Exercise 7 | Oxidation-Reduction Reactions

Balance the following redox reaction in basic solution:

H3AsO3 (aq) + I2 (aq) → H3AsO4 (aq) + I- (aq)

1) Divide the reaction into half-reactions:

H3AsO3 (aq) → H3AsO4 (aq)

I2 (aq) → 2 I- (aq)

2) Balance O atoms by adding H2O molecules:

H3AsO3 (aq) + H2O (l) → H3AsO4 (aq)

I2 (aq) → 2 I- (aq)

3) Balance H atoms by adding H+ ions:

H3AsO3 (aq) + H2O (l) → H3AsO4 (aq) + 2 H+ (aq)

I2 (aq) → 2 I- (aq)

4) Balance charge by adding electrons:

H3AsO3 (aq) + H2O (l) → H3AsO4 (aq) + 2 H+ (aq) + 2 e-

I2 (aq) + 2 e- → 2 I- (aq)

5) Multiply each half-reaction by an integer to have: number of electrons lost = number of electrons gained

(in this case, you don't have to do this step):

H3AsO3 (aq) + H2O (l) → H3AsO4 (aq) + 2 H+ (aq) + 2 e-

I2 (aq) + 2 e- → 2 I- (aq)

6) Add the half-reactions together:

H3AsO3 (aq) + H2O (l) + I2 (aq) → H3AsO4 (aq) + 2 H+ (aq) + 2 I- (aq)

7) Add HO- ions to react with all H+ ions: HO- + H+ à H2O:

H3AsO3 (aq) + H2O (l) + I2 (aq) + 2 HO- (aq) → H3AsO4 (aq) + 2 H+ (aq) + 2 HO- (aq) + 2 I- (aq)

 

The balanced equation in basic solution is:

H3AsO3 (aq) + I2 (aq) + 2 HO- (aq) → H3AsO4 (aq) + H2O (l) + 2 I- (aq)