JEE MAIN - Chemistry (2025 - 22nd January Morning Shift - No. 9)
A solution of aluminium chloride is electrolysed for 30 minutes using a current of 2 A . The amount of the aluminium deposited at the cathode is __________ .
[Given : molar mass of aluminium and chlorine are $27 \mathrm{~g} \mathrm{~mol}^{-1}$ and $35.5 \mathrm{~g} \mathrm{~mol}^{-1}$ respectively. Faraday constant $\left.=96500 \mathrm{C} \mathrm{~mol}^{-1}\right]$
Explanation
$$ \text{Al}^{3+} + 3e^- \rightarrow \text{Al} $$
The mass of aluminium deposited can be calculated using Faraday’s laws of electrolysis. The steps are as follows:
Compute the total charge passed:
$$ Q = I \cdot t = 2\,\text{A} \cdot (30 \times 60\,\text{s}) = 3600\,\text{C} $$
Determine the moles of aluminium deposited. Since three electrons are required to deposit one mole of aluminium, the number of moles is given by:
$$ n(\text{Al}) = \frac{Q}{3F} = \frac{3600}{3 \times 96500} \approx 0.01242\,\text{mol} $$
Finally, calculate the mass of aluminium using its molar mass ($M = 27\,\text{g/mol}$):
$$ m(\text{Al}) = n(\text{Al}) \times M = 0.01242 \times 27 \approx 0.335\,\text{g} $$
Thus, the amount of aluminium deposited at the cathode is approximately
$$ 0.336\, \text{g} $$
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