Exercise 5 | Oxidation-Reduction Reactions

Balance the following redox reaction in acidic solution:

Mn (s) + HNO3 (aq) → Mn2+ (aq) + NO2 (g)

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)