When KMnO_4 acts as an oxidizing agent and ul | vedclass

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Question

(C) The oxidation state of $Mn$ in $KMnO_4$ is $+7$. The number of electrons transferred is the difference in oxidation states of $Mn$ in the reactant

Vedclass · Accepted Answer

$1, 3, 4, 5$

Vedclass · Answer

(C) The oxidation state of $Mn$ in $KMnO_4$ is $+7$. The number of electrons transferred is the difference in oxidation states of $Mn$ in the reactant and product.$1$. For $MnO_4^{2-}$: $Mn$ is in $+6$ state. Electrons transferred = $7 - 6 = 1$.$2$. For $MnO_2$: $Mn$ is in $+4$ state. Electrons transferred = $7 - 4 = 3$.$3$. For $Mn_2O_3$: $Mn$ is in $+3$ state. Since there are two $Mn$ atoms,total change = $2 	imes (7 - 3) = 8$. However,per $Mn$ atom basis or standard reaction stoichiometry for $KMnO_4$ reduction,the change is $4$ electrons per $Mn$ atom ($7$ to $3$).$4$. For $Mn^{2+}$: $Mn$ is in $+2$ state. Electrons transferred = $7 - 2 = 5$.Thus,the sequence is $1, 3, 4, 5$.

When $KMnO_4$ acts as an oxidizing agent and ultimately forms $MnO_4^{2-}$,$MnO_2$,$Mn_2O_3$,and $Mn^{2+}$,what will be the number of electrons transferred in each case,respectively?

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