JEE Chemistry 2023 Question with Solution

Given: The reaction involves $\mathrm{KMnO_4}$ and potassium iodide in acidic medium.

Find: The total change in the oxidation state of manganese.

In acidic medium, potassium permanganate is reduced according to the half-reaction:

$$\mathrm{MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O}$$

The oxidation state of manganese in $\mathrm{MnO_4^-}$ is $+7$.

The oxidation state of manganese in $\mathrm{Mn^{2+}}$ is $+2$.

So, the change in oxidation state is:

$$+7 \rightarrow +2$$

Hence, the total decrease is:

$$7 - 2 = 5$$

Therefore, the total change in the oxidation state of manganese involved in the reaction is $5$.

Given: Manganese is present in $\mathrm{KMnO_4}$ in acidic medium.

Find: The total change in oxidation state of manganese.

In acidic medium, manganese in permanganate always changes from $+7$ to $+2$.

So directly:

$$\Delta \text{oxidation state} = 7 - 2 = 5$$

Therefore, the required numerical value is $5$.