Electrolysis of a solution containing HSO4− ions produces S2O82− ions. Only 75% of the current passed is used for this reaction (current efficiency 75%). What current must be passed to produce 1 mol of S2O82− per hour? (1 F = 96500 C mol−1)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 71.5 A
By Faraday’s first law, the amount of product is proportional to the charge passed, and one faraday (F) is the charge on one mole of electrons. At the anode, hydrogensulphate ions are oxidised to peroxodisulphate ions:
\( \displaystyle \ce{2HSO4^- -> S2O8^2- + 2H^+ + 2e^-} \)
So 1 mol of \( \ce{S2O8^2-} \) needs 2 mol of electrons (2 F). Charge needed in one hour:
\( \displaystyle Q=2\times96500\,\mathrm{C}=193000\,\mathrm{C} \)
Current that must go into this reaction (t = 1 h = 3600 s):
\( \displaystyle I_{used}=\frac{Q}{t}=\frac{193000\,\mathrm{C}}{3600\,\mathrm{s}}=53.6\,\mathrm{A} \)
Only 75% of the current passed does this work, so the current passed must be larger:
\( \displaystyle I=\frac{I_{used}}{0.75}=\frac{53.6\,\mathrm{A}}{0.75} \)
\( \displaystyle I=71.5\,\mathrm{A} \)