Question

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

• A. $71.47 \,A$
• B. $35.7 \,A$
• C. $142.96\, A$
• D. $285.93 \,A$

Answer 1

Answer: (A)

$2 HSO _{4}^{-} \rightarrow S _{2} O _{8}^{2-}+2 H ^{+}+2 e^{-}$

So, required rate $=1\, mol \,S _{2} O _{8}^{2-} / hr =2\, mol\, HSO _{4}^{-} / hr$

$=\frac{2 \times 96500\, C }{3600 s }=\frac{2 \times 965}{36} A \simeq 53.6 \,A$

So, required current $=\frac{4}{3} \times 53.6 \,A =71.47\, A$