In the reaction betwee.n lead sulphide and hydrogen peroxide which substance is reduced?
`PbS + 4H_2O_ 2 to PbSO_4 + 4H_2O`

Options

A. Lead sulphide
B. Hydrogenperoxide
C. Lead sulphate
D. Water

Explanation

To determine which substance is reduced in the reaction between lead sulfide (PbS) and hydrogen peroxide (H₂O₂), we can follow these steps:

### Step 1: Write down the reaction
The given reaction is:
\[ \text{PbS} + 4 \text{H}_2\text{O}_2 \rightarrow \text{PbSO}_4 + 4 \text{H}_2\text{O} \]

### Step 2: Identify the oxidation states
- **For PbS**:
- Lead (Pb) typically has an oxidation state of +2.
- Sulfide (S) has an oxidation state of -2.

- **For H₂O₂**:
- Each hydrogen (H) has an oxidation state of +1.
- The oxidation state of oxygen (O) in hydrogen peroxide is -1.

- **For PbSO₄**:
- Lead (Pb) has an oxidation state of +2.
- Sulfate (SO₄) has a total oxidation state of -2 (with sulfur at +6 and oxygen at -2).

### Step 3: Analyze the changes in oxidation states
- In the reaction, PbS is converted to PbSO₄.
- The sulfur in PbS goes from -2 to +6 in PbSO₄, indicating that it is oxidized (loss of electrons).
- The oxygen in H₂O₂ goes from -1 to -2 in water (H₂O), indicating that it is reduced (gain of electrons).

### Step 4: Determine which substance is reduced
- Since reduction is defined as the gain of electrons or the removal of oxygen, we can conclude that:
- H₂O₂ is reduced because the oxidation state of oxygen decreases from -1 to -2, indicating a gain of electrons.

### Conclusion
In the reaction between lead sulfide and hydrogen peroxide, the substance that is reduced is **hydrogen peroxide (H₂O₂)**.

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