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# In acidic medium,$MnO_4^-$ is converted to $Mn^{2+}$ when acts as an oxidizing agent.The quantity of electricity required to reduce 0.05 mol of $MnO_4^-$ would be

$\begin{array}{1 1}0.01F\\0.05F\\0.25F\\0.15F\end{array}$

The reaction is $MnO_4^-+8H^++5e^-\rightarrow Mn^{2+}+4H_2O$
The quantity of electricity of electricity needed=(1 F $mol^{-1}$)(5$\times$ 0.05mol)=0.25 F