Question

Electrolysis of a solution of $MnS{O}_4$ in aqueous sulphuric acid is a method for the preparation of $Mn{O}_2$ as the per reaction,
$M{n}^{2+}(aq.)+2H_{2}O\to Mn{O}_2+2H^{+}(aq.)+H_2$
Passing a current of $27A$ for $24$ hours gives one kg of $Mn{O}_3$. What is the value of current efficiency? Write the reactions taking place at the cathode and at the anode.

Answer 1

Apply $w=\frac{E\times i\times t}{96500}$
$1000=\frac{87\times i\times 24\times 60\times 60}{2\times 96500}$
$i=25.6ampere$
Current efficiency $=\frac{25.6}{27.0}\times 100=94.8$%
Reactions:
$Anode:M{n}^{2+}\to M{n}^{4+}+2e^{-}$
$Cathode:2H^{+}+2e^{-}\to H_2$.