Balancing redox reaction in alkaline medium

Steps:

• Reduction half-reaction: MnO₄⁻ + 2H₂O + 3e⁻ → MnO₂ + 4OH⁻
• Oxidation half-reaction: SO₃²⁻ + 2OH⁻ → SO₄²⁻ + H₂O + 2e⁻
• Equalize electrons (LCM of 3 and 2 is 6): multiply reduction by 2, oxidation by 3
• Combine: 2MnO₄⁻ + 3SO₃²⁻ + 2H₂O + 6OH⁻ → 2MnO₂ + 3SO₄²⁻ + 3H₂O + 4OH⁻ (simplifies to 2:3 ratio)

Why A is correct:

• Electron transfer requires 2 MnO₄⁻ (gaining 6e⁻) to oxidize 3 SO₃²⁻ (losing 6e⁻), per redox balancing rules.

Why the others are wrong:

• B: Reverses the 2:3 ratio, ignoring reduction gains 3e⁻ per MnO₄⁻ vs. oxidation loses 2e⁻ per SO₃²⁻.
• C: 4:7 ratio mismatches LCM of 6 electrons needed for balance.
• D: 7:4 inverts unrelated coefficients, not from electron equalization.

Final answer: A