculate E 0 Cu + |Cu . 7. Find emf of cell.303 RT/F = 0. how many day the dry cell will continue to give a current of 4 × 10–3 amp? 12. Calculate the potential of the cell: Cu | Mn(s) | MnCl2(0.799V. Two weak acid solutions HA1 and HA2 each with the same concentration and having pKa values 3 and 5 are placed in contact with hydrogen electrode (1 atm. At 250C.1191 + 1.06) A cell contains two hydrogen electrodes. immediately a pink colour is noticed.25 bar) | Pt | Cu. Cr3+.6 mA current to reach the end point. Given Ag+ + e → Ag has Eº = 0. Calculate electrode potential of Ag2S + 2e → 2Ag + S2– in a solution buffered at PH = 3 and which is also saturated with 0. H+ couple. NH4Cl and ZnCl2 in aqueous base. The negative electrode is in contact with a solution of 10–6M hydrogen ions. Given: Fe3+ + e → Fe2+. 28 . Sri Chaitanya IIT-JEE Academy. H+ | Pt in which [Fe2+] = 0. The emf of one cell is 0. how many gm of arsenic are present in the solution. 29 .ELECTRO CHEMISTRY 13. What should be the minimum mass of lead in the free and combined states in a battery designed to deliver 50 amp-hr? What is the standard free energy change for the reaction? Average voltage of lead storage battery is 2. the free iodine combines with the starch to give a deep blue colour.3% Ni and 22. Jhalawar Road. Sri Chaitanya IIT-JEE Academy. HAsO42–. The electrolysis of a solution containing Cu(I). A battery available for the electrode reaction involves the use of 25% of the Pb and PbO2 available in it for reaction. Assume that no other element was released and calculate the number of coulomb passed through the solution.324005. In an analytical determination of arsenic.9% Zn.8% by weight of Cu. If during a particular run. The deposit contained 72.3 sec for a current of 10. 10-A. 4. nickel and zinc as complex cyanides produced a deposit weighing 0.0V. H3AsO4. (As = 75) 14.175 gm. a solution containing arsenious acid. 15. it takes 65. (Raj). KI and a small amount of starch is electrolysed. Talwandi.5 mA to give an end point (indicated by the blue colour). Kota. I2 + H3AsO3 + H2O → 2I– + HAsO42– + 4H+ When the oxidation of arsenic is complete. The electrolysis produces free I2 from I– ions and the I2 immediately oxidises the arseneous acid to hydrogen arsenate ion. ELECTRO CHEMISTRY ENJOY THE VICTORY (LEVEL # III) 1. 3 Marks] calculate the standard reduction potential of the I– / AgI / Ag electrode. Kota. Also [1994. 2. Explain.3 × 10–13 [1992. How long will it take to produce 1 kg of Cl2? What will be the molarity of the solution with respect to hydroxide ion? (Assume no loss due to evaporation) [1992. Zinc granules are added in excess to a 500 ml of 1. The Edison storage cells is represented as: Fe(s) | FeO(s) | KOH(aq) | Ni2O3(s) | Ni(s) [1994. (Raj). E° = + 0.75V and –0. 30 .70 ampere is passed through 300. Calculate (i) (ii) how many grams of chromium will be plated out by 24. [1991.000 coulombs and how long will it take to plate out 1. [1991.8 × 10–10.7 × 10–17.24 V respectively. 10-A. The dependence of electrode potential for the electrode Mn+/M with concentration under STP conditions is given by the expression: E = E° + 6. Find out the molarity of Zn2+ after the deposition of Zn. find out the concentration of Ni2+ in solution at equilibrium. 0. Chromium metal can be plated out from an acidic solution containing CrO3 according to the following equation: CrO3(aq) + 6H+(aq) + 6e– → Cr(s) + 3H2O. Although aluminium is above hydrogen in the electrochemical series. Jhalawar Road. it is stable in air and water. KCl(0. calculate the EMF generated and assign correct polarity to each electrode for a spontaneous process after taking into account the cell reaction at 25°C.5A current? [1993. 1 Mark] The standard reduction potential for half-cell: NO3−(aq) + 2H+(aq) + e → NO2(g) + H2O(l) is 0.160 M solution of a ZnSO4 for 230 sec.? How does it depend on the concentration of KOH? (iii) What is the maximum amount of electrical energy that can be obtained from one mole of Ni2O3? 9. 4 Marks] The half-cell reactions are: Ni2O3(s) + H2O(l) + 2e– ⇌ 2NiO(s) + 2OH–.f.78V.2M)|| KBr (0.0M nickel nitrate solution at 25°C until the equilibrium is reached.324005.m. 3. evaluate the potential of the Ag+ / Ag electrode in a saturated solution of AgI. E° = – 0. 4. Write down the reactions taking place at the anode and the cathode. Ksp(AgCl) = 2.799V. If the standard reduction potential of Zn2+|Zn and Ni2+|Ni are –0. 2 Marks] 8.001M).0591 log10[Mn+] (True/false) n [1993.5g of chromium by using 12. Ag | AgCl(s). 3 Marks] 5. [1994. 10. 4 Marks] 7. Ksp = 8. 1 Mark] Sri Chaitanya IIT-JEE Academy. with a current efficiency of 90%. Talwandi. Given that for AgI. [1993. Cl2(g) and NaOH according to the reaction: 2Cl–(aq) + 2H2O → 2OH–(aq) + H2(g) + Cl2(g) A direct current of 25 amperes with a current efficiency of 62% is passed through 20 litres of NaCl solution (20% by weight). 2 Marks] A current of 1. (i) (ii) Calculate the reduction potential in 8M-H+ What will be the reduction potential of the half-cell in a neutral solution? Assume all the other species to be at unit concentration. Ksp(AgBr) = 3. The standard reduction potential of the Ag+/Ag electrode at 298K is 0. 4 Marks] For the galvanic cell. AgBr(s)| Ag. 4 Marks] An aqueous solution of NaCl on electrolysis gives H2(g).40V FeO(s) + H2O(l) + 2e– ⇌ Fe(s) + 2OH–. Assume the volume of the solution to retain constant during the electrolysis.0 ml of 0.87V (i) What is the cell reaction? (ii) What is the cell e. 0 × 10–3 M Fe3+.153 V respectively. assuming that the only reaction that + 2+ occurs is 2Hg + 2 Fe3+ → Hg 2 2 + 2Fe .337 V and 0. Calculate the reduction potential at pH = 14 for the above couple. 6 Marks] A gas X at 1 atm is bubbled through a solution containing a mixture of 1M Y– and 1M Z– at 25°C.0 × 10–19.77 V) [1995. [1997. Talwandi.54V respectively for Fe3+ | Fe2+ and I 3 | I– couples. –3.03 and –1.00254 cm? Density of silver is 10. 2Fe3+ + 3I– ⇌ 2Fe2+ + I 3 . X. Y and Z are 0. 3 Marks] 21. 10-A.)||Ag + (0.68V) [1997. 1 Mark] standard electrode potential of Cu+|Cu half cell is (A) 0. (Raj). The [1997. Na 2− (D) S2O 8 . (Given E Fe 3 + |Fe 2 + = 0. 3 Marks] Calculate the equilibrium constant for the reaction: Fe2+ + Ce4+ ⇌ Fe3+ + Ce3+. 1 Mark] 15. Ce 4 + /Ce 3 + = = 0.324005. 14. What is the value of current efficiency? Write the reactions taking place at the cathode and at the anode. The standard reduction potential values of three metallic cations. Electrolysis of a solution of MnSO4 in aqueous sulphuric acid is a method for the preparation of MnO2 as per the reaction: Mn2+ (aq) + 2H2O → MnO2 (s) + 2H+ (aq) + H2 (g). The standard reduction − potentials in acidic conditions are 0. 3 Marks] A standard hydrogen electrode has zero electrode potential because (A) hydrogen is easiest to oxidize (B) this electrode potential is assumed to be zero (C) hydrogen atom has only one electron (D) hydrogen is the lightest element [1997(C). [1998. Sri Chaitanya IIT-JEE Academy. Passing a current of 27 ampere for 2 hours gives only 1 kg of MnO2. The order of reducing power of the corresponding metals is [1998.521V (D) 0.184V (B) 0.52.490V 16. [1996. 2 Marks] (A) Y will oxidize X and not Z (C) Y will oxidize both X and Z (B) Y will oxidize Z and not X (D) Y will reduce both X and Z 22.5 g/cm3. 17. An excess of liquid mercury is added to an acidified solution of 1. Jhalawar Road. The standard reduction potential for Cu2+ | Cu is +0. 1 Mark] (A) O2.18 V respectively. 5 Marks] The standard reduction potentials of Cu2+|Cu and Cu2+|Cu+ are 0. Ag2CrO4 soln. [1998. [1997(C). A dilute aqueous solution of Na2SO4 is electrolyzed using platinum electrodes.1M) | Ag is 0.44V. Ksp of Cu(OH)2 is 1. then. 2 Marks] 19.77V and 0.164V at 298K.ELECTRO CHEMISTRY 11. [1999. 2 Marks] (A) Y > Z > X (B) X > Y > Z (C) Z > Y > X (D) Z > X > Y 20. Na (C) O2. 4 Marks] 12. − Calculate the equilibrium constant for the reaction. H2 2− (B) S2O 8 . E Fe /Fe 3+ 2+ 18. If the reduction potential of Z > Y > X. Find the solubility product of a saturated solution of Ag2CrO4 in water at 298K if the emf of the cell Ag|Ag+ (satd. Kota.0 hours at a current of 8.827V (C) 0. How many grams of silver could be plated out on a serving tray by electrolysis of a solution containing silver in +1 oxidation state for a period of 8. Calculate E Hg 2 2 |Hg . (Given E 1.34 V. H2 13.46 amperes? What is the area of the tray if the thickness of the silver plating is 0. 31 . It is found that 5% + of Fe3+ remains at equilibrium at 25°C. The products at the anode and cathode are [1996. ELECTRO CHEMISTRY 23.) + 8H (aq. Calculate the concentration of copper sulphate in the solution to begin with. 2 Marks] Saturated solution of KNO3 is used to make ‘salt-bridge’ because − (A) velocity of K+ is greater than that of NO 3 − (B) velocity of NO 3 is greater than that of K+ − (C) velocities of both K+ and NO 3 are nearly the same 26. A cell. [2001(S). (B) KCl > NaCl > LiCl (D) LiCl > KCl > NaCl [2001(S). Kota. 3 Marks] The standard potential of the following cell is 0. Standard electrode potential data are useful for understanding the suitability of an oxidant in a redox titration. 3 Marks] The following electrochemical cell has been set up: Pt(l) | Fe3+. 3 Marks] (A) M + X → M+ + X– is the spontaneous reaction (B) M+ + X– → M + X is the spontaneous reaction (C) Ecell = 0. 3 Marks] + – 2+ MnO − E° = 1.) E° = 1. Ce3+ (a = 1) | Pt(II) E° (Fe3+/Fe2+) = 0. (i) (ii) 30.80V. 3 Marks] (D) KNO3 is highly soluble in water 28. Pt | H2 (g) | HCl (aq) | AgCl(s) | Ag (s) Write the cell reaction Calculate ∆H° and ∆S° for the cell reaction by assuming that these quantities remain unchanged in the range 15°C to 35°C.61V. [1999.51 V 4 (aq. Ag | Ag+ || Cu2+ | Cu. A constant current of 2 mA was passed for 16 minutes. 10-A. Some half cell reactions and their standard potentials are given below: [2002(S). Copper sulphate solution (250 mL) was electrolysed using a platinum anode and a copper cathode.) + 4H2O(l) 2− Cr2O 7 (aq.) + 14H+ (aq. From this data one can deduce that [2000(S). Calculate the change in the cell potential after the passage of 9. M | M+ || X– | X. Fe2+ (a = 1) | Ce4+. (iii) Calculate the solubility of AgCl in water at 25°C.21V at 35°C. 6 Marks] For the electrochemical cell. (Raj). (A) MnO − 4 can be used in aqueous HCl (C) MnO − 4 can be used in aqueous H2SO4 2− (B) Cr2O 7 can be used in aqueous HCl 2− (D) Cr2O 7 can be used in aqueous H2SO4 Sri Chaitanya IIT-JEE Academy.65 ampere of current of 1 hour. NaCl and KCl is (A) LiCl > NaCl > KCl (C) NaCl > KCl > LiCl 29.77V. [2001M. 10 Marks] The standard reduction potential of Ag+/Ag couple is 0.324005.23V at 15°C and 0.38 V Fe3+ (aq. [2000M. initially contains 1 M Ag+ and 1 M Cu2+ ions.) + 5e → Mn (aq. 24.77 V – – Cl2 (g) + 2e → 2Cl (aq. Talwandi.) + e– → Fe2+ (aq.) + 6e– → 2Cr3+ (aq. predict the direction of flow of current.40 V Identify the only incorrect statement regarding the quantitative estimation of aqueous Fe(NO3)2.) + 7H2O(l) E° = 1.33V. Jhalawar Road. If an ammeter is connected between the two platinum electrodes. The correct order of equivalent conductance at infinite dilution of LiCl. It was found that after electrolysis the absorbance of the solution was reduced to 50% of its original value. Will the current increase or decrease with time? [2000M.77V (D) Ecell = – 0. 32 .77V 25. E0(M+/M) = 0. 27. E° (Ce4+/Ce3+) = 1. at 25°C.) E° = 0.44V and E0(X/X–) = 0. 65 from E0red (D) Ered will decrease by a factor of 0. Talwandi. (Raj).ELECTRO CHEMISTRY 31. Find out the conc. E° = – 0.324005. E° = 0. find conductivity (specific conductance) of this solution in terms of 10-7 S.06) [2003M.03 V higher than the other. Which of the following must be incorrect? (A) NH3 combines with Ag+ to form a complex (B) Ag(NH3)2+ is a stroger oxidising agent than Ag+ (C) In absence of ammonia. Ag+ + e → Ag. (B) 10 0.13 (B) 58.15V.01M) || Fe2+ (0.26 0. [2004M. Find lnK of this reaction. 3 Marks] (B) Cathode to anode through external supply (D) Anode to cathode through internal supply 32.80 V Zn2+ + 2e– → Zn.40V. Sri Chaitanya IIT-JEE Academy. Two students use same stock solution of ZnSO4 and a solution of CuSO4. E0red = 0. E 0 In 3+ /In + = – 0. In the electrolytic cell.32 0. Which half cell reaction is affected by pH and by how much? (A) Eoxd will increase by a factor of 0.m-1 units.295V then the value of equilibrium constant for the cell reaction is [2004(S).65 from E0oxd (B) Eoxd will decrease by a factor of 0.42V.65 from E0oxd (C) Ered will increase by a factor of 0. Ksp of AgBr is 12 × 10-14. The emf of one cell is 0. pH is raised to 11. If 10 moles of AgNO3 are added to 1 liter of this solution. The conc. Also find the solubility product of AgCl. λ0(Br−) = 8 × 10-3 Sm2mol-1. 2 Marks] The emf of the cell: Zn | Zn2+ (0.83 (C) 28.29 37. ∆G f (kJ/mol) +77 –129 –109 (b) If 6. 33 .0591 34.8V C6H12O6 + H2O → C6H12O7 (Gluconic acid) + 2H+ + 2e. 3 Marks] (A) e 0.539×10–2g of metallic zinc is added to 100 ml saturated solution of AgCl.5 M.76 V 36. E 0 In 2 + /In + = – 0. [2006. (A) 66. of CuSO4 in the other cell (2. 6 Marks] Tollen’s reagent is used for the detection of aldehyde when a solution of AgNO3 is added to glucose with NH4OH then gluconic acid is formed. 10-A. Jhalawar Road.32 0. flow of electrons is from (A) Cathode to anode in solution (C) Cathode to anode through internal supply [2003(S).303 RT/F = 0.65 from E0red Ammonia is always added in this reaction.0295 (C) 10 0.0295 (D) 10 0. λ0(NO3−) = 7 × 10-3 Sm2mol-1.0295 33. 3 x 5 = 15 Marks] 2Ag+ + C6H12O6 + H2O → 2Ag(s) + C6H12O7 + 2H+. -7 [2005M. Given λ0(Ag+) = 6 × 10-3 Sm2mol-1. Given: E Cu 2 + /Cu + = 0. 38.05V Ag(NH3)2+ + e → Ag(s) + 2 NH3. silver salt of gluconic acid is formed (D) NH3 has affected the standard reduction potential of glucose-gluconic acid electrode 39.337V [Use 2. Kota.303 RT/F = 0. E0red = 0.92 at 298K] [2006. Given: Species Ag (aq) Cl– (aq) AgCl (s) + 35. E0oxd = -0. Find the equilibrium constant for the reaction: In2+ + Cu2+ ⇌ In3+ + Cu+ at 298K. Given that [Ag + ]2 Ag+ + e– → Ag.0592 and F/RT = 38.How many moles of Ag will be precipitated in the above reaction.30 (D) 46. of CuSO4 in the cell with higher emf value is 0. Write the cell representation of above reaction and calculate E cell at 298K.001M) | Fe at 298K is 0. When ammonia is added to the solution. 6 Marks] We have taken a saturated solution of AgBr. 4 Marks] (a) For the reaction: Ag+ (aq) + Cl– (aq) ⇌ AgCl(s).32 0. Find the value of log10 [Zn 2 + ] . A 5.44 x 10-8 0. A 17.324005. A 7. 9. D 14. C 4. B Sri Chaitanya IIT-JEE Academy. 5.AgCl|Cl ||Fe3+|Fe2+.72 KJ. 6. 6. 13. B 14. 965 sec (a) +3 (b) -3 85. 15.137 V 15. A 10. B 2. B 18. B 4. 10-A. 14. 9.A PATHRILA RASTA (DPP – IV) 1. PbSO4|SO4-||Pb2+|Pb.697V GALVANIC CELL HASEEN SHURUAT (DPP – III) 1.3759 V 10. 1.952 V (d) Ag.71 V -10 1.8 x 10-4 5.35 amp CONDUCTANCE HASEEN SHURUAT (DPP – VII) 1. A 3.D 13. C 12.521 V x = 1. 13. B 19. D 4. C C D 2. A 6.233 V - (c) Ag. C C 3.8 J/K ELECTROLYSIS HASEEN SHURUAT (DPP – V) 1.458 V 7. D 11. 1. 0.2 x 10-5 4.1 % M(NO3)3 2. A 7. ∆Hº = −162. 2.10 V + - (b) Pb. C 2. Cathode. 230 12. E0 = 0.14 V. Kota. A 6.332 KJ.005. E0 = 0. 6. 99. 2.96% 8. 4. 12. C 3. 8. D 9.02. B A 2. 15. 20. 1.356 x 10-5 3. A A 3.295 4. E0 = 0.00 hr 2.188 x 105 amp/day Yes 11.264 gm. D 6. C 4. Talwandi. A 5. 9. A 11.55 V 1. 8. 7. 13. 8.95 1. 15. AgI|I ||Ag+|Ag.ELECTRO CHEMISTRY ANSWERS ELECTRODE POTENTIAL HASEEN SHURUAT (DPP – I) 1.154V 5. 43. D 10. D 12. 34 . A 5. y =2 14. 6. 0.63 amp 315.456 gm Anode. 15. B 7. 9. A 7. B 0. 0. B C C 2. 2. 9. B D D PATHRILA RASTA (DPP – VI) 1. 662 sec 3. 1. Jhalawar Road. 8. A 0. B 0. (Raj).34V ∆Gº = 130. B 10. 10.728V 4. (a) (Pt)H2|H ||Br |AgBr|Br2. A PATHRILA RASTA (DPP – II) 1. 1.1 hr 7. 6. 5. 0. C 16. E0 = 0. ∆Sº = −562. A 0.0 x 10-14 3. 8.695V 10. 0. D 5. B 6.7 x10 11. –0. 31.45 × 105 J 11. 0. 45. 27. C B B D CD B A 6. 10-A. 41. 6.137 V 4. 272. 48.66 x 10-4 gm 30. 0.0 x 10-14 CHEMISTRY ARENA (LEVEL # I) OBJECTIVE QUESTIONS 1. 49. C 10. 16.452 V 10. Kota.165V 14. 0. 2. 43. 0. 2. B 15. 5. 5. B D A A A B A 3. 18. B 10 –10 –6 34. 1.27 V (iii) 2. (ii) 1. (i) 2.26 × 10 21. 1. 1.44 × 10–12 22. 0. B 13. 8.45 x 10 cm s . 6. 42. A D A A B D A AGNI PARICHHA (LEVEL # II) 1.02 × 10 cm 18.154 M 3. -386 KJ ENJOY THE VICTORY (LEVEL # III) 1. 47.324005. C 2. 44.08 x 10-4 cm s-1 6. 2. 48. B 29.69 hr. 11.059V 25. 34.0 KJ 10-4M 9.ELECTRO CHEMISTRY 8.1673% 15.22 V 14. Na = 2. (ii) –22195 J mole. 13. 15. (i) 0. (a) 0. Talwandi.56 V 7. . 22.95 × 10 mol L 26. A B A B AC B B 5. 0. A 20.0. 46.0189V 3.52V 4.m-1 37. 23. -409. C 17.675 day 13. –0. 29.128 × 10–18 mol L–1 2. 10 35.8 KJ.1554 g (ii) 1336. 8. C A B C CD A D 7.833 V (ii) 0. C 32.59 V. 2. 19.544 Kg. Ce electrode to iron electrode. 17. –0. 0.010 V 24. A PATHRILA RASTA (DPP – VIII) 1. 9. 5. (Raj). 229 coul 0. 8. –49987 J mole–1 (iii) 1. 10 36. 39.325 V. D.5 J/K. C 3. 12. 21. 55 S. 2. A 39. False 6.9. A 31. decrease. 28. 94. 30. 27. C 28. B A D C C B B C 2. 1. 35.8% 4 2 12 16. 10 (b) 52. 36.76 x 10 1. 25. B 38. 0.367 V 7.408 M 5. B –5 –1 25.127 V 2. 50. 3. 0. 38.15 sec 8.29 cm 10.0353 7. 10.. -398.72 × 10 1.792 V 12 .37 x 105 litre 4. 14. A 23. 26.1456 amp + -3 -1 + H = 1. 32. 7.05M 11. 37. A C D A C C A HAVING ONLY ONE CORRECT OPTION 4. A 9. 40.05 M 33. 0. 12. 7. 20. 0.18 g. 39. Jhalawar Road. 24. 35 .6 × 10 7 19. -5 -10 9. Sri Chaitanya IIT-JEE Academy.149 V 9.476 × 10–5 mole / L 30. 33.