Is the redox equation MnO4- + 8 H+ + 5 Fe2+ → Mn2+ + 5 Fe3+ + 4 H2O balanced?

• A. No, atoms balanced but charges not
• B. Oxygen not balanced
• C. No, too many Fe3+ produced
• D. Yes

Answer: (D)

Balancing redox reactions in acidic solution hinges on equal electrons being transferred between the oxidation and reduction half-reactions. Here, permanganate is reduced from MnO4– (Mn in +7) to Mn2+ (Mn in +2), which requires 5 electrons. Each Fe2+ is oxidized to Fe3+ (Fe2+ → Fe3+ + e−), releasing one electron. To match the 5 electrons needed, five Fe2+ ions are oxidized, providing 5 e−.

Write the two half-reactions and combine them:

• Reduction: MnO4– + 8 H+ + 5 e− → Mn2+ + 4 H2O

• Oxidation: 5 Fe2+ → 5 Fe3+ + 5 e−

Adding them cancels the electrons and gives:

MnO4– + 8 H+ + 5 Fe2+ → Mn2+ + 5 Fe3+ + 4 H2O

Check atoms and charges: Mn and Fe balance (1 Mn, 5 Fe on each side). O balance with 4 in MnO4– and 4 H2O on the right. H balance with 8 H+ on the left and 4 H2O containing 8 H on the right. Net charge left is -1 + 8 + 10 = +17; right is +2 + 15 = +17. Everything matches, so the equation is balanced.