`Hg_(2)CI_(2)` is produced by the electrolytic reduction of `Hg^(2+)` ion in presence of `CI^(-)` ion is `2Hg^(2+) +2CI^(-) +2e^(Theta) rarr Hg_(2)CI_(2)`. Calculate the current required to have a rate production of 44g per hour of `Hg_(2)CI_(2)`.
[Atomic weight of `Hg = 200.6]`:-

Options

A. 5 ampere
B. 4 amapere
C. 6.5 ampere
D. 3.5 ampere

Explanation

To solve the problem of calculating the current required to produce 44 grams of \( \text{Hg}_2\text{Cl}_2 \) per hour through the electrolytic reduction of \( \text{Hg}^{2+} \) ions in the presence of \( \text{Cl}^- \) ions, we will follow these steps:

### Step 1: Calculate the equivalent weight of \( \text{Hg}_2\text{Cl}_2 \)
The formula for equivalent weight is given by:
\[ \text{Equivalent weight} = \frac{\text{Molar mass}}{\text{n}} \]
where \( n \) is the number of electrons transferred per formula unit.

1. The molar mass of \( \text{Hg}_2\text{Cl}_2 \):
- Atomic weight of \( \text{Hg} = 200.6 \)
- Atomic weight of \( \text{Cl} = 35.5 \)
- Molar mass of \( \text{Hg}_2\text{Cl}_2 = 2 \times 200.6 + 2 \times 35.5 = 401.2 + 71 = 472.2 \, \text{g/mol} \)

2. In the reaction \( 2 \text{Hg}^{2+} + 2 \text{Cl}^- + 2 e^- \rightarrow \text{Hg}_2\text{Cl}_2 \), 2 electrons are transferred. Thus, \( n = 2 \).

3. Therefore, the equivalent weight of \( \text{Hg}_2\text{Cl}_2 \):
\[ \text{Equivalent weight} = \frac{472.2}{2} = 236.1 \, \text{g/equiv} \]

### Step 2: Use Faraday's law of electrolysis
According to Faraday's law:
\[ \frac{\text{Weight}}{\text{Equivalent weight}} = \frac{I \cdot t}{F} \]
where:
- \( I \) = current in amperes
- \( t \) = time in seconds
- \( F \) = Faraday's constant \( = 96500 \, \text{C/mol} \)

### Step 3: Convert time from hours to seconds
Given that the production rate is 44 g per hour, we convert the time:
\[ t = 1 \, \text{hour} = 3600 \, \text{seconds} \]

### Step 4: Substitute the values into Faraday's law
We know:
- Weight of \( \text{Hg}_2\text{Cl}_2 = 44 \, \text{g} \)
- Equivalent weight = 236.1 g/equiv
- Time = 3600 s
- Faraday's constant = 96500 C/mol

Substituting these values into the equation:
\[ \frac{44}{236.1} = \frac{I \cdot 3600}{96500} \]

### Step 5: Solve for current \( I \)
Rearranging the equation gives:
\[ I = \frac{44 \cdot 96500}{236.1 \cdot 3600} \]

Calculating the right-hand side:
\[ I = \frac{4258000}{849960} \approx 5.01 \, \text{A} \]

### Conclusion
Thus, the current required to produce 44 g of \( \text{Hg}_2\text{Cl}_2 \) per hour is approximately:
\[ \boxed{5 \, \text{A}} \]