# Consider the following reaction

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.