b
The reaction given is: \(MnO _4^{-}+8 H ^{+}+5 e ^{-} \rightarrow Mn ^{2+}+4 H _2 O\)

In this reaction, \(1\, mole\) \(MnO _4{ }^{-}\)reduces to form \(Mn ^{2+}\) and for this reaction \(5 F e^{-}\)are required.

\(MnO _4{ }^{-} \rightarrow Mn ^{2+}\)

\(+7 \quad\quad+2\)

Now,

\(1\, mole \,MnO _4^{-}=5 \,F\)

\(5\, mole \,MnO _4^{-}=(5 \times 5) \,F\)

\(=25 \,F\)

Therefore, the quantity of electricity required to reduce five moles of \(MnO _4{ }^{-}\) is \(25 \,F\).