During electrolysis of an aqueous solution of `CuSO_(4)` using copper electrodes, if `2.5g` of `Cu` is deposited at cathode, then at anode

Options

A. 890 mL of `O_(2)` at STP is liberated
B. 445 mL of `O_(2)` at STP is liberated
C. a decrease of 2.5 g of mass takes place
D. 2.5 g of copper is deposited

Explanation

To solve the problem of how much copper is lost at the anode during the electrolysis of an aqueous solution of CuSO₄ using copper electrodes, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Electrolysis Process**:
- During electrolysis, copper ions (Cu²⁺) in the solution are reduced at the cathode, while copper from the anode is oxidized to maintain the balance of copper ions in the solution.

2. **Identify the Reaction at the Cathode**:
- At the cathode, the reduction reaction occurs:
\[ \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu (s)} \]
- This means that for every 2 moles of electrons, 1 mole of copper is deposited.

3. **Calculate Moles of Copper Deposited**:
- The molar mass of copper (Cu) is approximately 63.5 g/mol.
- To find the moles of copper deposited:
\[ \text{Moles of Cu} = \frac{\text{Mass of Cu deposited}}{\text{Molar mass of Cu}} = \frac{2.5 \text{ g}}{63.5 \text{ g/mol}} \approx 0.0394 \text{ mol} \]

4. **Determine the Moles of Electrons Used**:
- Since 1 mole of Cu requires 2 moles of electrons:
\[ \text{Moles of electrons} = 2 \times \text{Moles of Cu} = 2 \times 0.0394 \text{ mol} \approx 0.0788 \text{ mol} \]

5. **Identify the Reaction at the Anode**:
- At the anode, copper is oxidized:
\[ \text{Cu (s)} \rightarrow \text{Cu}^{2+} + 2e^- \]
- This means that for every mole of copper oxidized, 1 mole of Cu is lost.

6. **Calculate the Mass of Copper Lost at the Anode**:
- Since the same amount of copper that is deposited at the cathode is lost at the anode:
\[ \text{Mass of Cu lost} = \text{Mass of Cu deposited} = 2.5 \text{ g} \]

### Final Answer:
The mass of copper that is lost at the anode is **2.5 grams**.