Balance Redox Equations: Half Reaction concentrate on electrons, balance each ½ rx. separately Ex: Cr 2 O 7 2- (aq) + Cl - (aq)  Cr 3+ (aq) + Cl 2 (g)

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Balance Redox Equations: Half Reaction concentrate on electrons, balance each ½ rx. separately Ex: Cr 2 O 7 2- (aq) + Cl - (aq)  Cr 3+ (aq) + Cl 2 (g) IN ACID Ox: Cl - (aq)  Cl 2 (g) Red: : Cr 2 O 7 2- (aq)  Cr 3+ (aq) 1. Balance except H and O 2 Cl - (aq)  Cl 2 (g) Cr 2 O 7 2- (aq)  2 Cr 3+ (aq) 2. Balance O with H 2 O then add H + 2 Cl - (aq)  Cl 2 (g) Cr 2 O 7 2- (aq) + 14 H + (aq)  2 Cr 3+ (aq) + 7H 2 O(l) 3. Balance charge by add e- to side with > # + charges, then X so # e- same in both equations Cr 2 O 7 2- (aq) + 14 H + (aq) + 6e -  2 Cr 3+ (aq) + 7H 2 O(l) 2 Cl - (aq)  Cl 2 (g) + 2e - X3 6 Cl - (aq)  3Cl 2 (g) + 6e - 4. Add ½ rx. Cr 2 O 7 2- (aq) + 14 H + (aq) + 6 Cl - (aq) + 6e -  2 Cr 3+ (aq) + 7H 2 O(l) + 3Cl 2 (g) + 6e - Cr 2 O 7 2- (aq) + 14 H + (aq) + 6 Cl - (aq)  2 Cr 3+ (aq) + 7H 2 O(l) + 3Cl 2 (g) Ex: Cr(OH) 4 - (aq) + ClO - (aq)  CrO 4 2- (aq) + Cl - (aq) IN BASE 1. Balance except H and O 2. Balance O with H 2 O then add H + 4. Add ½ rx Ox: Cr(OH) 4 - (aq)  CrO 4 2- (aq)Red: ClO - (aq)  Cl - (aq) Cr(OH) 4 - (aq)  CrO 4 2- (aq) + 4H + (aq) Cr(OH) 4 - (aq)  CrO 4 2- (aq) + 4H + (aq) + 3e - ClO - (aq) + 2H + (aq)  Cl - (aq) + H 2 O(l) ClO - (aq) + 2H + (aq) + 2e -  Cl - (aq) + H 2 O(l) X2 X3 2Cr(OH) 4 - (aq)  2CrO 4 2- (aq) + 8H + (aq) + 6e - 3ClO - (aq) + 6H + (aq) + 6e -  3Cl - (aq) + 3H 2 O(l) 3ClO - (aq) + 6H + (aq) + 6e - + 2Cr(OH) 4 - (aq)  3Cl - (aq) + 3H 2 O(l) + 2CrO 4 2- (aq) + 8H + (aq) + 6e - 2H + (aq) 3ClO - (aq) + 2Cr(OH) 4 - (aq)  3Cl - (aq) + 3H 2 O(l) + 2CrO 4 2- (aq) + 2H + (aq) +2OH - (aq) 3ClO - (aq) + 2Cr(OH) 4 - (aq) + 2OH - (aq)  3Cl - (aq) + 5H 2 O(l) + 2CrO 4 2- (aq) 3. Balance charge by add e- to side with > # + charges, then X so # e- same in both equations

BALANCING REDOX EQUATIONS – ½ RX METHOD MASS AND CHARGE CONSERVATION [Cr(OH) 4 ] -  CrO H +  + MASS BALANCE (CONSERVATION) CHARGE BALANCE (CONSERVATION)  SO ADD 3 e – to this side!  Ex: an oxidation reaction already balanced for O and H

Problem: Using the half-reaction methods, balance the following equation showing the reaction of permanganate and ethanol to form acetic acid. Assume the reaction takes place under ACIDIC conditions. MnO 4 - (aq) + C 2 H 5 OH (aq)  Mn 2+ + CH 3 CO 2 H