Electrolysis of a solution of HSO4^- ions produces S2O8^2-. Assuming 75% current efficiency, what cu — Electrochemistry Chemistry Question

💡 Solution & Explanation

**Step 1 - Write the balanced oxidation half-reaction** During electrolysis, the oxidation of hydrogen sulfate ions (\ce{HSO4-}) to peroxydisulfate ions (\ce{S2O8^{2-}}) occurs at the anode according to the following half-reaction: $$\ce{2HSO4- -> S2O8^{2-} + 2H+ + 2e-}$$ From the stoichiometry of this reaction, the production of $1\text{ mole}$ of \ce{S2O8^{2-}} requires $2\text{ moles}$ of electrons. Therefore, the change in oxidation number ($n$) per mole of product is: $$n = 2$$ **Step 2 - Determine the theoretical charge required** We calculate the theoretical charge ($Q_{\text{theoretical}}$) required to produce $1\text{ mole}$ of \ce{S2O8^{2-}} using Faraday's constant: $$\text{Formula: } Q_{\text{theoretical}} = n \times F$$ $$\text{Substitution: } Q_{\text{theoretical}} = 2\text{ mol} \times 96500\text{ C mol}^{-1}$$ $$\text{Calculation: } Q_{\text{theoretical}} = 193000\text{ C}$$ **Step 3 - Calculate the actual charge required accounting for efficiency** Because the process has a current efficiency ($\eta$) of $75\%$, the actual charge ($Q_{\text{actual}}$) that needs to be supplied is greater than the theoretical charge: $$\text{Formula: } Q_{\text{actual}} = \frac{Q_{\text{theoretical}}}{\eta}$$ $$\text{Substitution: } Q_{\text{actual}} = \frac{193000\text{ C}}{0.75}$$ $$\text{Calculation: } Q_{\text{actual}} = 257333.33\text{ C}$$ **Step 4 - Calculate the required current** Since the production rate is specified per hour, the time ($t$) of electrolysis is $1\text{ hour}$ or $3600\text{ seconds}$. We find the current ($I$) using the relationship between charge, current, and time: $$\text{Formula: } I = \frac{Q_{\text{actual}}}{t}$$ $$\text{Substitution: } I = \frac{257333.33\text{ C}}{3600\text{ s}}$$ $$\text{Calculation: } I = 71.48\text{ A} \approx \boxed{71.5\text{ A}}$$ This corresponds to **Option (A)**.