Exercise 6 | Oxidation-Reduction Reactions

Balance the following redox reaction in basic solution:

N2H4 (aq) + Cu(OH)2 (s) → N2 (g) + Cu (s)

1) Divide the reaction into half-reactions:

N2H4 (aq) → N2 (g)

Cu(OH)2 (s) → Cu (s)

2) Balance O atoms by adding H2O molecules:

N2H4 (aq) → N2 (g)

Cu(OH)2 (s) → Cu (s) + 2 H2O (l)

3) Balance H atoms by adding H+ ions:

N2H4 (aq) → N2 (g) + 4 H+ (aq)

Cu(OH)2 (s) + 2 H+ (aq) → Cu (s) + 2 H2O (l)

4) Balance charge by adding electrons:

N2H4 (aq) → N2 (g) + 4 H+ (aq) + 4 e-

Cu(OH)2 (s) + 2 H+ (aq) + 2 e- → Cu (s) + 2 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):

N2H4 (aq) → N2 (g) + 4 H+ (aq) + 4 e-

2 Cu(OH)2 (s) + 4 H+ (aq) + 4 e- → 2 Cu (s) + 4 H2O (l)

6) Add the half-reactions together:

N2H4 (aq) + 2 Cu(OH)2 (s) → N2 (g) + 2 Cu (s) + 4 H2O (l)

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

There is no H+ ions.

The balanced equation in basic solution is: N2H4 (aq) + 2 Cu(OH)2 (s) → N2 (g) + 2 Cu (s) + 4 H2O (l)