In basic solution, balance MnO4- → MnO2 with H2O and OH-; which is the correct half-reaction?

In basic solution, balance MnO4- → MnO2 with H2O and OH-; which is the correct half-reaction?

• A. MnO4- + 4 OH- -> MnO2 + 2 H2O + 3 e-
• B. MnO4- + 3 e- + H2O -> MnO2 + OH-
• C. MnO4- + H2O -> MnO2 + OH- + e-
• D. MnO4- + 2 H2O + 3 e- -> MnO2 + 4 OH-

Answer: (D)

In basic solution, balancing a reduction half-reaction uses water to supply oxygen and hydroxide to balance hydrogen, with electrons added to balance the charge. Here, manganese goes from MnO4- (+7) to MnO2 (+4), so three electrons must be gained. To balance oxygen, add two water molecules to the side that needs oxygen balance; the permanganate side has more oxygen than MnO2, so two H2O are placed on the manganese side. That gives two extra oxygens and four hydrogens to work with. To account for those hydrogens in basic conditions, add hydroxide on the opposite side, yielding four OH- on the product side to match the four hydrogens from the two water molecules. Finally, add three electrons to the manganese-dioxidation side to balance the charge. This leads to the balanced half-reaction: MnO4- + 2 H2O + 3 e- -> MnO2 + 4 OH-. It checks out atom-for-atom and charge-for-charge, and it reflects the reduction of MnO4- to MnO2 in basic solution.

In basic solution, balancing a reduction half-reaction uses water to supply oxygen and hydroxide to balance hydrogen, with electrons added to balance the charge. Here, manganese goes from MnO4- (+7) to MnO2 (+4), so three electrons must be gained.

To balance oxygen, add two water molecules to the side that needs oxygen balance; the permanganate side has more oxygen than MnO2, so two H2O are placed on the manganese side. That gives two extra oxygens and four hydrogens to work with. To account for those hydrogens in basic conditions, add hydroxide on the opposite side, yielding four OH- on the product side to match the four hydrogens from the two water molecules. Finally, add three electrons to the manganese-dioxidation side to balance the charge.

This leads to the balanced half-reaction: MnO4- + 2 H2O + 3 e- -> MnO2 + 4 OH-. It checks out atom-for-atom and charge-for-charge, and it reflects the reduction of MnO4- to MnO2 in basic solution.