When KMnO_4 acts as a strong oxidizing agent in an acidic medium,...

In an acidic medium, potassium permanganate ($\mathrm{KMnO}_4$) acts as a strong oxidizing agent and gets reduced to manganese(II) ion ($\mathrm{Mn}^{2+}$).

The half-reaction is:
$\mathrm{MnO}_4^- + 8\mathrm{H}^+ + 5\mathrm{e}^- \rightarrow \mathrm{Mn}^{2+} + 4\mathrm{H}_2\mathrm{O}$

The oxidation state of manganese in $\mathrm{KMnO}_4$ is $+7$. In the product $\mathrm{Mn}^{2+}$, the oxidation state is $+2$.

Thus, the oxidation state of manganese changes from $+7$ to $+2$.

Answer: $+2$