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(C) In an acidic medium,the permanganate ion $(MnO_4^{-})$ acts as a strong oxidizing agent and reacts with hydrogen peroxide $(H_2O_2)$.The balanced

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$7 \rightarrow 2$

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(C) In an acidic medium,the permanganate ion $(MnO_4^{-})$ acts as a strong oxidizing agent and reacts with hydrogen peroxide $(H_2O_2)$.The balanced chemical equation for this reaction is: $2MnO_4^{-} + 5H_2O_2 + 6H^{+} \rightarrow 2Mn^{2+} + 5O_2 + 8H_2O$.In $MnO_4^{-}$,the oxidation state of $Mn$ is calculated as: $x + 4(-2) = -1$,which gives $x = +7$.In the product $Mn^{2+}$,the oxidation state of $Mn$ is $+2$.Therefore,the oxidation state of $Mn$ changes from $+7$ to $+2$.

What is the change in the oxidation state of $Mn$ in the reaction of $MnO_4^{-}$ with $H_2O_2$ in acidic medium?

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