Question

A solution of copper sulphate is electrolysed for \(10\) minutes with a current of \(1.5\) ampere. The mass of copper deposited at the cathode is: \[ \text{Given: Molar mass of Cu}=63.5\text{ g mol}^{-1} \] \[ F=96487\text{ C mol}^{-1} \]

• A. \(0.296\text{ g}\)
• B. \(1.7018\text{ g}\)
• C. \(2.4036\text{ g}\)
• D. \(0.5876\text{ g}\)

Hint:

For electrolysis, use \(m=\frac{ItM}{nF}\), where \(n\) is the number of electrons involved.

Answer 1

The Correct Option is A

Step 1: Write the cathode reaction.
In copper sulphate solution, copper ions get deposited at cathode. \[ \ce{Cu^{2+} + 2e^- -> Cu} \] So: \[ n=2 \]

Step 2: Calculate charge passed.
Current: \[ I=1.5\text{ A} \] Time: \[ t=10\text{ min}=10\times60=600\text{ s} \] Charge: \[ Q=It \] \[ Q=1.5\times600=900\text{ C} \]

Step 3: Use Faraday's law.
Mass deposited is: \[ m=\frac{Q\times M}{nF} \] Substitute: \[ m=\frac{900\times63.5}{2\times96487} \] \[ m=\frac{57150}{192974} \] \[ m\approx0.296\text{ g} \] Therefore, the mass of copper deposited is: \[ 0.296\text{ g}. \]