What is the balanced half-reaction for Al(s) + MnO4^-(aq) -> Al^3+(aq) + Mn^2+(aq) in acidic conditions?

• A. Al(s) + 4H2O(l) + MnO4^-(aq) + 8H^+(aq) -> Al^3+(aq) + Mn^2+(aq) + 4H2O(l)
• B. Al(s) + 3H2O(l) + MnO4^-(aq) -> Al^3+(aq) + Mn^2+(aq) + 3OH^-(aq)
• C. Al(s) + 2H2O(l) + MnO4^-(aq) + 4H^+(aq) -> Al^3+(aq) + Mn^2+(aq) + 2H2O(l)
• D. Al(s) + 4H2O(l) + MnO4^-(aq) -> Al^3+(aq) + Mn^2+(aq) + 4OH^-(aq)

Answer: (A)

Verified step by step guidance

1. Identify the oxidation and reduction components in the reaction. Aluminum (Al) is oxidized to Al³⁺, and permanganate ion (MnO₄⁻) is reduced to Mn²⁺.
2. Write the oxidation half-reaction for aluminum: Al(s) → Al³⁺(aq) + 3e⁻. This shows aluminum losing three electrons.
3. Write the reduction half-reaction for permanganate: MnO₄⁻(aq) + 8H⁺(aq) + 5e⁻ → Mn²⁺(aq) + 4H₂O(l). This shows permanganate gaining five electrons in acidic conditions.
4. Balance the electrons between the oxidation and reduction half-reactions. Multiply the oxidation half-reaction by 5 and the reduction half-reaction by 3 to balance the electrons exchanged.
5. Combine the balanced half-reactions, ensuring that the electrons cancel out, and verify that all atoms and charges are balanced in the final equation.