(D) The oxidation state of $Mn$ in $MnO_4^-$ is $+7$ and in $Mn^{2+}$ is $+2$. The change in oxidation state is $7 - 2 = 5$. Since the oxidation state

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(D) The oxidation state of $Mn$ in $MnO_4^-$ is $+7$ and in $Mn^{2+}$ is $+2$. The change in oxidation state is $7 - 2 = 5$. Since the oxidation state decreases,it is a reduction process involving $5$ electrons. The balanced half-reaction is: $MnO_4^- + 8H^+ + 5e^- \to Mn^{2+} + 4H_2O$.