How much charge in faraday is required for the reduction of 1 mole of `Cu^(2+)` ions to Cu ?

Explanation

To determine the charge in Faraday required for the reduction of 1 mole of \( \text{Cu}^{2+} \) ions to copper (Cu), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reaction**:
The reduction of copper ions can be represented by the half-reaction:
\[ \text{Cu}^{2+} + 2 \text{e}^- \rightarrow \text{Cu} \]
This equation shows that 1 mole of \( \text{Cu}^{2+} \) ions requires 2 moles of electrons (2 electrons) to be reduced to 1 mole of copper.

2. **Determine the Number of Moles of Electrons**:
From the balanced equation, we see that for every mole of \( \text{Cu}^{2+} \), 2 moles of electrons are needed. Therefore, for 1 mole of \( \text{Cu}^{2+} \):
\[ \text{Moles of electrons} = 2 \]

3. **Use Faraday's Constant**:
Faraday's constant (F) is the amount of charge per mole of electrons, which is approximately \( 96500 \, \text{C/mol} \). To find the total charge required, we multiply the number of moles of electrons by Faraday's constant:
\[ \text{Charge (Q)} = \text{Moles of electrons} \times F \]
Substituting the values:
\[ Q = 2 \, \text{mol} \times 96500 \, \text{C/mol} = 193000 \, \text{C} \]

4. **Convert Charge to Faraday**:
Since 1 Faraday is equivalent to the charge of 1 mole of electrons (96500 C), the charge in Faraday can be calculated as:
\[ \text{Charge in Faraday} = \frac{Q}{F} = \frac{193000 \, \text{C}}{96500 \, \text{C/mol}} \approx 2 \, \text{Faraday} \]

### Final Answer:
The charge required for the reduction of 1 mole of \( \text{Cu}^{2+} \) ions to Cu is approximately **2 Faraday** or **193000 Coulombs**.