Balancing redox reaction for water production

Steps:

• Oxidation half-equation: HCOOH → CO2 + 2H⁺ + 2e⁻ (loses 2 electrons per molecule).
• For five HCOOH: 5HCOOH → 5CO₂ + 10H⁺ + 10e⁻.
• Reduction half-equation (acidic): MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O; multiply by 2 for 10e⁻: 2MnO₄⁻ + 16H⁺ + 10e⁻ → 2Mn²⁺ + 8H₂O.
• Combine and simplify: net reaction produces 8H₂O from reduction, unaffected by H⁺ cancellation.

Why B is correct:

• Two MnO₄⁻ ions required to accept 10 electrons from five HCOOH, yielding 8H₂O per the reduction half-equation.

Why the others are wrong:

• A: Assumes 1H₂O per HCOOH, ignoring reduction stoichiometry.
• C: Might double-count H₂O from oxidation protons, but water comes only from reduction.
• D: Overcounts by assuming 4H₂O per MnO₄⁻ without electron balancing.

Final answer: B