Balance the following redox reaction in basic solution:

P₄ (s) + NO₃^- (aq) → PO₄^- (aq) + NO (g)

Question

Balance the following redox reaction in basic solution:

P₄ (s) + NO₃^- (aq) → PO₄^- (aq) + NO (g)

| General Chemistry 3

Accepted Answer

1) Divide the reaction into half-reactions:

P₄ (aq) → 4 PO₄^- (aq)

NO₃^- (aq) → NO (g)

2) Balance O atoms by adding H₂O molecules:

P₄ (aq) + 16 H₂O (l) → 4 PO₄^- (aq)

NO₃^- (aq) → NO (g) + 2 H₂O (l)

3) Balance H atoms by adding H⁺ ions:

P₄ (aq) + 16 H₂O (l) → 4 PO₄^- (aq) + 32 H⁺ (aq)

NO₃^- (aq) + 4 H⁺ (aq) → NO (g) + 2 H₂O (l)

4) Balance charge by adding electrons:

P₄ (aq) + 16 H₂O (l) → 4 PO₄^- (aq) + 32 H⁺ (aq) + 28 e^-

NO₃^- (aq) + 4 H⁺ (aq) + 3 e^- → NO (g) + 2 H₂O (l)

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

3 P₄ (aq) + 48 H₂O (l) → 12 PO₄^- (aq) + 96 H⁺ (aq) + 84 e^-

28 NO₃^- (aq) + 112 H⁺ (aq) + 84 e^- → 28 NO (g) + 56 H₂O (l)

6) Add the half-reactions together:

3 P₄ (aq) + 28 NO₃^- (aq) + 16 H⁺ (aq) → 12 PO₄^- (aq) + 28 NO (g) + 8 H₂O (l)

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

3 P₄ (aq) + 28 NO₃^- (aq) + 16 H⁺ (aq) + 16 HO^- (aq) → 12 PO₄^- (aq) + 28 NO (g) + 8 H₂O (l) + 16 HO^- (aq)

The balanced equation in basic solution is:

3 P₄ (aq) + 28 NO₃^- (aq) + 8 H₂O (l) → 12 PO₄^- (aq) + 28 NO (g) + 16 HO^- (aq)

Final - Question 1 | General Chemistry 3