What is the number of electrons transferred c | vedclass

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Question

(C) In $KMnO_4$,the oxidation state of $Mn$ is calculated as: $1 + x + 4(-2) = 0$,so $x = +7$. In $Mn_2O_3$,the oxidation state of $Mn$ is calculated

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(C) In $KMnO_4$,the oxidation state of $Mn$ is calculated as: $1 + x + 4(-2) = 0$,so $x = +7$. In $Mn_2O_3$,the oxidation state of $Mn$ is calculated as: $2x + 3(-2) = 0$,so $2x = +6$,which means $x = +3$. The change in oxidation state per $Mn$ atom is $|7 - 3| = 4$. Since there are $2$ $Mn$ atoms in $Mn_2O_3$,the total number of electrons transferred is $2 	imes 4 = 8$. However,the question asks for the number of electrons transferred per $Mn$ atom involved in the reduction process from $KMnO_4$ to $Mn_2O_3$. Considering the stoichiometry $2KMnO_4 ightarrow Mn_2O_3$,the total change is $2 	imes (7 - 3) = 8$ electrons. Given the options provided,the change per $Mn$ atom is $4$.

What is the number of electrons transferred considering $Mn$ when $KMnO_4$ is converted into $Mn_2O_3$?

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