Question 4.15: Balancing Oxidation–Reduction Reactions: Half-Reaction Metho...
Balancing Oxidation–Reduction Reactions: half-Reaction Method
Silver is sometimes found in nature as large nuggets; more often it is found mixed with other metals and their ores. Cyanide ion is often used to extract the silver by the following reaction that occurs in basic solution:
Ag(s) + CN^{-}(aq) + O_{2} (g) \xrightarrow[]{Basic} Ag (CN)_{2}^{-}(aq)
Balance this equation by using the half-reaction method.
1. Balance the equation as if H^{+} ions were present.
Balance the oxidation half-reaction:
CN^{-}(aq) + Ag(s) → Ag (CN)_{2}^{-}(aq)
Balance carbon and nitrogen:
2 CN^{-}(aq) + Ag(s) → Ag (CN)_{2}^{-}(aq)
Balance the charge:
2 CN^{-}(aq) + Ag(s) → Ag (CN)_{2}^{-}(aq) + e ^{-}
Balance the reduction half-reaction:
O_{2}(g) →
Balance oxygen:
O_{2}(g) → 2 H_{2}O (l)
Balance hydrogen:
O_{2}(g) + 4 H^{+} (aq)→ 2 H_{2}O (l)
Balance the charge:
4e ^{-} + O_{2}(g) + 4 H^{+} (aq)→ 2 H_{2}O (l)
Multiply the balanced oxidation half-reaction by 4:
8CN^{-}(aq) + 4Ag(s) → 4Ag (CN)_{2}^{-}(aq) +4 e ^{-}
Add the half-reactions, and cancel identical species: Oxidation half-reaction:
8 CN^{-}(aq) + 4Ag(s) → 4Ag (CN)_{2}^{-}(aq) +4e ^{-}
Reduction half-reaction:
4 e ^{-} + O_{2}(g) + 4 H^{+} (aq)→ 2 H_{2}O (l)
8 CN^{-}(aq) + 4Ag(s) + O_{2}(g) + 4 H^{+} (aq) → 4Ag (CN)_{2}^{-}(aq) + 2 H_{2}O (l)
2. Add OH^{-} ions to both sides of the balanced equation.
We need to add 4 OH^{-} to each side:
| 8 CN^{-}(aq) + 4Ag(s) + O_{2}(g) + 4 H^{+} (aq) + 4 OH^{-} (aq) |
| 4H_{2}O(l) |
→ 4 Ag (CN)_{2}^{-}(aq) + 2 H_{2}O (l) + 4 OH^{-} (aq)
3. Eliminate as many H_{2}O molecules as possible.
8 CN^{-}(aq) + 4Ag(s) + O_{2}(g) + 2 H_{2}O (l) → 4 Ag (CN)_{2}^{-}(aq) + 4 OH^{-} (aq)
4. Check that elements and charges balance.