Question:
Consider the following reaction
$\mathrm{MnO}_{4}^{-}+8 \mathrm{H}^{+}+5 \mathrm{e}^{-} \rightarrow \mathrm{Mn}^{+2}+4 \mathrm{H}_{2} \mathrm{O}, \mathrm{E}^{\circ}=1.51 \mathrm{~V}$
The quantity of electricity required in Faraday to reduce five moles of $\mathrm{MnO}_{4}^{-}$is
Solution:
(25)
$\mathrm{MnO}_{4}^{-}+8 \mathrm{H}^{+}+5 \mathrm{e}^{-} \rightarrow \mathrm{Mn}^{+2}+4 \mathrm{H}_{2} \mathrm{O}$
1 mole of $\mathrm{MnO}_{4}^{-}$require 5 faraday charge
5 moles of $\mathrm{MnO}_{4}^{-}$will require 25 faraday charge.
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