2021 H2 Chem P1 Q27 — Electrolysis Solution

The answer is B (\(6.05 \times 10^{23}\)).

\(\text{Al}^{3+} + 3\text{e}^- \rightarrow \text{Al}\)
\(n_{\text{Al}} = \frac{0.27}{27.0} = 0.0100\) mol, so \(n_{\text{e}^-} = 3 \times 0.0100 = 0.0300\) mol
\(F = \frac{Q}{n_{\text{e}^-}} = \frac{2904}{0.0300} = 96\,800\) C mol\(^{-1}\)
Avogadro's constant, \(L = \frac{F}{e} = \frac{96\,800 \text{ C mol}^{-1}}{1.60 \times 10^{-19} \text{ C}} = 6.05 \times 10^{23}\) mol\(^{-1}\)

• A is wrong: \(6.02 \times 10^{23}\) is the accepted value from the Data Booklet, but the question asks for the value these experimental figures give, which uses \(F\) = 96 800 C mol\(^{-1}\), not 96 500 C mol\(^{-1}\).
• C is wrong: \(1.82 \times 10^{24}\) results from taking \(n_{\text{e}^-}\) = 0.0100 mol (one electron per Al): \(\frac{2904}{0.0100} = 290\,400\) C mol\(^{-1}\) and \(\frac{290\,400}{1.60 \times 10^{-19}} = 1.82 \times 10^{24}\). Each \(\text{Al}^{3+}\) needs 3 electrons.
• D is wrong: \(2.02 \times 10^{23}\) is one third of the correct value; it results from taking \(n_{\text{e}^-}\) = 0.0900 mol (multiplying by 3 twice): \(\frac{2904}{0.0900} = 32\,270\) C mol\(^{-1}\) and \(\frac{32\,270}{1.60 \times 10^{-19}} = 2.02 \times 10^{23}\).