A certain current liberates `0.5g` of hydrogen in 2 hours. How many grams of copper can be liberated by the same current flowing for the same time in a copper sulphate solution ?

Options

A. `12.7g`
B. `15.9g`
C. `31.8g`
D. `63.5g`

Explanation

To solve the problem, we will use Faraday's laws of electrolysis, which state that the amount of substance liberated during electrolysis is directly proportional to the quantity of electricity passed through the electrolyte.

### Step-by-Step Solution:

1. **Determine the moles of hydrogen liberated:**
- The molar mass of hydrogen (H₂) is approximately 2 g/mol.
- Given that 0.5 g of hydrogen is liberated, we can calculate the number of moles of hydrogen:
\[ \text{Moles of } H_2 = \frac{\text{mass}}{\text{molar mass}} = \frac{0.5 \, \text{g}}{2 \, \text{g/mol}} = 0.25 \, \text{mol} \]

2. **Calculate the charge (Q) using Faraday's constant:**
- The reaction for the liberation of hydrogen gas at the cathode is:
\[ 2H^+ + 2e^- \rightarrow H_2 \]
- From the equation, 2 moles of electrons (2 Faradays) are needed to liberate 1 mole of hydrogen.
- Therefore, for 0.25 moles of hydrogen, the number of moles of electrons required is:
\[ \text{Moles of } e^- = 0.25 \, \text{mol} \times 2 = 0.5 \, \text{mol} \]
- The total charge (Q) can be calculated using Faraday's constant (F = 96500 C/mol):
\[ Q = \text{moles of } e^- \times F = 0.5 \, \text{mol} \times 96500 \, \text{C/mol} = 48250 \, \text{C} \]

3. **Determine the equivalent weight of copper:**
- The molar mass of copper (Cu) is approximately 63.5 g/mol.
- The reaction for the deposition of copper is:
\[ Cu^{2+} + 2e^- \rightarrow Cu \]
- From this, we see that 2 moles of electrons are needed to deposit 1 mole of copper. Therefore, the equivalent weight of copper is:
\[ \text{Equivalent weight of Cu} = \frac{\text{Molar mass of Cu}}{2} = \frac{63.5 \, \text{g/mol}}{2} = 31.75 \, \text{g/equiv} \]

4. **Calculate the grams of copper deposited:**
- Since the same amount of charge (Q) is used, we can find out how many grams of copper can be deposited using the equivalent weight:
\[ \text{Grams of Cu} = \frac{Q}{F} \times \text{Equivalent weight of Cu} \]
- First, we calculate the moles of copper deposited:
\[ \text{Moles of Cu} = \frac{Q}{96500} = \frac{48250}{96500} = 0.5 \, \text{mol} \]
- Now, we can calculate the mass of copper:
\[ \text{Grams of Cu} = 0.5 \, \text{mol} \times 63.5 \, \text{g/mol} = 31.75 \, \text{g} \]

### Final Answer:
The amount of copper that can be liberated by the same current flowing for the same time in a copper sulfate solution is **31.75 grams**.