Balancing redox half-reactions in acidic medium Steps:

• Oxidation half-reaction: ClO⁻ + H₂O → ClO₂⁻ + 2H⁺ + 2e⁻ (Cl from +1 to +3).
• Reduction half-reaction: MnO₄⁻ + 4H⁺ + 3e⁻ → MnO₂ + 2H₂O (Mn from +7 to +4).
• Multiply oxidation by 3 and reduction by 2 for 6e⁻ transfer: 3ClO⁻ + 3H₂O → 3ClO₂⁻ + 6H⁺ + 6e⁻; 2MnO₄⁻ + 8H⁺ + 6e⁻ → 2MnO₂ + 4H₂O.
• Combine and simplify by canceling 6H⁺ and 3H₂O: 3ClO⁻ + 2MnO₄⁻ + 2H⁺ → 3ClO₂⁻ + 2MnO₂ + H₂O (w=3, x=2, y=2). Why D is correct:
• w=3, x=2, y=2 equalizes electrons (3×2=6 from oxidation; 2×3=6 from reduction) per conservation of charge in redox reactions. Why the others are wrong:
• A: 1×2=2e⁻ vs. 1×3=3e⁻, electrons unequal.
• B: 2×2=4e⁻ vs. 2×3=6e⁻, electrons unequal.
• C: 2×2=4e⁻ vs. 3×3=9e⁻, electrons unequal.

Final answer: D