`0.5F` of electricity is passed through `500mL` of copper sulphate solution. The amount of copper which can be deposited will be
a. 63.5 g
b.31.75 g
c.15.8 g
d.Unpredictable

Options

A. `63.5g`
B. `31.75g`
C. `15.8g`
D. Unpredictable

Explanation

To solve the problem of how much copper can be deposited when 0.5 Faraday of electricity is passed through 500 mL of copper sulfate solution, we can follow these steps:

### Step 1: Understand the electrochemical reaction
The reaction for the deposition of copper from copper sulfate (CuSO₄) is:
\[ \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \]
This indicates that 1 mole of copper ions (Cu²⁺) requires 2 moles of electrons (2 Faradays) to deposit 1 mole of copper.

### Step 2: Determine the amount of copper deposited per Faraday
From the reaction, we know that:
- 2 Faradays of electricity deposit 1 mole of copper.
- The molar mass of copper (Cu) is 63.5 g.

### Step 3: Calculate the amount of copper deposited for 0.5 Faraday
Since 2 Faradays deposit 63.5 g of copper, we can find out how much copper is deposited with 0.5 Faraday:
\[ \text{Amount of copper deposited} = \left(\frac{63.5 \, \text{g}}{2 \, \text{Faradays}}\right) \times 0.5 \, \text{Faraday} \]

### Step 4: Perform the calculation
Now, we can calculate:
\[ \text{Amount of copper deposited} = \frac{63.5}{2} \times 0.5 = 31.75 \, \text{g} \times 0.5 = 15.875 \, \text{g} \]
Rounding this to two decimal places gives us approximately 15.8 g.

### Step 5: Conclusion
Thus, the amount of copper that can be deposited is approximately **15.8 g**.

### Final Answer
The correct answer is **c. 15.8 g**.

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