Question

Amount of copper deposited on the cathode of an electrolytic cell containing copper sulphate solution by the passage of 2 amperes for 30 minutes (At. mass of Cu = 63.5):

• A. 1.184 gm
• B. 0.2214 gm
• C. 2.214 gm
• D. 0.1184 gm

Hint:

To find the amount of a substance deposited during electrolysis, use Faraday's law and remember to convert all units into the SI system.

Answer 1

The Correct Option is B

We use Faraday's law of electrolysis, which is given by the formula: \[ m = \frac{M \times I \times t}{n \times F} \] Where: - \( m \) = mass of copper deposited - \( M = 63.5 \, \text{gm/mol} \) = molar mass of copper - \( I = 2 \, \text{A} \) = current - \( t = 30 \, \text{minutes} = 1800 \, \text{seconds} \) - \( n = 2 \) (because copper has a valency of 2 in the electrolyte) - \( F = 96500 \, \text{C/mol} \) = Faraday's constant Substitute the values: \[ m = \frac{63.5 \times 2 \times 1800}{2 \times 96500} = \frac{228600}{193000} = 0.2214 \, \text{gm} \] Thus, the mass of copper deposited is 0.2214 gm.