JEE Main 2025 — Electrochemistry Question with Solution

$${\text{Al}}^{3+}+3e^{-}\to \text{Al}$$

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

$$0.336\text{g}$$