After partial roasting, the sulphide of copper can be reduced by

Options

A. Carbon
B. Self-reduction
C. Cyanide process
D. Heating

Explanation

To solve the question "After partial roasting, the sulphide of copper can be reduced by," we need to understand the process of reduction of copper sulphide (Cu₂S) after it has undergone partial roasting.

### Step-by-Step Solution:

1. **Understanding Partial Roasting**:
- Partial roasting of copper sulphide (Cu₂S) involves heating it in the presence of limited oxygen. This process converts some of the sulphide into copper(I) oxide (Cu₂O) and sulfur dioxide (SO₂).
- The reaction can be represented as:
\[ 2Cu_2S + 3O_2 \rightarrow 2Cu_2O + 2SO_2 \]

2. **Reduction of Copper(I) Oxide**:
- After partial roasting, we have copper(I) oxide (Cu₂O) present. This oxide can be reduced back to metallic copper (Cu).
- The reduction can occur using the remaining copper sulphide (Cu₂S) itself. This is known as self-reduction.
- The reaction is:
\[ Cu_2O + Cu_2S \rightarrow 6Cu + SO_2 \]

3. **Identifying the Reducing Agent**:
- In this case, the reducing agent is the copper sulphide (Cu₂S) that reduces the copper(I) oxide (Cu₂O) to metallic copper.
- This process is referred to as self-reduction because the sulphide of copper is reducing the oxide of copper.

4. **Conclusion**:
- Therefore, after partial roasting, the sulphide of copper can be reduced by its own sulphide.

### Final Answer:
The correct answer is **self-reduction**.