In balancing the basic equation MnO4− + NO2− → MnO2 + NO3−, what is the coefficient of NO3−?

• A. 1
• B. 5
• C. 7
• D. 3

Answer: (D)

Balancing redox in basic solution uses half-reactions, balancing atoms and charges with water and hydroxide, and then combining them so the electrons cancel.

Identify the two half-reactions. In this couple, manganese is reduced: MnO4− becomes MnO2, gaining 3 electrons. Nitrite is oxidized: NO2− becomes NO3−, releasing 2 electrons.

Half-reaction for the manganese species in base:

MnO4− + 2 H2O + 3 e− → MnO2 + 4 OH−

Half-reaction for the nitrogen species in base:

NO2− + 2 OH− → NO3− + 2 e− + H2O

To balance electrons, multiply the manganese half-reaction by 2 and the nitrite half-reaction by 3, then add:

2 MnO4− + 4 H2O + 6 e− → 2 MnO2 + 8 OH−

3 NO2− + 6 OH− → 3 NO3− + 6 e− + 3 H2O

Cancel the electrons and simplify by canceling waters and OH− where possible, yielding:

2 MnO4− + NO2− + H2O → 2 MnO2 + 3 NO3− + 2 OH−

Thus, the coefficient of NO3− is 3.