A current of 2 ampere was passed through solutions of `CuSO_4` and `AgNO_3` in series. 0.635 g of copper was deposited. Then the weight of silver deposited will be-

Options

A. 0.59g
B. 3.24g
C. 1.08g
D. 2.16g

Answer

To solve the problem, we will use Faraday's second law of electrolysis, which states that the amount of substance deposited during electrolysis is directly proportional to its equivalent weight when the same amount of current is passed through different electrolytes.

### Step-by-Step Solution:

1. **Identify the Given Data**:
- Current (I) = 2 A
- Weight of copper deposited (W_Cu) = 0.635 g
- Atomic weight of copper (Cu) = 63.5 g/mol
- Atomic weight of silver (Ag) = 107.9 g/mol

2. **Calculate the Equivalent Weight of Copper**:
- The equivalent weight of a substance is calculated using the formula:
\[ \text{Equivalent Weight} = \frac{\text{Atomic Weight}}{n} \]
- For copper (Cu), which is reduced from Cu²⁺ to Cu, the n-factor (number of electrons involved) is 2:
\[ \text{Equivalent Weight of Cu} = \frac{63.5}{2} = 31.75 \text{ g/equiv} \]

3. **Calculate the Equivalent Weight of Silver**:
- For silver (Ag), which is reduced from Ag⁺ to Ag, the n-factor is 1:
\[ \text{Equivalent Weight of Ag} = \frac{107.9}{1} = 107.9 \text{ g/equiv} \]

4. **Apply Faraday's Second Law**:
- According to Faraday's second law:
\[ \frac{W_{Cu}}{E_{Cu}} = \frac{W_{Ag}}{E_{Ag}} \]
- Where:
- \( W_{Cu} \) = weight of copper deposited = 0.635 g
- \( E_{Cu} \) = equivalent weight of copper = 31.75 g/equiv
- \( W_{Ag} \) = weight of silver deposited (unknown)
- \( E_{Ag} \) = equivalent weight of silver = 107.9 g/equiv

5. **Rearranging the Equation**:
- Rearranging the equation to find \( W_{Ag} \):
\[ W_{Ag} = W_{Cu} \times \frac{E_{Ag}}{E_{Cu}} \]

6. **Substituting the Values**:
- Substitute the known values into the equation:
\[ W_{Ag} = 0.635 \times \frac{107.9}{31.75} \]

7. **Calculating the Weight of Silver Deposited**:
- Performing the calculation:
\[ W_{Ag} = 0.635 \times 3.397 = 2.16 \text{ g} \]

### Final Answer:
The weight of silver deposited will be **2.16 g**.

---