Q.Balance the following redox reactions by ion - electron method: MnO4-(aq) + SO2(g) -> Mn2+(aq) + HSO4-(aq) MnO4-(aq) + SO2(g) -> Mn2+(aq) + HSO4-(aq) (in acidic solution)
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Start your 14-day free trial to unlock the full solution →Split the reaction into a reduction half-reaction (MnO₄⁻ → Mn²⁺) and an oxidation half-reaction (SO₂ → HSO₄⁻), balance each for atoms and charge separately, equalise the electrons transferred, and add them: 2MnO₄⁻ + 5SO₂ + 2H₂O + H⁺ → 2Mn²⁺ + 5HSO₄⁻.
Reaction to balance (acidic medium): MnO₄⁻(aq) + SO₂(g) → Mn²⁺(aq) + HSO₄⁻(aq)
Step 1 — Identify the two half-reactions and oxidation number changes.
- Mn goes from +7 (in MnO₄⁻) to +2 (in Mn²⁺): reduction, gain of 5 electrons.
- S goes from +4 (in SO₂) to +6 (in HSO₄⁻): oxidation, loss of 2 electrons.
Step 2 — Balance the reduction half-reaction.
MnO₄⁻ → Mn²⁺
Balance O by adding H₂O: MnO₄⁻ → Mn²⁺ + 4H₂O
Balance H by adding H⁺ (acidic medium): MnO₄⁻ + 8H⁺ → Mn²⁺ + 4H₂O
Balance charge by adding electrons (left charge = −1+8 = +7; right = +2; add 5e⁻ to the left):
MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O ... (i)
Step 3 — Balance the oxidation half-reaction.
SO₂ → HSO₄⁻
Balance O by adding H₂O: SO₂ + 2H₂O → HSO₄⁻ (left O = 2+2 = 4; right O = 4 ✓)
Balance H by adding H⁺ (left H = 4; right H so far = 1, so add 3H⁺ to the right): SO₂ + 2H₂O → HSO₄⁻ + 3H⁺
Balance charge by adding electrons (left charge = 0; right charge = −1+3 = +2; add 2e⁻ to the right to make it 0):
SO₂ + 2H₂O → HSO₄⁻ + 3H⁺ + 2e⁻ ... (ii)
Step 4 — Equalise the number of electrons transferred.
LCM of 5 and 2 is 10. Multiply (i) by 2 and (ii) by 5:
2×(i): 2MnO₄⁻ + 16H⁺ + 10e⁻ → 2Mn²⁺ + 8H₂O
5×(ii): 5SO₂ + 10H₂O → 5HSO₄⁻ + 15H⁺ + 10e⁻
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