ElectrochemistryHard

Question

Electrolysis of a solution of HSO4− ions produces S2O82−. Assuming 75% current efficiency, what current should be employed to achieve a production rate of 1 mole of S2O82− per hour?

Options

A.71.5 A
B.35.7 A
C.53.0 A
D.143 A

Solution

$2HSO_{4}^{-} \rightleftharpoons S_{2}O_{8}^{2 -} + 2H^{+} + 2e^{-}$

$\text{Now, }n_{eq} = \frac{Q}{F} \Rightarrow 1 \times 2 = \frac{i \times 3600 \times 0.75}{96500} $$$\therefore i = 71.48\text{ A}$$

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