Balancing redox half-reactions in alkaline medium

Steps:

• Reduction: MnO₄⁻ + 2H₂O + 3e⁻ → MnO₂ + 4OH⁻ (Mn⁺⁷ to Mn⁺⁴, 3e⁻ gain)
• Oxidation: SO₃²⁻ + 2OH⁻ → SO₄²⁻ + H₂O + 2e⁻ (S⁺⁴ to S⁺⁶, 2e⁻ loss)
• Balance electrons: multiply reduction by 2, oxidation by 3 (6e⁻ total)
• Overall: 2MnO₄⁻ + 3SO₃²⁻ → products, ratio 2:3

Why A is correct:

• Ratio 2:3 equalizes 6 electrons (3 per MnO₄⁻ × 2 = 2 per SO₃²⁻ × 3) per redox balance rule

Why the others are wrong:

• B. 3:2 reverses the required coefficients for electron equality
• C. 4:7 does not match least common multiple of 3 and 2 electrons
• D. 7:4 inverts C, ignoring half-reaction stoichiometry

Final answer: A