Electrolysis of a solution of HSO₄^– ions produces S₂O₈^–2. Assuming 75% current efficiency, what current should be employed to achieve a production rate of 1 mole S₂O₈ per hour?

71.5 A

The correct answer is option 2. 71.5 A.

To determine the required current to produce \( \text{S}_2\text{O}_8^{2-} \) at a rate of 1 mole per hour with 75% current efficiency, we need to follow these steps:

The electrochemical reaction for the formation of \( \text{S}_2\text{O}_8^{2-} \) from \( \text{HSO}_4^- \) is:

\(2\text{HSO}_4^- \rightarrow \text{S}_2\text{O}_8^{2-} + 2\text{H}^+ + 2e^-\)

From the reaction, 2 moles of electrons are required to produce 1 mole of \( \text{S}_2\text{O}_8^{2-} \).

Using Faraday's constant \( (F = 96485 \text{ C/mol}) \):

The charge required for 1 mole of \( \text{S}_2\text{O}_8^{2-} \) is:

\(Q = n \times F = 2 \times 96485 \text{ C/mol} = 192970 \text{ C}\)

Given the current efficiency is 75%, only 75% of the applied current is used effectively for the reaction. Therefore, the actual charge required \( (Q_{\text{actual}}) \) is:

\(Q_{\text{actual}} = \frac{192970 \text{ C}}{0.75} = 257293.33 \text{ C}\)

The production rate is 1 mole per hour, so we need to calculate the current \( (I) \):

\(I = \frac{Q_{\text{actual}}}{\text{time}}\)

Since the time is 1 hour \( = 3600 \text{ seconds} \):

\(I = \frac{257293.33 \text{ C}}{3600 \text{ s}} = 71.47 \text{ A}\)

Rounding to the nearest option, the required current is: \(71.5 \text{ A}\)

Therefore, the correct answer is \( 71.5 \text{ A} \).