An electrolytic cell is set up to purify a piece of copper that contains silver and zinc as impurities - NSC Physical Sciences - Question 9 - 2023 - Paper 2

At electrode Q, copper ions ($ext{Cu}^{2+}$) in the electrolyte gain electrons and are reduced to form solid copper. The half-reaction can be written as:

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

The electrons flow from Electrode Q to Electrode R in the external circuit.

To find the current, we need to follow these steps:

1. Calculate the number of moles of copper: The molar mass of copper (Cu) is approximately 63.5 g/mol. $n = \frac{m}{M} = \frac{16 g}{63.5 g/mol} \approx 0.2524 \text{ mol}$

2. Calculate the number of electrons needed: Each mole of copper requires 2 moles of electrons for reduction (from $ext{Cu}^{2+}$ to $ext{Cu}$): $\text{Electrons} = n \times 2 = 0.2524 \text{ mol} \times 2 = 0.5048 \text{ mol}$

3. Convert moles of electrons to number of electrons: Using Avogadro's number, $N_A = 6.02 \times 10^{23} \text{ atoms/mol}$: $n(e^-) = 0.5048 \text{ mol} \times 6.02 \times 10^{23} \text{ electrons/mol} \approx 3.04 \times 10^{23} \text{ electrons}$

4. Calculate total charge (Q) passed: Each electron has a charge of approximately $1.6 \times 10^{-19} \text{ C}$: $Q = n(e^-) \times e = 3.04 \times 10^{23} \times 1.6 \times 10^{-19} \approx 48,640 \text{ C}$

5. Calculate the current (I): The time (t) the cell operates is 5 hours, or 18,000 seconds: $I = \frac{Q}{t} = \frac{48640 \text{ C}}{18000 \text{ s}} \approx 2.70 \text{ A}$

Silver is not oxidised during the electrolysis because it is a weaker reducing agent compared to copper and zinc. The voltage applied may not be sufficient to oxidise silver ions ($ext{Ag}^+$) in the presence of more reactive ions like $ext{Cu}^{2+}$ and $ext{Zn}^{2+}$.