Exercise 8 | Oxidation-Reduction Reactions

General Chemistry 3 - Exercise 8

Balance the following equation under basic conditions by using oxidation and reduction half-reactions:

Pb(OH)42- (aq) + ClO- (aq) → PbO2 (s) + Cl- (aq)

What is the oxidizing and the reducing agent?

Pb(OH)42- (aq) → PbO2 (s) + 2 H2O (l) + 2 e-

Under basic conditions: Pb(OH)42- (aq) → PbO2 (s) + 2 H2O (l) + 2 e-

ClO- (aq) + 2 H+ (aq) + 2 e- → Cl- (aq) + H2O (l)

Under basic conditions: ClO- (aq) + H2O (l) + 2 e- → Cl- (aq) + 2 HO- (aq)


Pb(OH)42- (aq) + ClO- (aq) → PbO2 (s) + H2O (l) + Cl- (aq) + 2 HO- (aq)


Oxidizing agent: ClO-

Reducing agent: Pb(OH)42-