The anodic half-cell of lead-acid battery is recharged using electricity of 0.05 Faraday. The amount of `PbSO_(4)` electrolyzed in g during the process is : (Molar mass of `PbSO_(4)=303gmol^(-1)`)

Options

A. 22.8
B. 15.2
C. 7.6
D. 11.4

Answer

To solve the problem, we need to determine the amount of PbSO₄ that is electrolyzed when 0.05 Faraday of electricity is passed through the anodic half-cell of a lead-acid battery.

### Step-by-Step Solution:

1. **Understanding the Reaction**:
The reaction at the anode during the recharging of the lead-acid battery can be represented as:
\[ \text{PbSO}_4 + 2 \text{e}^- \rightarrow \text{Pb} + \text{H}_2\text{SO}_4 \]
This shows that 1 mole of PbSO₄ requires 2 moles of electrons (2 Faraday) to be reduced.

2. **Calculating Moles of Electrons**:
Given that 0.05 Faraday of electricity is used, we can calculate the number of moles of electrons:
\[ \text{Number of moles of electrons} = \text{Faraday} = 0.05 \text{ mol e}^- \]

3. **Finding Moles of PbSO₄**:
Since 2 moles of electrons are required to reduce 1 mole of PbSO₄, we can find the moles of PbSO₄ that can be electrolyzed using the formula:
\[ \text{Moles of PbSO}_4 = \frac{\text{Moles of electrons}}{2} = \frac{0.05}{2} = 0.025 \text{ moles} \]

4. **Calculating Mass of PbSO₄**:
To find the mass of PbSO₄ that corresponds to 0.025 moles, we use the molar mass of PbSO₄, which is given as 303 g/mol:
\[ \text{Mass of PbSO}_4 = \text{Moles} \times \text{Molar mass} = 0.025 \text{ moles} \times 303 \text{ g/mol} \]
\[ \text{Mass of PbSO}_4 = 7.575 \text{ g} \]

5. **Final Answer**:
Rounding off, the amount of PbSO₄ electrolyzed is approximately 7.6 grams.

### Conclusion:
The amount of PbSO₄ electrolyzed during the process is **7.6 grams**.