On passing 0.5 mole of electrons through `CuSO_(4)` and `Hg_(2)(NO_(3))_(2)` solutions in series using inert electrodes

Options

A. `0.5 mol ` of `Cu` is deposited
B. `0.5 mol` of `Hg` is deposited
C. `0.125 mol` of `O_(2)` is produced
D. `0.5 mol ` of `O_(2)` is produced

Explanation

To solve the problem, we will analyze the electrochemical reactions occurring in the solutions of `CuSO4` and `Hg2(NO3)2` when 0.5 moles of electrons are passed through them.

### Step-by-Step Solution:

1. **Understanding the Electrochemical Reactions**:
- For `Hg2(NO3)2` (mercuric nitrate), the reduction reaction is:
\[ \text{Hg}^{2+} + 2e^- \rightarrow \text{Hg} \]
This indicates that 1 mole of mercury requires 2 moles of electrons for deposition.

- For `CuSO4` (copper(II) sulfate), the reduction reaction is:
\[ \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \]
This indicates that 1 mole of copper also requires 2 moles of electrons for deposition.

2. **Calculating the Amount of Mercury Deposited**:
- Since we have 0.5 moles of electrons, we can determine how much mercury can be deposited:
\[ \text{Moles of Hg deposited} = \frac{0.5 \text{ moles of electrons}}{2 \text{ moles of electrons per mole of Hg}} = 0.25 \text{ moles of Hg} \]

3. **Calculating the Amount of Copper Deposited**:
- Similarly, for copper:
\[ \text{Moles of Cu deposited} = \frac{0.5 \text{ moles of electrons}}{2 \text{ moles of electrons per mole of Cu}} = 0.25 \text{ moles of Cu} \]

4. **Calculating the Amount of Oxygen Evolved**:
- The oxidation of hydroxide ions to produce oxygen can be represented as:
\[ 4 \text{OH}^- \rightarrow \text{O}_2 + 2 \text{H}_2O + 4e^- \]
This indicates that 4 moles of electrons are required to produce 1 mole of oxygen.
- Therefore, using 0.5 moles of electrons:
\[ \text{Moles of O}_2 evolved = \frac{0.5 \text{ moles of electrons}}{4 \text{ moles of electrons per mole of O}_2} = 0.125 \text{ moles of O}_2 \]

5. **Summarizing the Results**:
- From the calculations:
- Moles of mercury deposited = 0.25 moles
- Moles of copper deposited = 0.25 moles
- Moles of oxygen evolved = 0.125 moles

### Conclusion:
Based on the calculations, the correct answers are:
- 0.25 moles of mercury are deposited (not listed in options).
- 0.25 moles of copper are deposited (not listed in options).
- 0.125 moles of oxygen are produced.

### Final Answer:
- The correct statements from the options provided are:
- B: 0.5 mole of mercury is deposited (incorrect, should be 0.25).
- C: 0.125 mole of oxygen is produced (correct).