Question

Electrolysis of aqueous solution of $\text{CuSO}_4$ is carried out, where $300 \text{ mg}$ of copper is deposited (atomic mass of $\text{Cu} = 63.54$). After this $600 \text{ milli amp}$. current is further passed for $28 \text{ minutes}$. Calculate total volume of $\text{O}_2$ released (in $\text{ml}$). (Given $1 \text{ mole of gas occupy } 22.4 \text{ litre}$)

• A. 111
• B. 100
• C. 90
• D. 122

Hint:

Faraday's laws state that the quantity of substance discharged at an electrode is proportional to the equivalents of electricity passed. $Q_{\text{total}}$ is summed for sequential processes.

Answer 1

The Correct Option is A

1. Moles of electrons passed in Part 1 ($Q_1$):
$Q_1 = \frac{\text{Mass of } \text{Cu}}{\text{Equivalent mass of } \text{Cu}} = \frac{300 \times 10^{-3} \text{ g}}{63.54/2 \text{ g}} \approx 9.443 \times 10^{-3} \text{ F}$.
2. Moles of electrons passed in Part 2 ($Q_2$):
$Q_2 = \frac{I t}{F} = \frac{0.6 \text{ A} \times (28 \times 60) \text{ s}}{96500 \text{ C/F}} \approx 10.446 \times 10^{-3} \text{ F}$.
Total electrons $Q_{\text{total}} = Q_1 + Q_2 \approx 19.889 \times 10^{-3} \text{ F}$.
3. Moles of $\text{O}_2$ released: $2\text{H}_2\text{O} \rightarrow \text{O}_2 + 4\text{H}^+ + 4\text{e}^-$. ($z=4$).
$n_{\text{O}_2} = \frac{Q_{\text{total}}}{4} \approx \frac{19.889 \times 10^{-3}}{4} \approx 4.972 \times 10^{-3} \text{ mol}$.
Volume of $\text{O}_2$ (at $\text{STP}$): $V = n_{\text{O}_2} \times 22.4 \text{ L/mol}$.
$V = 4.972 \times 10^{-3} \times 22.4 \text{ L} \approx 0.11137 \text{ L} = 111.37 \text{ ml}$.
$V \approx 111 \text{ ml}$.