An electrolytic cell is composed of Cu and Zn. A current of 9.65A is drawn from a cell for 1 hour. Then the loss in mass at anode and gain in mass at cathode, respectively could be

Options

A. 11.77g, 11.43g
B. 11.77g, 10g
C. 22.86g, 23.54g
D. 23.54g, 22.86g

Answer

To solve the problem, we need to calculate the loss in mass at the anode (zinc) and the gain in mass at the cathode (copper) when a current of 9.65 A is drawn for 1 hour in an electrolytic cell.

### Step-by-Step Solution:

1. **Identify the reactions at the electrodes:**
- At the anode (oxidation):
\[ \text{Zn} \rightarrow \text{Zn}^{2+} + 2e^- \]
- At the cathode (reduction):
\[ \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \]

2. **Determine the total charge (Q) passed through the cell:**
- The formula for charge is:
\[ Q = I \times T \]
- Where:
- \( I = 9.65 \, \text{A} \)
- \( T = 1 \, \text{hour} = 3600 \, \text{seconds} \)
- So:
\[ Q = 9.65 \, \text{A} \times 3600 \, \text{s} = 34,740 \, \text{C} \]

3. **Calculate the mass of copper deposited at the cathode:**
- Use the formula:
\[ W = \frac{I \times T \times M}{n \times F} \]
- Where:
- \( M \) (molar mass of copper) = 63.5 g/mol
- \( n \) (number of electrons transferred) = 2
- \( F \) (Faraday's constant) = 96,500 C/mol
- Plugging in the values:
\[ W_{\text{Cu}} = \frac{9.65 \times 3600 \times 63.5}{2 \times 96500} \]
- Calculating:
\[ W_{\text{Cu}} = \frac{34,740 \times 63.5}{193000} \approx 11.43 \, \text{g} \]

4. **Calculate the mass of zinc lost at the anode:**
- Using the same formula for zinc:
- \( M \) (molar mass of zinc) = 65.38 g/mol
- Plugging in the values:
\[ W_{\text{Zn}} = \frac{9.65 \times 3600 \times 65.38}{2 \times 96500} \]
- Calculating:
\[ W_{\text{Zn}} = \frac{34,740 \times 65.38}{193000} \approx 11.77 \, \text{g} \]

5. **Final Results:**
- Loss in mass at the anode (zinc): **11.77 g**
- Gain in mass at the cathode (copper): **11.43 g**

### Summary:
The loss in mass at the anode is 11.77 g and the gain in mass at the cathode is 11.43 g.