What is the reduction half-reaction in the MnO4- / C2O4^2- redox reaction in acidic solution?

• A. MnO4- + 8 H+ + 5 e- → Mn2+ + 4 H2O
• B. MnO4- + 4 H+ + e- → Mn3+ + H2O
• C. MnO4- + 2 e- → MnO2
• D. MnO4- + H+ + e- → MnO2

Answer: (A)

In acidic solution, permanganate (MnO4-) is reduced to Mn2+. That means the manganese oxidation state goes from +7 down to +2, so it must gain 5 electrons. To balance atoms and charge, you add protons to supply hydrogen and water to balance oxygen. You end up with MnO4- + 8 H+ + 5 e- → Mn2+ + 4 H2O. Check the balance: left side has MnO4- (Mn with +7), eight H+ provides eight hydrogens, five electrons bring in negative charge, totaling a +2 charge; right side has Mn2+ (+2) plus four neutral water molecules, which also balances to +2. The four water molecules provide the necessary hydrogens and oxygen balance.

This is the standard reduction product for permanganate in strongly acidic conditions; other forms would imply a different manganese product (like Mn3+ or MnO2) and don’t balance correctly for the acidic medium.