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(D) The reaction of hydrogen peroxide $(H_2O_2)$ with potassium permanganate $(KMnO_4)$ in an alkaline medium is given by the equation: $2KMnO_4 + 3H_

Vedclass · Accepted Answer

$+4$

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(D) The reaction of hydrogen peroxide $(H_2O_2)$ with potassium permanganate $(KMnO_4)$ in an alkaline medium is given by the equation: $2KMnO_4 + 3H_2O_2 \rightarrow 2MnO_2 + 2KOH + 2H_2O + 3O_2$ In this reaction,the product containing manganese is manganese dioxide $(MnO_2)$. To find the oxidation state of $Mn$ in $MnO_2$,let it be $x$. Since the oxidation state of oxygen is $-2$,we have: $x + 2(-2) = 0$ $x - 4 = 0$ $x = +4$ Therefore,the oxidation state of $Mn$ is $+4$.

In alkaline medium,the reaction of hydrogen peroxide with potassium permanganate produces a compound in which the oxidation state of $Mn$ is $....$

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