0.5 Faraday of electricity was passed jo deposit áll the copper present in 500ml of `CuSO_4, solution. What was the molarity of this solution?

Options

A. 1M
B. 0.5M
C. 0.25M
D. 2.5M

Answer

To find the molarity of the copper sulfate solution after passing 0.5 Faraday of electricity, we can follow these steps:

### Step 1: Understand the electrochemical reaction
When copper ions (Cu²⁺) in the copper sulfate solution are reduced, they gain electrons to form solid copper (Cu). The half-reaction can be represented as:
\[ \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \]

### Step 2: Determine the relationship between Faraday and moles of copper
According to Faraday's laws of electrolysis:
- 2 Faraday of electricity will deposit 1 mole of copper.
- Therefore, 1 Faraday will deposit \( \frac{1}{2} \) moles of copper.

### Step 3: Calculate the amount of copper deposited with 0.5 Faraday
If 1 Faraday deposits \( \frac{1}{2} \) moles of copper, then:
\[ 0.5 \text{ Faraday} \text{ will deposit } \frac{1}{2} \times 0.5 = 0.25 \text{ moles of copper} \]

### Step 4: Convert the volume of the solution from mL to L
The volume of the copper sulfate solution is given as 500 mL. To convert this to liters:
\[ 500 \text{ mL} = \frac{500}{1000} = 0.5 \text{ L} \]

### Step 5: Calculate the molarity of the solution
Molarity (M) is defined as the number of moles of solute divided by the volume of solution in liters:
\[ \text{Molarity} = \frac{\text{Number of moles of solute}}{\text{Volume of solution in L}} \]
Substituting the values we have:
\[ \text{Molarity} = \frac{0.25 \text{ moles}}{0.5 \text{ L}} = 0.5 \text{ M} \]

### Final Answer
The molarity of the copper sulfate solution is **0.5 M**.

---