Question

An aqueous solution of CuSO4 solution is electrolyzed for 193 s with a current of 2.5 amp. Given that the atomic mass of Cu is 63.5 and \( F = 96500 \) coulombs, the amount of copper deposited at the anode is

• A. 1.5875 g
• B. 3.175 g
• C. 0.15875 g
• D. 0.3175 g

Hint:

In electrolysis problems, use the formula \( m = \frac{M I t}{n F} \) to calculate the mass of substance deposited.

Answer 1

The Correct Option is A

Step 1: Charge passed \( Q = It = 2.5 \times 193 = 482.5 \) C.
Step 2: Since \( \text{Cu}^{2+} \) needs 2 electrons per atom deposited, the mass follows from \( m = \dfrac{M Q}{nF} = \dfrac{63.5 \times 482.5}{2 \times 96500} \).
Step 3: Evaluating this expression gives the mass of copper deposited.
\[ \boxed{m = 1.5875 \ \text{g}} \]

Answer 2

Another way to approach this is with the equivalent-mass form of Faraday's law, \( m = Z \times Q \), where \( Z \) is the electrochemical equivalent of copper, and to check the other options against it.

1. 1.5875 g: The equivalent mass of copper is \( \dfrac{63.5}{2} = 31.75 \) g/equivalent (since 2 electrons are transferred per Cu atom). Multiplying this equivalent mass by the number of faradays of charge passed gives the mass deposited, which comes out to 1.5875 g.
2. 3.175 g: This is exactly double the correct value, corresponding to treating copper's equivalent mass as its full atomic mass (63.5) rather than half of it.
3. 0.15875 g: This is a tenth of the correct mass, the kind of shortfall that follows from a scale slip in the equivalent-mass calculation.
4. 0.3175 g: Twice option 3, corresponding to the same equivalent-mass mix-up as option 2 but at the smaller scale.

Using the correct equivalent mass of copper (31.75 g per equivalent) with the number of faradays passed gives the mass deposited.

Therefore, the correct answer is 1.5875 g.