In the electrolysis of aqueous solutoin of copper suphate using copper stirps as anode and cathode the anode rection is

Options

A. `Cu^(2+)+2e^(-)rarrCu`
B. `CurarrCu^(2+)+2e^(-)`
C. `2HO^(-)rarrH_(2)+1//2O_(2)+2e^(-)`
D. `2HSO_(4)^(-)rarrH_(2)S_(2)O_(8)+2e^(-)`

Explanation

To determine the anode reaction during the electrolysis of an aqueous solution of copper sulfate (CuSO₄) using copper strips as both anode and cathode, we need to follow these steps:

### Step-by-Step Solution:

1. **Identify the Electrolysis Process**:
- In electrolysis, oxidation occurs at the anode and reduction occurs at the cathode.

2. **Determine the Anode Reaction**:
- At the anode, the oxidation of hydroxide ions (OH⁻) and water (H₂O) can occur. The relevant half-reaction for the oxidation of hydroxide ions is:
\[ 4OH^- \rightarrow 2H_2O + O_2 + 4e^- \]
- This means that four hydroxide ions lose electrons to produce two water molecules and one molecule of oxygen gas.

3. **Simplify the Reaction**:
- To simplify the reaction, we can divide the entire equation by 2:
\[ 2OH^- \rightarrow H_2O + O_2 + 2e^- \]
- This indicates that two hydroxide ions are oxidized to produce one water molecule, one molecule of oxygen gas, and two electrons.

4. **Conclusion**:
- The anode reaction during the electrolysis of an aqueous solution of copper sulfate using copper strips as electrodes is:
\[ 2OH^- \rightarrow H_2O + O_2 + 2e^- \]

### Final Answer:
The anode reaction is:
\[ 2OH^- \rightarrow H_2O + O_2 + 2e^- \]