1. A redox reaction is one in which the reactants’ oxidation numbers change. What are the oxidation numbers of the metals in the reaction below? +3+2 +4 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 0 +2 +4 +6 +8 -2 -6 -4 0 +2 +4 +6 +8 -2 -6 -4 Fe Sn 2 FeCl 3 + SnCl 2  2 FeCl 2 + SnCl 4

2. Which reactant is oxidized and which is reduced in the following? 2) Cu + 2AgNO 3  Cu(NO 3 ) 2 + 2Ag 0+1+20 reduced oxidized 1) Cl 2 + SnCl 2  SnCl 4 0 reduced oxidized +2+4 3) 2KClO 3  2KCl + 3O 2 -20 +5 reduced oxidized Synthesis Single-replacement Decomposition Note that a redox reaction can also be another type of reaction. KClO 3 +1 + x + 3(–2) = 0 x = +5

3. +7+3+4 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 2+ 3. Balance the oxygens with H 2 O and the hydrogens with H +. C 2 O 4 2–  2 CO 2 + 4 H 2 O5e – +8 H + + + 2e– 4. Multiply both half-reactions by a factor that will make the #e – transferred equal. 2 5 5. Add the two half-reactions to get the final balanced redox rxn. + 16 H + + 2 MnO 4 – + 5 C 2 O 4 2–  2 Mn 2+ + 10 CO 2 + 8 H 2 O (acidic conditions)

4. Balancing Redox Reactions Under Basic Conditions 1. Balance reaction using acidic conditions. 16 H + + 2 MnO 4 – + 5 C 2 O 4 2–  2 Mn 2+ + 10 CO 2 + 8 H 2 O 2. Add OH – to both sides to neutralize the H +. +16 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 2+ + 10 CO 2 + 16 OH –

5. Voltaic (alt. Galvanic) Electrochemical Cells Salt bridge Reduction takes place at the cathode. Oxidation takes place at the anode. E A = +0.763 VE C = +0.153 V E cell = E A + E C E cell = 0.763V + 0.153 V 0.916 V 0.92 Zn  Zn 2+  Cu 2+  Cu Anode  Cathode

6. Example: What are the standard cell potentials for the folllowing? 1. Al  Al 3+  Ag +  Ag Al 3+ + 3e–  AlE o = –1.66 VAg+ + e–  AgE o = +0.80 V Br 2 + 2e–  2Br– E o = +1.06 V Ox: Al  Al 3+ + 3e– Red: Ag + + e–  Ag +1.66 V +0.80 V +2.46 V 2. Ag  Ag +  Br 2  Br– Ox: Ag  Ag + + e– Red: Br 2 + e–  2 Br– –0.80 V +1.06 V +0.26 V Note: Voltaic cells ALWAYS have positive cell emf’s (voltages).

7. 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

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

9. 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 = 8.314 J/mol·K, T = temperature in Kelvin, F = 96485 C/mol n = # electrons transferred in reaction 20.58 What are  G and K for: 2VO 2 + + 4H + + 2Ag  VO 2+ + 2H 2 O + 2Ag + E cell = E C + E A = 1.00 V + (–0.799 V) = 0.201 V n = 2 e–  G = -nFE = -(2 e–)(96485 C/mol e–)(0.201 V) = -38787 J/mol (Remember 1 V = 1 J/C) K = e –  G/RT = e (38787 J/mol)/(8.314 Jmol*298K) = e15.7 = 6294875

10. Operating under non-standard conditions: Nernst Eq. = 20.50 What is the cell potential for the following if [Ce 4+ ]=2.0M, [Ce 3+ ] = 0.010 M and [Cr 3+ ] = 0.010 M? 3Ce 4+ + Cr(s)  3Ce 3+ + Cr 3+ n =3e– RED OX E o = E C + E A = +1.61 V + +0.74 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.

11. Electrolysis:Decomposition of a compound by passing electricity through it. 20.80 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+ = 567000 C Calculate mol Mg = Calculate grams Mg = Ans: 143 g of Mg will be formed 5.25 C/s * 216000 s = 1134000 C 567000 C /(96485 C/mol) = 5.877 mol Mg(s) 5.877 mol Mg(s) * 24.305 g/mol = 142.8 g