Question

Aqueous CuSO₄ solution was electrolysed by passing 2 amp of current for 10 min. What is the weight (in g) of copper deposited at cathode? (Cu = 63 u; F = 96500 C mol^-1)

• A. 0.195
• B. 0.39
• C. 0.78
• D. 1.57

Hint:

The weight of copper deposited can be calculated using the formula.

Answer 1

The Correct Option is B

To find the weight of copper deposited at the cathode, we can use Faraday's laws of electrolysis.

According to Faraday's First Law, the mass of a substance deposited or liberated at an electrode is directly proportional to the quantity of electricity passed through the electrolyte. The relationship is given by:

m = (Q × M) / (n × F)

Where:

• m = mass of the substance deposited (in grams)
• Q = total charge passed (in Coulombs)
• M = molar mass of the substance (in g/mol)
• n = number of electrons transferred in the reaction
  (for Cu²⁺ + 2e⁻ → Cu, n = 2)
• F = Faraday’s constant = 96500 C/mol

Step 1: Calculate the total charge (Q) 

Q = I × t

• I = current = 2 A
• t = time = 10 minutes = 10 × 60 = 600 seconds

Q = 2 A × 600 s = 1200 C

Step 2: Calculate the mass (m) of copper deposited

m = (1200 × 63) / (2 × 96500)

m = 75600 / 193000

m ≈ 0.3912 g

Conclusion:
The weight of copper deposited at the cathode is approximately 0.39 g.

Answer 2

The weight of copper deposited can be calculated using the formula:

W = (I × t × M) / (F × n)

Where:

• I = Current = 2 A
• t = Time = 10 minutes = 600 seconds
• M = Molar mass of Cu = 63 g/mol
• F = Faraday's constant = 96500 C/mol
• n = Valency of Cu = 2

Substituting the values:

W = (2 × 600 × 63) / (96500 × 2) = 0.39 g