The total change in the oxidation state of manganese involved in the reaction of KMnO₄ and potassium iodide in the acidic medium is ____________.

Solution

Chemistry ma’am · 4 steps

Acidic medium means permanganate goes all the way to Mn²⁺, so track one Mn from +7 to +2.

1. Start in permanganate. In MnO₄⁻, let Mn be x. Then x + 4(-2) = -1, so x = +7.
2. Product in acid. In acidic medium, MnO₄⁻ is reduced to Mn²⁺, and the iodide ions are oxidised to I₂.

   2MnO₄⁻ + 10I⁻ + 16H⁺ → 2Mn²⁺ + 5I₂ + 8H₂O

3. Change per Mn. Each Mn goes from +7 to +2. That is a fall of 7 - 2 = 5 in oxidation state, which matches the 5 electrons it gains.
4. So. The change in the oxidation state of manganese is 5.

The common slip: 10. Adding up the change for both Mn atoms in the balanced equation gives 10. The question asks for the change in the oxidation state of manganese, which is 5 per Mn.

The idea to keep. Permanganate's product depends on the medium: acidic gives Mn²⁺ (change 5), neutral or weakly basic gives MnO₂ (change 3), strongly basic gives MnO₄²⁻ (change 1).

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