How much charge is present on 1 mole of `Cu^(+2)` ion in faraday. (1 Faraday `= 96500` coulomb)

Explanation

To find the charge present on 1 mole of \( \text{Cu}^{2+} \) ions in Faraday, we can follow these steps:

### Step 1: Understand the concept of charge
The charge (Q) on an ion can be calculated using the formula:
\[ Q = n \times F \]
where:
- \( n \) is the number of moles of electrons transferred per mole of ion (which is equal to the charge number of the ion).
- \( F \) is Faraday's constant, approximately \( 96500 \, \text{C/mol} \).

### Step 2: Identify the charge of \( \text{Cu}^{2+} \)
The \( \text{Cu}^{2+} \) ion has a charge of +2, which means it will gain 2 moles of electrons during the reduction process.

### Step 3: Calculate the total charge for 1 mole of \( \text{Cu}^{2+} \)
Using the formula from Step 1:
\[ Q = n \times F \]
Substituting the values:
- \( n = 2 \) (since \( \text{Cu}^{2+} \) gains 2 electrons)
- \( F = 96500 \, \text{C/mol} \)

So,
\[ Q = 2 \times 96500 \]
\[ Q = 193000 \, \text{C} \]

### Step 4: Convert charge to Faraday
Since 1 Faraday is defined as the charge of 1 mole of electrons (which is \( 96500 \, \text{C} \)), we can express the charge in Faraday:
\[ \text{Charge in Faraday} = \frac{Q}{F} = \frac{193000}{96500} = 2 \]

Thus, the charge present on 1 mole of \( \text{Cu}^{2+} \) ions is **2 Faraday**.

### Summary of the Solution
The charge on 1 mole of \( \text{Cu}^{2+} \) ions is 2 Faraday.

---