A solution of aluminium chloride is electrolysed for 30 minutes using a current of 2A. The amount of the aluminium deposited at the cathode is ___ .
[Given : molar mass of aluminium and chlorine are 27 g mol–1 and 35.5 g mol–1 respectively, Faraday constant = 96500 C mol–1].
- AOption (a)
1.660 g
- BOption (b)
1.007 g
- COption (c)
0.336 g
- DOption (d)
0.441 g
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Correct answer
0.336 g
Why this answer is correct
CONCEPT:
Faraday's Laws of Electrolysis
- Faraday's first law of electrolysis states that the amount of substance deposited at an electrode is directly proportional to the quantity of electricity passed through the electrolyte.
- The relationship can be expressed as:
m = (ItM)/(nF)
where m is the mass of the substance deposited, I is the current, t is the time, M is the molar mass of the substance, n is the number of electrons transferred per ion, and F is the Faraday constant.
EXPLANATION:
- For the electrolysis of aluminium chloride (AlCl3), the reaction at the cathode is:
Al3+ + 3e- → Al
- Given data:
- Current (I) = 2A
- Time (t) = 30 minutes = 30 × 60 = 1800 seconds
- Molar mass of Aluminium (M) = 27 g mol-1
- Faraday constant (F) = 96500 C mol-1
- In the reaction, the number of electrons (n) transferred per ion is 3.
- Using the formula:
m = (ItM)/(nF)
m = (2 × 1800 × 27)/(3 × 96500)
m = (97200)/(289500)
m = 0.336 g
Therefore, the amount of aluminium deposited at the cathode is 0.336 g.

