Which of the following statements about the following reaction is wrong?
`2Cu_2O(s)+Cu_2S(s)rarr6Cu(s)+SO_2(g)`
(I)both `Cu_2O and Cu_2S` are reduced
(II)`Cu_2S` is reduced while `Cu_2O` is oxidised
(III)`Cu_2S` is the oxidant
(IV)`Cu_2O` is reduced whilw `Cu_2S` is oxidiseduij

Options

A. I and II
B. III only
C. II,III,and Iv
D. II and IV

Answer

To determine which statement about the reaction \( 2Cu_2O(s) + Cu_2S(s) \rightarrow 6Cu(s) + SO_2(g) \) is wrong, we need to analyze the oxidation states of the elements involved in the reaction.

### Step-by-Step Solution:

1. **Identify the oxidation states of the reactants and products:**
- In \( Cu_2O \), copper (Cu) has an oxidation state of +1 (since oxygen is -2, and there are two copper atoms).
- In \( Cu_2S \), copper (Cu) also has an oxidation state of +1 (sulfur is -2).
- In the product \( Cu \), copper has an oxidation state of 0.
- In \( SO_2 \), sulfur has an oxidation state of +4 (since oxygen is -2 and there are two oxygen atoms).

2. **Determine changes in oxidation states:**
- For \( Cu_2O \): The oxidation state of Cu changes from +1 to 0 (reduction).
- For \( Cu_2S \): The oxidation state of Cu remains +1, but sulfur is oxidized from -2 to +4 (oxidation).

3. **Identify which species are oxidized and reduced:**
- \( Cu_2O \) is reduced (gains electrons).
- \( Cu_2S \) is oxidized (loses electrons).

4. **Evaluate the statements:**
- **Statement (I):** Both \( Cu_2O \) and \( Cu_2S \) are reduced. **(False)**
- **Statement (II):** \( Cu_2S \) is reduced while \( Cu_2O \) is oxidized. **(False)**
- **Statement (III):** \( Cu_2S \) is the oxidant. **(True)**
- **Statement (IV):** \( Cu_2O \) is reduced while \( Cu_2S \) is oxidized. **(True)**

5. **Conclusion:**
The wrong statements are (I) and (II). However, since the question asks for one wrong statement, we can conclude that statement (I) is the most incorrect as it states both are reduced.

### Final Answer:
The wrong statement is **(I) both \( Cu_2O \) and \( Cu_2S \) are reduced.**