2019 H2 Chem P1 Q30 — Electrolysis Solution

The answer is C (49.4 h).

Reading 'a total depth of 1000 atoms' as 1000 atoms deep on each side:
Number of Cu atoms along 0.1 m \(= \dfrac{0.1}{3.0 \times 10^{-12}} = 3.333 \times 10^{10}\)
Number of Cu atoms in one layer on one face \(= (3.333 \times 10^{10})^2 = 1.111 \times 10^{21}\)
Number of atoms deposited \(= 1.111 \times 10^{21} \times 1000 \times 2 \text{ sides} = 2.222 \times 10^{24}\)
Amount of Cu \(= \dfrac{2.222 \times 10^{24}}{6.02 \times 10^{23}} = 3.691\) mol
\(\text{Cu}^{2+}\text{(aq)} + 2\text{e}^- \rightarrow \text{Cu(s)}\), so amount of electrons \(= 2 \times 3.691 = 7.382\) mol
\(Q = It = n_\text{e}F\): \(4.0 \times t = 7.382 \times 9.65 \times 10^4\)
\(t = 1.781 \times 10^5 \text{ s} = \dfrac{178110}{3600} \text{ h} = 49.5\) h to 3 significant figures. The paper prints this value as 49.4 h (cut short rather than rounded), so the answer is option C.

• B (24.7 h) is not the keyed answer, but it follows if 'a total depth of 1000 atoms' is read as 1000 atoms in total over the two sides (500 on each side): the number of atoms, the charge and the time all halve, \(t = 8.906 \times 10^4 \text{ s} = 24.7\) h. Because the wording is ambiguous, some published solutions (NYJC) accept B as well.
• A is wrong: 178 s is the time to cover both sides with a single layer of atoms; it leaves out the depth of 1000 atoms.
• D is wrong: 98.9 h (98.95 h unrounded) is twice the answer, i.e. 2000 atoms deep on each side, more copper than the question asks for.