Question

In the following reaction, which component acts as an oxidizing agent? 10 I− (aq) + 2 MnO4− (aq) + 16 H+ (aq) → 5 I2 (s) + 2 Mn2+ (aq) + 8 H2O (l)

Answer 1

Answer:

The oxidizing agent is the MnO₄⁻

Explanation:

This is the redox reaction:

10 I⁻ (aq) + 2 MnO₄⁻ (aq) + 16 H⁺ (aq) → 5 I₂ (s) + 2 Mn²⁺ (aq) + 8 H2O (l)

Let's determine the oxidation and the reduction.

I⁻ acts with -1 in oxidation state and changes to 0, at I₂.

All elements in ground state has 0 as oxidation state.

As the oxidation state has increased, this is the oxidation, so the iodide is the reducing agent.

In the permanganate (MnO₄⁻), Mn acts with +7 in oxidation state and decreased to Mn²⁺. As the oxidation state is lower, we talk about the reduction. Therefore, the permanganate is the oxidizing agent because it oxidizes iodide to iodine