Exercise 7 | Oxidation-Reduction Reactions

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)