JEE Chemistry 2025 Question with Solution

Given: Current is $2 \, \text{A}$, time is $30 \, \text{minutes} = 1800 \, \text{s}$, molar mass of aluminium is $27 \, \text{g mol}^{-1}$, and Faraday constant is $96500 \, \text{C mol}^{-1}$.

Find: The mass of aluminium deposited at the cathode.

For deposition of aluminium:

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

So, the number of electrons involved is $n = 3$.

Using Faraday's law of electrolysis:

$$m = \frac{MIt}{nF}$$

Substituting the values:

$$m = \frac{27 \times 2 \times 1800}{3 \times 96500}$$ $$m = \frac{97200}{289500} = 0.336 \, \text{g}$$

Therefore, the mass of aluminium deposited at the cathode is $0.336 \, \text{g}$. The correct option is C.

Given: Current is $2 \, \text{A}$, time is $30 \times 60 \, \text{s}$, molar mass of aluminium is $27 \, \text{g mol}^{-1}$, and Faraday constant is $96500 \, \text{C mol}^{-1}$.

Find: The amount of aluminium deposited.

The gram equivalent of aluminium deposited is calculated by:

$$w = \frac{It}{96500} \times \frac{27}{3}$$

where $I = 2 \, \text{A}$ and $t = 30 \times 60 \, \text{s}$.

Substituting the values:

$$w = \frac{2 \times 30 \times 60}{96500} \times \frac{27}{3}$$ $$w = 0.336 \, \text{g}$$

Thus, the correct answer is $0.336 \, \text{g}$, so the correct option is C.