1. Write the cathode reaction

At the cathode, aluminium ions gain electrons:

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

So, 3 moles of electrons are required to deposit 1 mole of aluminium.

2. Calculate total charge passed

Given:

• Current, $I = 2\,\text{A}$
• Time, $t = 30\,\text{min} = 30 \times 60 = 1800\,\text{s}$

Using:

$Q = It$

$Q = 2 \times 1800 = 3600\,\text{C}$

3. Calculate moles of electrons passed

Using Faraday's constant, $F = 96500\,\text{C mol}^{-1}$:

$\text{moles of } e^- = \frac{Q}{F} = \frac{3600}{96500}$

$\approx 0.0373\,\text{mol}$

4. Calculate moles of aluminium deposited

Since 3 moles of electrons deposit 1 mole of Al:

$\text{moles of Al} = \frac{0.0373}{3} \approx 0.01243\,\text{mol}$

5. Calculate mass of aluminium deposited

Molar mass of aluminium = $27\,\text{g mol}^{-1}$

$m = nM = 0.01243 \times 27$

$m \approx 0.336\,\text{g}$

6. Match with options

The deposited mass is:

$\boxed{0.336\,\text{g}}$

So the correct option is C.