Verified Solution • 2025–26 Session moderate

QJEEM25Cha4ea: JEE Main 2025 Chemistry - Class 12 JEEM25Cha4ea Solution: A solution of aluminium chloride is electrolysed for 30 minutes …

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 g mol−127 \mathrm{~g} \mathrm{~mol}^{-1} and 35.5 g mol−135.5 \mathrm{~g} \mathrm{~mol}^{-1} respectively. Faraday constant =96500C mol−1]\left.=96500 \mathrm{C} \mathrm{~mol}^{-1}\right]

Multiple Choice Options

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

Step-by-Step Verified Solution

Al3++3e−→Al\text{Al}^{3+} + 3e^- \rightarrow \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⋅t=2 A⋅(30×60 s)=3600 CQ = 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(Al)=Q3F=36003×96500≈0.01242 moln(\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 g/molM = 27\,\text{g/mol} ):

m(Al)=n(Al)×M=0.01242×27≈0.335 gm(\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 g0.336\, \text{g}

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