A redox reaction is one in which the reactants’ oxidation numbers change. What are the oxidation numbers of the metals in the reaction below? The iron’s charge has become more negative, so it is reduced. The tin’s charge has become more positive, so it is oxidized. reduced oxidized Fe Sn 2 FeCl 3 + SnCl 2  2 FeCl 2 + SnCl 4

Which reactant is oxidized and which is reduced in the following? 2) Cu + 2AgNO 3  Cu(NO 3 ) 2 + 2Ag reduced oxidized 1) Cl 2 + SnCl 2  SnCl 4 0 reduced oxidized ) 2KClO 3  2KCl + 3O reduced oxidized Synthesis Single-replacement Decomposition Note that a redox reaction can also be another type of reaction. KClO x + 3(–2) = 0 x = +5

Balancing Redox Reactions MnO 4 – + C 2 O 4 2–  Mn 2+ + CO 2 Rxn: 1. Determine the oxidation number of the redox active species. 2. Split the redox reaction into two half reactions (remember OIL RIG) and add in the #e– being transferred. Red: Ox: MnO 4 –  Mn Balance the oxygens with H 2 O and the hydrogens with H +. C 2 O 4 2–  2 CO H 2 O5e – +8 H e– 4. Multiply both half-reactions by a factor that will make the #e – transferred equal Add the two half-reactions to get the final balanced redox rxn H MnO 4 – + 5 C 2 O 4 2–  2 Mn CO H 2 O (acidic conditions)

Balancing Redox Reactions Under Basic Conditions 1. Balance reaction using acidic conditions. 16 H MnO 4 – + 5 C 2 O 4 2–  2 Mn CO H 2 O 2. Add OH – to both sides to neutralize the H OH– 16 H 2 O 3. Cancel out excess water and rewrite reaction equation. 8 8 H 2 O + 2 MnO 4 – + 5 C 2 O 4 2–  2 Mn CO OH –

Voltaic (alt. Galvanic) Electrochemical Cells Salt bridge Reduction takes place at the cathode. Oxidation takes place at the anode. E A = VE C = V E cell = E A + E C E cell = 0.763V V V 0.92 Zn  Zn 2+  Cu 2+  Cu Anode  Cathode

Example: What are the standard cell potentials for the folllowing? 1. Al  Al 3+  Ag +  Ag Al e–  AlE o = –1.66 VAg+ + e–  AgE o = V Br 2 + 2e–  2Br– E o = V Ox: Al  Al e– Red: Ag + + e–  Ag V V V 2. Ag  Ag +  Br 2  Br– Ox: Ag  Ag + + e– Red: Br 2 + e–  2 Br– –0.80 V V V Note: Voltaic cells ALWAYS have positive cell emf’s (voltages).

Some terminology emf–Electromotive force; force causing e– to move SHE–Standard hydrogen electrode; E o = 0 V by definition. Standard conditions–Solution concentrations = 1 M and gas pressures = 1 atm Oxidizing agent–is reduced during redox rxn Reducing agent–is oxidized during redox rxn Faraday’s constant =96485 C/mol e– transferred R =8.314 J/mol·K 1 A =1 C/s (C  Coulomb = a quantity of charge) 1 V =1 J/C

Table of Standard Reduction Potentials  SHE Good oxidizing agent = large, positive reduction potential Good reducing agent = large, negative reduction potential

A voltaic cell generates current as a result of a spontaneous redox reaction. The equations relating E,  G and K are:  G = -nFE lnK = -  G/RT = nFE/RT R = J/mol·K, T = temperature in Kelvin, F = C/mol n = # electrons transferred in reaction What are  G and K for: 2VO H + + 2Ag  VO H 2 O + 2Ag + E cell = E C + E A = 1.00 V + (–0.799 V) = V n = 2 e–  G = -nFE = -(2 e–)(96485 C/mol e–)(0.201 V) = J/mol (Remember 1 V = 1 J/C) K = e –  G/RT = e (38787 J/mol)/(8.314 Jmol*298K) = e15.7 =

Operating under non-standard conditions: Nernst Eq. = What is the cell potential for the following if [Ce 4+ ]=2.0M, [Ce 3+ ] = M and [Cr 3+ ] = M? 3Ce 4+ + Cr(s)  3Ce 3+ + Cr 3+ n =3e– RED OX E o = E C + E A = V V = 2.35 V E = 2.35 V – (0.0592/3)log [0.010 M] 3 [0.010 M] [2.0 M] 3 = 2.53 V Because [reactants]>>[products] AND the reaction has a large positive E o, the emf has increased to produce more product.

Electrolysis:Decomposition of a compound by passing electricity through it Metallic magnesium can be made by the electrolysis of molten MgCl 2. What mass of Mg is formed by passing a current of 5.25 A through molten MgCl 2 for 2.50 days? Calculate total time: Calculate total charge passed = Divide by 2 because it takes 2e–/equiv Mg 2+ = C Calculate mol Mg = Calculate grams Mg = Ans: 143 g of Mg will be formed 5.25 C/s * s = C C /(96485 C/mol) = mol Mg(s) mol Mg(s) * g/mol = g