Question

An aqueous solution of CuSO$_4$ solution is electrolysed 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:

Always ensure time is in seconds for Faraday's law calculations. Double-check the valency (n-factor) for the ion being deposited. Remember that cations deposit at the cathode (negative electrode) and anions (or less reactive substances like O$_2$) are liberated at the anode (positive electrode).

Answer 1

The Correct Option is C

Step 1: Find the total charge passed: \(Q = I \times t = 2.5 \times 193 = 482.5\) C. Since 1 mole of electrons carries a charge of \(F = 96500\) C, the moles of electrons passed is \(n_e = \frac{482.5}{96500} = 0.005\) mol.

Step 2: Copper is deposited by the reduction \(Cu^{2+} + 2e^- \rightarrow Cu\), so every 2 moles of electrons deposit 1 mole of copper. Moles of \(Cu\) deposited \(= \frac{n_e}{2} = \frac{0.005}{2} = 0.0025\) mol.

Step 3: Convert moles of copper to mass using its atomic mass of 63.5 g/mol: \(W = 0.0025 \times 63.5 = 0.15875\) g.
\[ \boxed{W \approx 0.15875 \text{ g}} \]