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

Tardigrade Admin · Accepted Answer

2 HSO 4- longrightarrow S 2 O 8 2-+2 H ++2 e - so required rate =1 mol / hr =2 mole of e - / hr =(2 × 96500 C /3600 sec )=(2 × 965/36) A ≃ 53.6 A so required current =(4/3) × 53.6 A =71.4 A

Q. Electrolysis of a solution of $H S O_{4}^{-}$ 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?

$2 H S O_{4}^{-} ⟶ S_{2} O_{8}^{2 -} + 2 H^{+} + 2 e^{-}$
so required rate $= 1 m o l / h r$
$= 2$ mole of $e^{-} / h r$
$= \frac{2 \times 96500 C}{3600 s e c} = \frac{2 \times 965}{36} A ≃ 53.6 A$
so required current $= \frac{4}{3} \times 53.6 A = 71.4 A$

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

2HSO4−​⟶S2​O8​2−+2H++2e− so required rate =1mol/hr =2 mole of e−/hr =3600sec2×96500C​=362×965​A≃53.6A so required current =34​×53.6A=71.4A

Q. Electrolysis of a solution of $H S O_{4}^{-}$ 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?

$2 H S O_{4}^{-} ⟶ S_{2} O_{8}^{2 -} + 2 H^{+} + 2 e^{-}$
so required rate $= 1 m o l / h r$
$= 2$ mole of $e^{-} / h r$
$= \frac{2 \times 96500 C}{3600 s e c} = \frac{2 \times 965}{36} A ≃ 53.6 A$
so required current $= \frac{4}{3} \times 53.6 A = 71.4 A$