Justify that the reaction
`2Cu_(2)O_(s)+Cu_(2)S(s)rarr6Cu(s)+SO_(2)(g)` a redox reaction. Identify the species oxidised`//`reduced. Which acts as an oxidanat and which acts as a reductant?

Explanation

To determine if the reaction

\[ 2Cu_{2}O_{(s)} + Cu_{2}S_{(s)} \rightarrow 6Cu_{(s)} + SO_{2(g)} \]

is a redox reaction, we need to analyze the oxidation states of the elements involved in the reaction.

### Step 1: Assign Oxidation States
- In \( Cu_{2}O \), copper (Cu) has an oxidation state of +1 (since oxygen is -2, and the compound is neutral).
- In \( Cu_{2}S \), copper (Cu) also has an oxidation state of +1 (since sulfur is -2).
- In the product \( Cu \), copper has an oxidation state of 0 (elemental form).
- In \( SO_{2} \), sulfur (S) has an oxidation state of +4 (since oxygen is -2 and there are two oxygen atoms).

### Step 2: Identify Changes in Oxidation States
- Copper in \( Cu_{2}O \) changes from +1 to 0, indicating a reduction (gain of electrons).
- Copper in \( Cu_{2}S \) remains at +1, but sulfur changes from -2 to +4, indicating an oxidation (loss of electrons).

### Step 3: Determine Which Species is Oxidized and Which is Reduced
- **Oxidized Species**: Sulfur in \( Cu_{2}S \) (oxidation state increases from -2 to +4).
- **Reduced Species**: Copper in \( Cu_{2}O \) (oxidation state decreases from +1 to 0).

### Step 4: Identify Oxidizing and Reducing Agents
- **Oxidizing Agent**: \( Cu_{2}O \) (because it causes the oxidation of sulfur).
- **Reducing Agent**: \( Cu_{2}S \) (because it causes the reduction of copper).

### Conclusion
Since there is both a reduction (gain of electrons) and an oxidation (loss of electrons) occurring in the reaction, we can conclude that this is indeed a redox reaction.

### Summary of Findings
- The reaction is a redox reaction.
- Species oxidized: Sulfur in \( Cu_{2}S \).
- Species reduced: Copper in \( Cu_{2}O \).
- Oxidizing agent: \( Cu_{2}O \).
- Reducing agent: \( Cu_{2}S \).

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