Electrolysis of a solution of HSO 4- ions produces S 2 O 82- Assuming 75 % current efficiency, what current should be employed to achieve a production rate of 1 mol of S 2 O 82- per hour?

Question

Electrolysis of a solution of HSO 4- ions produces S 2 O 82- Assuming 75 % current efficiency, what current should be employed to achieve a production rate of 1 mol of S 2 O 82- per hour? (A) 71.47 A (B) 35.7 A (C) 142.96 A (D) 285.93 A

Tardigrade Admin · Accepted Answer

2 HSO 4- arrow S 2 O 82-+2 H ++2 e- So, required rate =1 mol S 2 O 82- / hr =2 mol HSO 4- / hr =(2 × 96500 C /3600 s )=(2 × 965/36) A ≃ 53.6 A So, required current =(4/3) × 53.6 A =71.47 A

Q. Electrolysis of a solution of $H S O_{4}^{-}$ ions produces $S_{2} O_{8}^{2 -}$ Assuming $75%$ current efficiency, what current should be employed to achieve a production rate of $1 m o l$ of $S_{2} O_{8}^{2 -}$ per hour?

$71.47 A$

$35.7 A$

$142.96 A$

$285.93 A$

$2 H S O_{4}^{-} \to S_{2} O_{8}^{2 -} + 2 H^{+} + 2 e^{-}$

So, required rate $= 1 m o l S_{2} O_{8}^{2 -} / h r = 2 m o l H S O_{4}^{-} / h r$

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

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

Q. Electrolysis of a solution of HSO4−​ ions produces S2​O82−​ Assuming 75% current efficiency, what current should be employed to achieve a production rate of 1mol of S2​O82−​ per hour?

71.47A

35.7A

142.96A

285.93A