Balancing redox reaction via half-reaction method in acidic medium Steps:

• Oxidation: ClO₃⁻ + H₂O → ClO₄⁻ + 2H⁺ + 2e⁻ (Cl from +5 to +7)
• Reduction: MnO₄⁻ + 4H⁺ + 3e⁻ → MnO₂ + 2H₂O (Mn from +7 to +4)
• Equalize electrons by multiplying oxidation by 3 and reduction by 2 (6e⁻ each)
• Combine and simplify: 3ClO₃⁻ + 2MnO₄⁻ + 2H⁺ → 3ClO₄⁻ + 2MnO₂ + H₂O (cancels 3H₂O and 6H⁺) Why D is correct:
• Coefficients 3, 2, 2 match 3×2e⁻ lost = 2×3e⁻ gained, conserving electrons per the redox law. Why the others are wrong:
• A: 1×2e⁻ ≠ 1×3e⁻, electrons imbalanced
• B: 2×2e⁻ ≠ 3×3e⁻, electrons imbalanced
• C: Identical to B, electrons imbalanced

Final answer: D