The anodic half-cell of lead-acid battery is recharged using electricity of 6 Faraday. The amount of `PbSo_(4)` electrolyzed in moles during the process is__________.

Explanation

To solve the problem of how many moles of PbSO₄ are electrolyzed when 6 Faraday of electricity is used in the anodic half-cell of a lead-acid battery, we can follow these steps:

### Step 1: Understand the Reaction
In a lead-acid battery, the anodic half-reaction involves the oxidation of lead (Pb) to lead ions (Pb²⁺) and the formation of lead sulfate (PbSO₄). The half-reaction can be represented as:

\[ \text{Pb (s)} + \text{SO}_4^{2-} \text{(aq)} \rightarrow \text{PbSO}_4 \text{(s)} + 2 \text{e}^- \]

This means that for every mole of PbSO₄ produced, 2 moles of electrons (or 2 Faraday) are required.

### Step 2: Calculate Moles of PbSO₄ from Faraday
Given that 1 mole of PbSO₄ requires 2 Faraday of electricity, we can set up a proportion to find out how many moles of PbSO₄ can be produced from 6 Faraday:

- 2 Faraday → 1 mole of PbSO₄
- 6 Faraday → \( x \) moles of PbSO₄

Using the ratio:

\[ \frac{6 \text{ Faraday}}{2 \text{ Faraday/mole}} = 3 \text{ moles of PbSO}_4 \]

### Step 3: Conclusion
Thus, the amount of PbSO₄ electrolyzed during the process when 6 Faraday of electricity is used is:

\[ \text{3 moles of PbSO}_4 \]

### Final Answer
The amount of PbSO₄ electrolyzed in moles during the process is **3 moles**.

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