Chemistry · Electrochemistry

JEE Main 2025 — 22 January, Morning Shift — Question 26

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 $\left.=96500 \mathrm{C} \mathrm{mol}^{-1}\right]$ .

Option A:
1.660 g
Option B:
1.007 g
Option C:
0.336 g
Correct
Option D:
0.441 g

Answer: C

Step-by-step solution

Correct option is: (3) $0.336\,\mathrm{g}$

\mathrm{Al^{3+} + 3e^- \rightarrow Al}

Mole of $e^- = \frac{It}{F} = \frac{2 \times 30 \times 60}{96500}$

Mass of Al deposited = $\frac{1}{3} \times \frac{2 \times 30 \times 60}{96500} \times 27 = 0.336\,\mathrm{g}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 22 January, Morning Shift
Subject: Chemistry
Chapter: Electrochemistry
Topic: Faraday's Laws