In alkaline medium, the reduction of permanganate anion involves a gain of __________ electrons.

Solution

Chemistry ma’am · 4 steps

In alkaline medium permanganate stops at MnO₂, so track Mn going from +7 down to +4.

1. Find the product. In acidic medium MnO₄⁻ goes all the way to Mn²⁺. In alkaline (and neutral) medium it is reduced only as far as the brown solid MnO₂.
2. Oxidation states. In MnO₄⁻, x + 4(-2) = -1, so Mn is +7. In MnO₂, x + 2(-2) = 0, so Mn is +4. The drop is 7 - 4 = 3.

   MnO₄⁻ + 2H₂O + 3e⁻ → MnO₂ + 4OH⁻

3. Check the balance. Oxygen is 6 on each side and hydrogen is 4 on each side. The charge is -1 - 3 = -4 on the left and -4 from 4OH⁻ on the right, so 3 electrons is correct.
4. So. Each permanganate ion gains 3 electrons in alkaline medium.

The common slip: 5. Using the acidic-medium product Mn²⁺ gives 5. That change from +7 to +2 happens only in acid, not in alkaline medium.

The idea to keep. The number of electrons MnO₄⁻ gains depends on the medium: 5 in acid (Mn²⁺), 3 in neutral or alkaline (MnO₂). Always fix the product first, then count the change in oxidation state.

Answer from the NTA final answer key. Solution by PrepWiser.