Question

A current of 2.5 amperes is passed through 800 ml of 0.48 M solution of CuSO$_4$ for 1.0 hour with a current efficiency of 80%. If the volume of the solution remains unchanged, what is the final molarity of the solution?

• A. 0.386
• B. 0.315
• C. 0.433
• D. 0.298

Hint:

In electrolysis problems, always use Faraday's law and adjust for current efficiency. Also remember stoichiometry of electron transfer to find substance consumed.

Answer 1

The Correct Option is C

Step 1: Understanding the electrolysis process.
When current is passed through CuSO$_4$ solution, copper ions are reduced at the cathode:
\[ \mathrm{Cu^{2+} + 2e^- \rightarrow Cu(s)} \] Thus, Cu$^{2+}$ ions are removed from the solution, decreasing its concentration.

Step 2: Calculating total charge passed.
Given:
\[ I = 2.5 \, A, \quad t = 1 \, \text{hour} = 3600 \, s \] \[ Q = I \times t = 2.5 \times 3600 = 9000 \, C \] Considering 80% efficiency:
\[ Q_{\text{effective}} = 0.8 \times 9000 = 7200 \, C \]

Step 3: Calculating moles of electrons transferred.
Using Faraday constant \( F = 96500 \, C/mol \):
\[ \text{Moles of electrons} = \frac{7200}{96500} \approx 0.0746 \, mol \]

Step 4: Calculating moles of Cu$^{2+$ reduced.}
From the reaction:
\[ 2 \, \text{mol electrons} \Rightarrow 1 \, \text{mol Cu}^{2+} \] \[ \text{Moles of Cu}^{2+} = \frac{0.0746}{2} \approx 0.0373 \, mol \]

Step 5: Initial moles of CuSO$_4$.
Given molarity and volume:
\[ M = 0.48, \quad V = 800 \, ml = 0.8 \, L \] \[ \text{Initial moles} = 0.48 \times 0.8 = 0.384 \, mol \]

Step 6: Remaining moles after electrolysis.
\[ \text{Remaining moles} = 0.384 - 0.0373 = 0.3467 \, mol \]

Step 7: Calculating final molarity.
Volume remains unchanged:
\[ M_{\text{final}} = \frac{0.3467}{0.8} \approx 0.433 \, M \] Final Answer:
\[ \boxed{0.433} \] Hence, the correct answer is option (C).