The value of x in the partial redox equation | vedclass

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Question

(A) To balance the charge in the half-reaction,we calculate the oxidation state of $Mn$ in $MnO_4^-$. Let the oxidation state be $y$. $y + 4(-2) = -1$

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(A) To balance the charge in the half-reaction,we calculate the oxidation state of $Mn$ in $MnO_4^-$. Let the oxidation state be $y$. $y + 4(-2) = -1$,so $y = +7$. In the product side,$Mn$ is in the $+2$ oxidation state. The change in oxidation state is $7 - 2 = 5$. Therefore,$5$ electrons are gained by $Mn$ to reduce from $+7$ to $+2$. The balanced equation is $MnO_4^- + 8H^{+} + 5e^- \rightleftharpoons Mn^{2+} + 4H_2O$. Thus,the value of $x$ is $5$.

The value of $x$ in the partial redox equation $MnO_4^- + 8H^{+} + xe^- \rightleftharpoons Mn^{2+} + 4H_2O$ is

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