1. Assign oxidation states:
- Mn in MnO₄⁻: +7
- S in S₂O₃²⁻: +2 (average)
- S in S₄O₆²⁻: +2.5 (average)
- Mn in Mn²⁺: +2
2. Identify half-reactions:
- Reduction: MnO₄⁻ → Mn²⁺ (Mn from +7 to +2, gain of 5e⁻)
- Oxidation: S₂O₃²⁻ → S₄O₆²⁻ (S from +2 to +2.5, loss of 1e⁻ per S atom)
3. Balance atoms other than O and H:
- Reduction: MnO₄⁻ → Mn²⁺ (Mn balanced)
- Oxidation: 2S₂O₃²⁻ → S₄O₆²⁻ (S balanced)
4. Balance oxygen by adding H₂O:
- Reduction: MnO₄⁻ → Mn²⁺ + 4H₂O
- Oxidation: 2S₂O₃²⁻ → S₄O₆²⁻ (no O imbalance)
5. Balance hydrogen by adding H⁺:
- Reduction: MnO₄⁻ + 8H⁺ → Mn²⁺ + 4H₂O
- Oxidation: 2S₂O₃²⁻ → S₄O₆²⁻ (no H involved)
6. Balance charge by adding electrons:
- Reduction: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O
- Oxidation: 2S₂O₃²⁻ → S₄O₆²⁻ + 2e⁻
7. Equalize electrons transferred:
- Multiply reduction by 2: 2MnO₄⁻ + 16H⁺ + 10e⁻ → 2Mn²⁺ + 8H₂O
- Multiply oxidation by 5: 10S₂O₃²⁻ → 5S₄O₆²⁻ + 10e⁻
8. Add half-reactions and cancel electrons:
2MnO₄⁻ + 16H⁺ + 10S₂O₃²⁻ → 2Mn²⁺ + 8H₂O + 5S₄O₆²⁻
9. Verify atom and charge balance:
- Atoms: Mn (2), O (8+30=38 on left; 8+30=38 on right), H (16 on left; 16 on right), S (20 on both sides)
- Charge: Left: 2(-1) + 16(+1) + 10(-2) = -2 + 16 - 20 = -6; Right: 2(+2) + 5(-2) = +4 - 10 = -6
Balanced equation:
2MnO₄⁻ + 16H⁺ + 10S₂O₃²⁻ → 2Mn²⁺ + 8H₂O + 5S₄O₆²⁻
Parent Tip: Review the logic above to help your child master the concept of balancing redox reactions worksheet.
