When a quantity of electricity is passed through `CuSO_(4)` solution, 0.16 g of copper gets deposited. If the same quantity of electricity is passed through acidulated water, then the volume of `H_(2)` liberated at STP will be : (given atomic weight of Cu=64)

Options

A. `4 cm^(3)`
B. `56 cm^(3)`
C. `604 cm^(3)`
D. `8 cm^(3)`

Explanation

To solve the problem step-by-step, we will use the principles of electrochemistry, particularly Faraday's laws of electrolysis.

### Step-by-Step Solution:

1. **Identify the Given Data**:
- Mass of copper deposited (m_Cu) = 0.16 g
- Atomic weight of copper (Cu) = 64 g/mol
- Molar mass of hydrogen (H₂) = 2 g/mol
- At STP, 1 mole of gas occupies 22.4 L.

2. **Calculate the Number of Moles of Copper Deposited**:
\[ \text{Number of moles of Cu} = \frac{\text{mass of Cu}}{\text{molar mass of Cu}} = \frac{0.16 \, \text{g}}{64 \, \text{g/mol}} = 0.0025 \, \text{mol} \]

3. **Determine the Charge (Q) Passed**:
According to Faraday's first law of electrolysis:
\[ m = Z \cdot Q \]
where \( Z \) is the electrochemical equivalent. For copper, the electrochemical equivalent can be calculated as:
\[ Z_{Cu} = \frac{\text{Molar mass of Cu}}{nF} \]
Here, \( n = 2 \) (since Cu²⁺ gains 2 electrons to deposit Cu) and \( F \) (Faraday's constant) is approximately 96500 C/mol.
\[ Z_{Cu} = \frac{64 \, \text{g/mol}}{2 \times 96500 \, \text{C/mol}} = \frac{64}{193000} \, \text{g/C} \]

4. **Calculate the Charge (Q)**:
Rearranging the equation \( m = Z \cdot Q \):
\[ Q = \frac{m}{Z} = \frac{0.16 \, \text{g}}{\frac{64}{193000}} = 0.16 \cdot \frac{193000}{64} \approx 482.5 \, \text{C} \]

5. **Calculate the Equivalent Weight of Hydrogen**:
For hydrogen, the equivalent weight is:
\[ Z_{H_2} = \frac{\text{Molar mass of H}_2}{nF} = \frac{2 \, \text{g/mol}}{2 \times 96500 \, \text{C/mol}} = \frac{2}{193000} \, \text{g/C} \]

6. **Calculate the Weight of Hydrogen Liberated**:
Using the same charge \( Q \):
\[ m_{H_2} = Z_{H_2} \cdot Q = \left(\frac{2}{193000}\right) \cdot 482.5 \approx 0.005 \, \text{g} \, \text{(or 5 mg)} \]

7. **Calculate the Number of Moles of Hydrogen**:
\[ \text{Number of moles of H}_2 = \frac{m_{H_2}}{\text{molar mass of H}_2} = \frac{0.005 \, \text{g}}{2 \, \text{g/mol}} = 0.0025 \, \text{mol} \]

8. **Calculate the Volume of Hydrogen at STP**:
\[ \text{Volume of H}_2 = \text{Number of moles} \times 22.4 \, \text{L/mol} = 0.0025 \, \text{mol} \times 22.4 \, \text{L/mol} = 0.056 \, \text{L} = 56 \, \text{cm}^3 \]

### Final Answer:
The volume of \( H_2 \) liberated at STP is **56 cm³**.