Exercise 7 | Chemical Reactivity

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)