What is the coefficient for the nitrate ion (NO2−) when the redox reaction is balanced in basic solution? MnO4− (aq) + NO2− (aq) → MnO2 (s) + NO3− (aq)

• A. 1
• B. 2
• C. 3
• D. 4

Answer: (A)

In basic solution, balancing a redox reaction uses the half-reaction method: split into oxidation and reduction parts, balance atoms and charges within each half (using H2O and OH− as needed in base), balance the electrons, then scale the halves so the electrons cancel and combine them.

For this reaction, the reduction half is MnO4− reducing to MnO2. Balancing in base gives:

MnO4− + 2 H2O + 3 e− → MnO2 + 4 OH−.

The oxidation half is nitrite to nitrate. In base, balancing leads to:

NO2− + 2 OH− → NO3− + H2O + 2 e−.

To cancel electrons, multiply the first half by 2 and the second by 3:

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

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

Add and cancel electrons, water, and OH− to obtain the overall balanced equation:

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

Thus, the coefficient for the nitrite ion is 3, and the coefficient for the nitrate ion is also 3.