To deposite 2 mol of Ca from `CaCl_(2)` ............... electricity is required.

Answer

To determine the amount of electricity required to deposit 2 moles of calcium from calcium chloride (CaCl₂), we can follow these steps:

### Step-by-Step Solution:

1. **Write the Dissociation Reaction**:
The dissociation of calcium chloride in solution can be represented as:
\[ \text{CaCl}_2 \rightarrow \text{Ca}^{2+} + 2\text{Cl}^- \]

2. **Identify the Reduction Reaction**:
To deposit calcium, we need to reduce the calcium ions (\(\text{Ca}^{2+}\)):
\[ \text{Ca}^{2+} + 2e^- \rightarrow \text{Ca} \]
This shows that 1 mole of calcium is deposited by using 2 moles of electrons.

3. **Calculate the Moles of Electrons Required**:
Since we want to deposit 2 moles of calcium, we need to calculate the moles of electrons required:
- For 1 mole of calcium, 2 moles of electrons are needed.
- Therefore, for 2 moles of calcium:
\[ \text{Moles of electrons} = 2 \text{ moles of Ca} \times 2 \text{ moles of e}^- = 4 \text{ moles of e}^- \]

4. **Convert Moles of Electrons to Faraday**:
We know that 1 mole of electrons is equivalent to 1 Faraday (F). Thus, the total electricity required in Faraday is:
\[ \text{Electricity required} = 4 \text{ moles of e}^- = 4F \]

5. **Conclusion**:
Therefore, the total electricity required to deposit 2 moles of calcium from calcium chloride is:
\[ \text{Electricity required} = 4F \]

### Final Answer:
To deposit 2 moles of calcium from CaCl₂, **4 Faraday** of electricity is required.

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