What is the change in the oxidation state of | vedclass

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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

Vedclass · Accepted Answer

$7 \rightarrow 2$

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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 $+2$.

Column-$I$	Column-$II$
$A$. $Cr_2O_7^{2-}$	$1$. $+3$
$B$. $MnO_4^{-}$	$2$. $+4$
$C$. $VO_3^{-}$	$3$. $+5$
$D$. $FeF_6^{3-}$	$4$. $+6$
	$5$. $+7$

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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Match the Column-$I$ with Column-$II$ for the oxidation states of the central atoms.

Column-$I$ Column-$II$

$A$. $Cr_2O_7^{2-}$ $1$. $+3$

$B$. $MnO_4^{-}$ $2$. $+4$

$C$. $VO_3^{-}$ $3$. $+5$

$D$. $FeF_6^{3-}$ $4$. $+6$

$5$. $+7$

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