Exercise 5 | Oxidation-Reduction Reactions

1) Divide the reaction into half-reactions:

Mn (s) → Mn²⁺ (aq)

HNO₃ (aq) → NO₂ (g)

2) Balance O atoms by adding H₂O molecules:

Mn (s) →Mn²⁺ (aq)

HNO₃ (aq) → NO₂ (g) + H₂O (l)

3) Balance H atoms by adding H⁺ ions:

Mn (s) → Mn²⁺ (aq)

HNO₃ (aq) + H⁺ (aq) → NO₂ (g) + H₂O (l)

4) Balance charge by adding electrons:

Mn (s) → Mn²⁺ (aq) + 2 e^-

HNO₃ (aq) + H⁺ (aq) + e^- → NO₂ (g) + H₂O (l)

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

(in this case, multiply by 2 the second half-reaction):

Mn (s) → Mn²⁺ (aq) + 2 e^-

2 HNO₃ (aq) + 2 H⁺ (aq) + 2 e^- → 2 NO₂ (g) + 2 H₂O (l)

6) Add the half-reactions together:

Mn (s) + 2 HNO₃ (aq) + 2 H⁺ (aq) → Mn²⁺ (aq) + 2 NO₂ (g) + 2 H₂O (l)