Question

A solution of aluminium chloride is electrolyzed for 30 minutes using a current of 2A. The amount of the aluminium deposited at the cathode is _________
 

• A. 1.660 g
• B. 1.007 g
• C. 0.336 g
• D. 0.441 g

Hint:

To find the amount of substance deposited during electrolysis, use the formula \( w = \frac{It}{96500} \times \frac{M}{n} \), where \( M \) is the molar mass and \( n \) is the valency.

Answer 1

The Correct Option is C

The gram equivalent of Al deposited is given by: \[ w = \frac{I t}{96500} \times \frac{27}{3} \] where \( I = 2 \, \text{A}, \, t = 30 \times 60 \, \text{s} \). Substituting the values: \[ w = \frac{2 \times 30 \times 60}{96500} \times \frac{27}{3} = 0.336 \, \text{g} \] Thus, the correct answer is \( \boxed{0.336 \, \text{g}} \).

Answer 2

Given: The current is 2A, time is 30 minutes, the molar mass of aluminium is 27 g/mol, Faraday constant is 96500 C/mol, and the number of electrons involved in the deposition of aluminium is 3 (since aluminium forms Al³⁺ ions in the reaction).

Formula: The formula to calculate the mass of the substance deposited is given by Faraday's law of electrolysis:

\( m = \frac{M \times I \times t}{n \times F} \)

Where:
- \( m \) is the mass of the substance deposited (in grams),
- \( M \) is the molar mass of the substance (in grams per mole),
- \( I \) is the current (in amperes),
- \( t \) is the time (in seconds),
- \( n \) is the number of electrons involved in the reaction,
- \( F \) is Faraday's constant (96500 C/mol).

Substitute the known values:

\( m = \frac{27 \times 2 \times 1800}{3 \times 96500} \)

Step-by-step calculation:

\( m = \frac{97200}{289500} = 0.336 \, \text{g} \)

Conclusion: The mass of aluminium deposited at the cathode is 0.336 g, which corresponds to option (3).