University of the Philippines Chemical Engineering Society, Inc.REVIEW MODULE – Chemistry 17, 3rd Exam Coverage Electrochemistry study of chemical reactions/changes accompanied by the production of current and the consumption of electrical energy electrochemical reactions are reductionoxidation reactions that involve the transfer of electrons Parts of an electrochemical cell: electrodes – where current enters or exits o inert electrodes – does not participate in electrochemical reaction anode – where oxidation (loss of electrons) occur cathode – where reduction (gain of electrons) occur Helpful mnemonics: AN OX ate a RED CAT (anode – oxidation, reduction – cathode) OILRIG – oxidation is loss, reduction is gain 2 kinds of electrochemical cells: Electrolytic cells – non-spontaneous electrochemical reactions (electrolysis), external energy source (direct current) needed o negative anode, positive cathode Voltaic or galvanic cells – spontaneous electrochemical reactions produce electric energy (e.g. batteries) o positive cathode, negative anode 2 kinds of electric conduction: metallic conduction – electron flow with no atomic motion ionic/electrolytic conduction – electron flow accompanied by ionic motion o cations negative electrode o anions positive electrode Electrolytic cell: Example: Molten NaCl (Down Cell) Overall reaction o Cathode(-) half – reaction (reduction) o Anode (+) half – reaction (oxidation) o Direction of electrons – from anode (+) to cathode (-) Example: Electrolysis of Aqueous IA salts (NaCl, KI, etc.) Overall reaction o Cathode(-) half – reaction (reduction) o Anode (+) half – reaction (oxidation) o Direction of electrons – from anode (+) to cathode (-) Exercise: What would be the anode half – reaction for the electrolysis of KI? University of the Philippines Chemical Engineering Society, Inc. REVIEW MODULE – Chemistry 17, 3rd Exam Coverage Example: Electrolysis of Aqueous Sodium Sulfate (Electrolysis of Water) Overall reaction o passge of 3.20 A of current through a solution of Pd(II) sulfate for 30.0 minutes. Solution: Electrochemical reaction: Electrical charge passed through the cell = (3.20 A)(30.0 min)(60 s/1 min)=5760 C moles of electrons = (5760 C)(1 mol e/96,485 C) = 5.97 X 10-2 mol moles of Pd metal formed = (5.97x10-2 mol e-) x (1mol Pd/2 mol e-) = 2.98x10-2 mol grams of Pd = 2.98x10-2 x 106.4 g/mol = 3.17 g o Cathode(-) half – reaction (reduction) o 2 ( )− Anode (+) half – reaction (oxidation) o Direction of electrons – from anode (+) to cathode (-) and ions are just supporting electrolytes Other sample problems: 1. An electrolytic cell contains 50.0 ml of a 0.152M FeCl3 solution. A current of 0.620 is passed through the cell, causing deposition of Fe(s) at the cathode.What is the concentration of the Fe+3 in the cell after this current has run for 20.0 min? 2. A current is passed through a 500.0ml solution of CaI2. The following electrode reactions occur: Anode: Cathode: After some time, analysis of the solution shows that 37.5 mmol of I2 has been formed. (a) How many faradays of charge have passed through the solution? (b) How many coulombs? (c) What volume of dry H2 at STP has been formed? d) What is the pH of the solution? Other notes about electrolytic cells: Molten salt or electrolyte solution serves as current-carrying medium Electrochemical reactions can be predicted by comparing reduction potentials Electrode may participate in electrochemical reactions Coulometry Faraday’s Law of Electrolysis: During electrolysis, one Farady of electricity reduces and oxidizes, respectively, one equivalent of the oxidizing agent and the reducing agent. 1 faraday of electricity = 6.022 X 1023 electrons 1 Ampere (A) = 1 Coulomb per second 1 Faraday = 96, 485 C ≈ 96,500 C Voltaic/Galvanic Cells Sample Problem: Calculate the mass of Pd produced by the reduction of Pd(II) ions during the University of the Philippines Chemical Engineering Society, Inc. REVIEW MODULE – Chemistry 17, 3rd Exam Coverage Notation example: Zn |Zn2+(1M) || Cu2+(1M) | Cu half cell 1 – bridge – half cell 2 Example: Zinc – Copper Cell (Daniel Cell) Zn |Zn2+(1M) || Cu2+(1M) | Cu Overall reaction o Cathode(+) half – reaction (reduction) o Anode (-) half – reaction (oxidation) o Direction of electrons – from anode (-) to cathode (+) 5% agar of KCl is usually used for the salt bridge (Li, K, Ca, Na) > Mg, Al, Mn, Zn, Cr, Fe, Cd, Co, Ni, Sn, Pb, H, Sb, Cu, Hg, Ag, Pt, Au Standard electrode potentials (Eo) : measurement of spontaneity of the redox reaction higher (more positive) Eo = greater driving force for the reaction as written, the greater tendency for it to occur Examples: 1.100 V for standard Zn – Cu cell and 0.462 V for standard Cu – Ag cell Standard Hydrogen Electrode (SHE) – arbitrary standard with Eo of 0.0000 V o o Anode: Cathode: Example: Copper - Silver Cell Cu |Cu2+(1M) || Ag+(1M) | Ag Overall reaction o Cathode(+) half – reaction (reduction) o Anode (-) half – reaction (oxidation) o Direction of electrons – from anode (-) to cathode (+) 5% agar of KCl is usually used for the salt bridge Sample Problem: Will permanganate ions, MnO4oxidize Fe (II) ions to Fe(III) ions, or will Fe(III) ions oxidize Mn(II) ions to MnO4- in acidic solution? Given: MnO4- Mn2+, Eo = 1.507 V Fe3+ Fe2+, Eo = 0.771 V Solution steps: Choose the appropriate half-reactions from a table of standard reduction potentials. Write the equation for the half-reaction with the more positive E0 value first, along with its E0 value. Write the equation for the other halfreaction as an oxidation with its oxidation potential, i.e. reverse the tabulated reduction half-reaction and change the sign of the tabulated E0. Balance the electron transfer. (Do not multiply the E0 value with the factor used for balancing the equation). Add the reduction and oxidation halfreactions and their potentials. Answer to sample problem: Eocell = 0.74 V, MnO4-, permanganate ions will oxidize Fe(II) ions to Insights from comparison of Zinc – Copper cell and Copper – Silver Cell Strength as oxidizing agent o Zn2+ < Cu2+ < Ag+ Strength as reducing agent o Ag < Cu < Zn Activity series: University of the Philippines Chemical Engineering Society, Inc. REVIEW MODULE – Chemistry 17, 3rd Exam Coverage Fe(III) and are reduced to Mn(II) ions in acidic solution. Nernst equation (for non-standard conditions): − = potential under condition of interest = potential under standard conditions = ideal gas constant = absolute temp in K = # of e- transferred = 96,487 J/V mol e= reaction quotient for the half – reaction: Relationship of Eocell to Go and K − − − − − − Sample Problem: Calculate the standard Gibbs free energy change, G0, at 250C for the following reaction. Pb + Cu2+ Pb2+ + Cu Solution: Eocell is calculated is calculated − Sample Problem: Calculate the initial potential of a cell that consists of an Fe3+/Fe2+ electrode in which [Fe3+]=1.0 x 10-2 M and [Fe2+]=0.1 M connected to a Sn4+/Sn2+ electrode in which [Sn4+]=1.0 M and [Sn2+]=0.10 M . A wire and salt bridge complete the circuit. Temperature is at a standard value of 25 oC. Given: Fe3+ + e- Fe2+, Eo = 0.771 V Sn4+ + 2e- Sn2+, Eo = 0.15 V Solution: Eocell is calculated − Apply Nernst equation Some applications of electrochemical cells: to determine concentration (e.g. pH meter) – recall Nernst equation corrosion (e.g. corrosion of iron) o Anode (oxidation): Fe Fe2+ + 2e manifested as pitting o Cathode (reduction): O2 + 2H2O + 4e- 4OH can be indicated using phenolphthalein (turns pink) o Cathode (reduction): O2 + 2H2O + manifested by the appearance of rust University of the Philippines Chemical Engineering Society, Inc. REVIEW MODULE – Chemistry 17, 3rd Exam Coverage Corrosion prevention o plate with less active metal o connect to more active metal (sacrificial anode) o coating o passivation layer (coating that is allowed to form naturally) o coat with more active metal (galvanizing) batteries (voltaic or galvanic cells) – as it discharges, its chemicals are consumed o primary voltaic cells – not rechargeable (e.g. dry cell or flashlight battery, alkaline cells, mercury battery) o secondary voltaic cells – reversible, rechargeable (e.g. lead storage / car battery, Ni – Cd battery, fuel cells) Maximum work done by the system. o The cell potential if each species in solution are 0.20M in concentration and the gases are at 1 atm pressure. (T cell = 298K). Draw the cell notation for the cell (Include polarity (sign) and the direction of the electron flow. o • Maximum useful work, wmax: for a voltaic cell, Eocell is (+), ΔG is (-), wmax is (-) work done by the system o wmax = -nFEocell for an electrolytic cell Eocell is (-), ΔG is (+), wmax is (+) work done on the system o wmax = nFEoexternal Overpotential/Overvoltage difference between the actual electrode potential and the thermodynamic or equilibrium potential 2. Electrolytic cell: A 1.00 M acidified aq. CaCl2 solution is electrolyzed using Pt. electrodes at 298K. A current of 1.50A passes through the solution for 50.0 hours. Write the half-reactions at the electrodes. What is the decomposition potential, Eocell? Calculate the mass (in grams) of the product formed in the cathode. What is the ΔGo ? Draw a simple schematic diagram of how the electrolysis setup should look like. Indicate the electrodes, sign, direction of electron flow, etc. Given: O2 (g) + 4H+ + 4e- 2H2O(l) E0= 1.229V Cl2 (g) + 2e- 2Cl-(aq) E0= 1.360V Ca2+ (aq) + 4e- 2Ca(s) Eo= -2.870V Coordination Compounds More excercises: 1. Voltaic/galvanic cell: Given the following standard reduction potentials: E° (MnO2(s) Mn2+) = 1.230V E°(Cl2Cl-) = 1.360V Write a balanced (net) cell equation for the spontaneous reaction. Calculate the o Standard cell potential, E0cell o Standard Gibbs Free Energy, ΔG° coordinate covalent bond pair of electrons shared from a donor to an acceptor usually formed in Lewis acid – base reactions Example: ammonia (donor) and borontrifluoride (acceptor with empty valence orbital) H3N: + BF3 H3N : BF3 University of the Philippines Chemical Engineering Society, Inc. REVIEW MODULE – Chemistry 17, 3rd Exam Coverage Coordination compound/complexes – metal ion bonded with ligands ligand - Lewis base that coordinates to a central metal atom or ion donor atom - atom in a ligand that donate a lone pair of electrons to form a coordinate covalent bond unidentate ligand – only one atom can serve as donor atom polydentate ligand (chelating agent) – has several donor atoms that can coordinate at the same time coordination number – number of donor atoms coordinated to the central atom / ion coordination sphere – central atom / ion + ligands coordinated to it Example: For the complex compound Na3[Co(Cl)6] the coordination number is 6, and the coordination sphere is 7. The oxidation state of Co is 3. Common monodentate ligands: Fluoro F-, Chloro Cl, Bromo Br-, Iodo I-, Azido N3-, Cyano CN-, Thiocyanato SCN- (S-bonded), Isothiocyanato NCS- (N-bonded), Hydroxo OH-, Aqua H20, Carbonyl CO, Thiocarbonyl CS, Nitrosyl NO+, Nitro NO2-, Nitrito ONO- O-bonded) Other common ligands: Methylisocyanide CH3NC, Phosphine (phosphane) PR3, Pyridine, Ammine NH3, Methylamine MeNH2, Amido NH2-, Sulfato SO42-, Oxalato C2O4-2 Common chelating ligands: • Ammine, methylamine NH3, CH3NH2 (monodentate), Ethylenediamine (en) NH2CH2CH2NH2 (bidentate), diethylenetriamine (dien) NH2CH2CH2NHCH2CH2NH2, Triethylenetetraamine (trien), Tetraethylenepentaamine, EDTA Nomenclature cations first, then anions ligands in alphabetical number prefixes used to specify number or each ligand (di, tri, tetra, penta, hexa, etc.) but prefixes are not included in alphabetization prefixes that are part of name of ligand (e.g. diethylamine) are included in alphabetization if ligands contain the prefixes di, tri, etc. in their names, the following prefixes are used to indicate number: bis, tris, tetrakis, pentakis, hexakis names of most ionic ligands end in –o names of most neutral ligands are unchanged [exceptions: NH3 (ammine), H2O (aqua), NO (nitrosyl), CO (carbonyl)] The oxidation number of a metal that exhibits more than one oxidation states is designated by a Roman numeral in parentheses following the name of the complex. If a complex is an anion, the suffix "ate" ends the name. (e.g. ferrate, plumbate) Examples: 1. Na3[Fe(Cl)6] sodium hexachloroferrate (III) 2. [Ni(NH3)4(OH2)2](NO3)2 tetraamminediaquanickel(II) nitrate More practice exercises: Give the names for the following: 1. [Cu(NH3)4](NO3)2 2. (NH4)4[Mn(CN)6] 3. Na[AlCl4] 4. [FeBrCl(en)2]Cl 5. [Fe(H2O)4(OH)2]Br Give the chemical formula for the following: University of the Philippines Chemical Engineering Society, Inc. REVIEW MODULE – Chemistry 17, 3rd Exam Coverage • Dichlorobis(ethylenediammine)cobalt(III) Nitrate • Potassium trioxalatochromate(III) • Ammonium diamminetetraisothiocyanochromate(III) • Tetraamminedinitroplatinum(IV) bromide • Tri-aqua-cis-dibromochlorochromium(III) Structure of coordination compounds: determined primarily by coordination number can be predicted using VSEPR theory (Chem 16) Coordination Number 2 4 4 5 5 6 Geometry linear tetrahedral square planar trigonal bipyramidal square pyramidal octahedral Metal Hybridization sp sp3 sp2d or dsp2 dsp3 d2sp2 d2sp3 or sp3d2 o trans- [Pt(NH3)2Cl2] coordination isomers – exchange of ligands between the coordination spheres of the cation and anion o [Pt(NH3)4][PtCl6] and [Pt(NH3)4Cl2][PtCl4] linkage isomers o [Co(NH3)5ONO]Cl2 and [Co(NH3)5NO2]Cl2 Stereoisomers Note: Complexes with only simple ligands can have stereoisomers only if they have coordination numbers equal to or greater than 4. geometrical or positional isomers – not optical isomers o cis-trans isomerism o cis- [Pt(NH3)2Cl2] Isomerism structural/constitutional isomers – same formula but different ligands stereoisomers – same ligands but different arrangement Structural/constitutional isomers: ionization or ion-ion exchange isomers o [Pt(NH3)4Cl2]Br2 (counter ions areBr) and [Pt(NH3)4Br2]Cl2 (counter ions are Cl-) hydrate isomers – water molecules may be changed from inside to outside of coordination molecule o [Fe(OH2)6]Cl3, [Fe(OH2)5Cl]Cl2·H2O, and [Fe(OH2)4Cl2]Cl2·2H2O o cis-diammine-cis-diaqua-transdichlorocobalt(III) ions optical isomers/enantiomers non – superimposable mirror images o cis-diammine-cis-diaqua-cisdichlorocobalt(III) ion University of the Philippines Chemical Engineering Society, Inc. REVIEW MODULE – Chemistry 17, 3rd Exam Coverage optical activity - separate equimolar solutions of two optical isomers / enantiomers rotate plane polarized light by equal angles but in opposite directions Electronic Spectra The energy of light absorbed as electrons are raised to higher levels is the difference in energy between the energy states Valence Bond Theory – uses hybridization concepts Example: Co(II) – may have 3, 1, or 0 unpaired electrons For octahedral complexes The five d orbitals are split into: eg (higher energy) orbitals directed along x, y, and z axes , t2g (lower energy) orbitals directed in between the axes or or crystal field splitting energy – energy separation between 2 sets of d orbitals – proportional to crystal field strength of ligands Crystal Field Theory Bonding between the metal and the ligands is considered to be largely electrostatic and thus the energy of the metal ion and coordinated ligands is lower than that of isolated ions because of electrostatic attraction. Ligands are treated as point charges and the effect of these point charges are considered on the relative energies of the d orbitals. The energies of the metal d-electrons are increased by the repulsive interaction between them and the electrons from the ligands give rise to the splitting of the metal d-orbitals Recall: σ-bonding and π-bonding Ligands with π-bonding capabilities: ligands with occupied p-orbitals are potential π-donors ligands with vacant d orbitals are potential π-acceptors Crystal field strength for some common ligands: I- < Br- < Cl- < F-, OH-, <H2O < NCS- < NH3 < en < NO3<CO, CN weak field, high spin π-donors, interact strongly with metal orbitals, large crystal field splitiing, large strong field, low spin π-acceptors, interact weakly with metal orbital, small crystal field splitting, small spectrochemical series colors of complexes, reflects the ligand arrangement based on crystal field strength University of the Philippines Chemical Engineering Society, Inc. REVIEW MODULE – Chemistry 17, 3rd Exam Coverage High spin complexes (w/ weak field ligand): Low spin complexes (w/ strong field ligand): Recall: Paramagnetism paramagnetic w/ unpaired electrons diamagnetic w/out unpaired electrons The visible range spectra is of the same order of magnitude as the energy of the photon of visible light absorbs light that excites e- from lower energy d orbitals to higher energy d orbitals spectral color color or wavelength of light absorbed complementary color color or wavelength of light transmitted Complexes with both strong field and weak field ligands: Example: [Cr(H2O)6]3+(light violet) weak field ligand, small , absorbs lower energy/larger wavelength light (ROY end), transmits higher energy/smaller wavelength light (BLUE end) [Cr(NH3)6]3+ (yellow) strong field ligand, large , absorbs higher energy/smaller wavelength light (BLUE end), transmits lower energy/larger wavelength light (ROY end) Consider the complex [FeX6]4If X = Cl- (weak field ligand)high spin complex For tetrahedral complexes splitting is opposite of that of octahedral If X = CN- (strong field ligand) low spin complex Notes: (1) is about 4/9 of (2) all tetrahedral complexes are paramagnetic For square planar complexes University of the Philippines Chemical Engineering Society, Inc. REVIEW MODULE – Chemistry 17, 3rd Exam Coverage d8 central metal almost always low spin (diamagnetic) example: heavy metals (46Pd2+, 78Pt2, 77Ir+ and 79Au3+) Comparison: Limitation of CFT: fails to consider the actual covalent character formed between the metal and the ligands Practice Exercise: K4[MnF6] Name : potassium hexafluoromanganate (II) Coordination sphere = 7 Coordination number = 6, Octahedral geometry Mn+2 d5 (Mn=25) F is a weak field, therefore, a high spin complex is formed where the e- will occupy eg orbitals before pairing in the t2g Paramagnetic. We expect the complex to absorbed in the red-orange region and hence, the compound would likely appear green to blue.