The charge required for the reduction of 1 t | vedclass

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(B) The reduction half-reaction for the conversion of $MnO_4^{-}$ to $MnO_2$ in an acidic medium is given by:$MnO_4^{-} + 4H^{+} + 3e^{-} \rightarrow

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$3$

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(B) The reduction half-reaction for the conversion of $MnO_4^{-}$ to $MnO_2$ in an acidic medium is given by:$MnO_4^{-} + 4H^{+} + 3e^{-} ightarrow MnO_2 + 2H_2O$From the stoichiometry of the balanced equation,$3 \ 	ext{moles}$ of electrons are required to reduce $1 \ 	ext{mole}$ of $MnO_4^{-}$.Since the charge of $1 \ 	ext{mole}$ of electrons is $1 \ F$,the total charge required is $3 \ F$.

The charge required for the reduction of $1 \ \text{mole}$ of $MnO_4^{-}$ to $MnO_2$ is (in $F$)

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