Which of the following statements about the following reaction is`//`are Wrong?
`2Cu_(2)O(s)+Cu_(2)S(s)rarr6Cu(s)+5O_(2)(g)`

Options

A. Both `Cu_(2)` and `Cu_(2)S` are reduced.
B. Only `Cu_(2)S` is reduced.
C. `Cu_(2)S` is the oxidant.
D. Only `Cu_(2)O` is reduced.

Explanation

To determine which statements about the reaction \( 2Cu_2O(s) + Cu_2S(s) \rightarrow 6Cu(s) + 5O_2(g) \) are wrong, we can analyze the oxidation states of the elements involved in the reaction.

### Step-by-Step Solution:

1. **Identify the Oxidation States:**
- In \( Cu_2O \), the oxidation state of Cu is +1 (since O is -2).
- In \( Cu_2S \), the oxidation state of Cu is also +1 (since S is typically -2 in sulfides).
- In elemental Cu (the product), the oxidation state is 0.
- In \( O_2 \), the oxidation state of O is 0.

2. **Determine Changes in Oxidation States:**
- For Cu in \( Cu_2O \) and \( Cu_2S \):
- The oxidation state changes from +1 to 0 (reduction).
- For O in \( O_2 \):
- The oxidation state changes from -2 (in \( Cu_2O \)) to 0 (oxidation).

3. **Identify Oxidation and Reduction:**
- **Oxidation**: The process where the oxidation state increases. Here, O is oxidized from -2 to 0.
- **Reduction**: The process where the oxidation state decreases. Here, Cu is reduced from +1 to 0.

4. **Analyze the Statements:**
- **Statement A**: Both \( Cu_2O \) and \( Cu_2S \) are oxidized.
- **Wrong**: Both \( Cu_2O \) and \( Cu_2S \) are reduced.
- **Statement B**: Only \( Cu_2S \) is reduced.
- **Wrong**: Both \( Cu_2O \) and \( Cu_2S \) are reduced.
- **Statement C**: \( Cu_2S \) is the oxidizing agent.
- **Wrong**: \( Cu_2O \) is also an oxidizing agent.
- **Statement D**: Only \( Cu_2O \) is reduced.
- **Wrong**: Both \( Cu_2O \) and \( Cu_2S \) are reduced.

5. **Conclusion:**
- The wrong statements are A, B, C, and D.

### Final Answer:
The wrong statements about the reaction are A, B, C, and D.