A-Level Chemistry May/June 2025 Q5(e): A solution containing [Cu(H2O)6]2+ is electrolysed for 5.00 hours using a constant elec…

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  Step 1: Calculate the moles of copper formed.

  The molar mass of copper ($A_r$ of Cu) is 63.5 g mol⁻¹.

  $\text{Moles of Cu} = \frac{\text{mass}}{A_r} = \frac{0.764 \text{ g}}{63.5 \text{ g mol}^{-1}} = 0.012031... \text{ mol}$

  (Moles of Cu = 0.0120 mol to 3 s.f.)

  Step 2: Calculate the total quantity of charge passed.

  The half-equation for the reduction of copper(II) ions is:

  $\text{Cu}^{2+}(\text{aq}) + 2\text{e}^- \rightarrow \text{Cu}(\text{s})$

  From the stoichiometry, 2 moles of electrons are required to produce 1 mole of copper.

  $\text{Moles of electrons} = 2 \times \text{moles of Cu} = 2 \times 0.012031... = 0.024062... \text{ mol}$

  The relationship between charge ($Q$), moles of electrons ($n(e^-)$), and the Faraday constant ($F$) is $Q = n(e^-) \times F$. The Faraday constant is 96500 C mol⁻¹.

  $Q = 0.024062... \text{ mol} \times 96500 \text{ C mol}^{-1} = 2322.08... \text{ C}$

  (Charge = 2322 C to 4 s.f.)

  Step 3: Calculate the electric current.

  The relationship between charge ($Q$), current ($I$), and time ($t$) is $Q = I \times t$.

  First, convert the time from hours to seconds:

  $t = 5.00 \text{ hours} \times 60 \text{ mins/hr} \times 60 \text{ s/min} = 18000 \text{ s}$

  Now, rearrange the formula to find the current, $I$:

  $I = \frac{Q}{t} = \frac{2322.08... \text{ C}}{18000 \text{ s}} = 0.12900... \text{ A}$

  Giving the answer to three significant figures as requested:

  $\text{Current, } I = 0.129 \text{ A}$

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  How the marks are awarded

  • M1 — For correctly calculating the moles of copper deposited (0.0120 mol) by dividing the given mass by the molar mass of copper.
  • M2 — For correctly calculating the total charge required (2322 C). This involves using the 2:1 mole ratio from the half-equation to find the moles of electrons, and then multiplying by the Faraday constant.
  • M3 — For correctly calculating the final current (0.129 A). This requires converting the time to seconds, rearranging Q = It, and giving the final answer to three significant figures.

  Common mistakes

  • Using an incorrect mole ratio of electrons to copper, such as 1:1 instead of 2:1, which leads to an answer of 0.0645 A.
  • Failing to convert the time from hours into seconds, for example by dividing the charge by 5.00, leading to a vastly incorrect current.
  • Incorrectly rounding the final answer or not giving it to the required three significant figures, e.g., writing 0.13 A or 0.1290 A.
  • Confusing the two main equations, for example by multiplying charge by time instead of dividing (I = Q × t).

  Examiner tip: Master electrolysis calculations by remembering to link the two fundamental charge equations, $Q = It$ and $Q = n(e^{-})F$, via the moles of electrons determined from the reaction stoichiometry.

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