Question

Electrolysis of a solution of $HSO _{4}{ }^{-}$ions produces $S _{2} O _{8}^{2-}$. Assuming $75 \%$ current efficiency, what current (in amp.) should be employed to achieve a production rate of 1 mole of $S _{2} O _{8}^{2-}$ per hour?

Answer 1

$2 HSO _{4}^{-} \longrightarrow S _{2} O _{8}{ }^{2-}+2 H ^{+}+2 e ^{-}$
so required rate $=1\, mol / hr$
$=2$ mole of $e ^{-} / hr$
$=\frac{2 \times 96500\, C }{3600\, \sec }=\frac{2 \times 965}{36} A \simeq 53.6\, A$
so required current $=\frac{4}{3} \times 53.6\, A =71.4\, A$