Exercise 6 | Oxidation-Reduction Reactions

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)