Balancing redox equation for MnO4- oxidizing I- to I2 in acid Steps:

• Write half-reactions: MnO4- + 8H+ + 5e- → Mn2+ + 4H2O; 2I- → I2 + 2e-.
• Balance electrons (LCM=10): 2(MnO4- + 8H+ + 5e- → Mn2+ + 4H2O); 5(2I- → I2 + 2e-).
• Combine: 2MnO4- + 16H+ + 10I- → 2Mn2+ + 5I2 + 8H2O.
• Coefficients (u=2, v=16, w=10, x=2, y=8, z=5) are whole numbers; u=x=2 satisfies condition.

Why D is correct:

• In minimal integer coefficients, H+ coefficient v=16 per balanced equation (from 8H+ per MnO4-, doubled for electron balance).

Why the others are wrong:

• A: v=2 requires fractional coefficients (e.g., z=0.25), violating whole numbers.
• B: v=8 requires fractional coefficients (e.g., u=0.8), violating whole numbers.
• C: v=10 fits I- coefficient w, not H+ coefficient v.

Final answer: D