How many faraday are needed to reduce a mole | vedclass

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(B) The reduction of $1 \, 	ext{mole}$ of $MnO_4^-$ to $Mn^{2+}$ is represented by the following half-reaction:$MnO_4^- + 8H^+ + 5e^- ightarrow Mn^

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$5$

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(B) The reduction of $1 \, 	ext{mole}$ of $MnO_4^-$ to $Mn^{2+}$ is represented by the following half-reaction:$MnO_4^- + 8H^+ + 5e^- ightarrow Mn^{2+} + 4H_2O$In $MnO_4^-$,the oxidation state of $Mn$ is $+7$.In $Mn^{2+}$,the oxidation state of $Mn$ is $+2$.The change in oxidation state is $7 - 2 = 5$.Since $5 \, 	ext{moles}$ of electrons are required for the reduction of $1 \, 	ext{mole}$ of $MnO_4^-$,the amount of electricity required is $5 \, 	ext{Faraday}$.

How many faraday are needed to reduce a mole of $MnO_4^-$ to $Mn^{2+}$?

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