The number of electrons involved in the reduction of permanganate to manganese dioxide in acidic medium is _____________.

Solution

Chemistry ma’am · 4 steps

Track manganese only: it goes from +7 in permanganate to +4 in MnO₂, so count that drop.

1. Oxidation states. In MnO₄⁻, Mn is +7, since x + 4(-2) = -1. In MnO₂, Mn is +4, since x + 2(-2) = 0.
2. Electrons gained. The drop from +7 to +4 is 3 units, and there is one Mn atom on each side. So each permanganate ion takes up 3 electrons.

   MnO₄⁻ + 4H⁺ + 3e⁻ → MnO₂ + 2H₂O

3. Check the balance. Charge on the left is -1 + 4 - 3 = 0, matching the neutral right side. Oxygen is 4 on each side, and hydrogen is 4 on each side.
4. So. The reduction of permanganate to manganese dioxide involves 3 electrons.

The common slip: 5. Using the familiar MnO₄⁻ to Mn²⁺ change gives 5, but here the product is MnO₂ with Mn at +4, not +2.

The idea to keep. For permanganate, the electron count is set by where Mn ends up: Mn²⁺ takes 5, MnO₂ takes 3, MnO₄²⁻ takes 1. Read the product, then subtract oxidation states.

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