Identify oxidising & Reducing Agent -
`PbS+4H_(2)O_(2) rarr PbSO_(4) + 4H_(2)O`

Options

A. `PbS,H_(2)O_(2)`
B. `H_(2)O_(2),PbS`
C. `H_(2)O_(2)`both
D. PbS both

Answer

To identify the oxidizing and reducing agents in the reaction \( \text{PbS} + 4 \text{H}_2\text{O}_2 \rightarrow \text{PbSO}_4 + 4 \text{H}_2\text{O} \), we will follow these steps:

### Step 1: Assign Oxidation States
First, we need to assign oxidation states to each element in the reactants and products.

- **In PbS**:
- Lead (Pb) is in the +2 oxidation state.
- Sulfur (S) is in the -2 oxidation state.

- **In H2O2**:
- Each hydrogen (H) is in the +1 oxidation state.
- Each oxygen (O) in hydrogen peroxide is in the -1 oxidation state.

- **In PbSO4**:
- Lead (Pb) is in the +2 oxidation state.
- Sulfur (S) is in the +6 oxidation state (calculated as follows: let x be the oxidation state of sulfur, then \( x + 4(-2) = 0 \) leads to \( x - 8 = 0 \) or \( x = +6 \)).
- Each oxygen (O) is in the -2 oxidation state.

- **In H2O**:
- Each hydrogen (H) is in the +1 oxidation state.
- Each oxygen (O) is in the -2 oxidation state.

### Step 2: Determine Changes in Oxidation States
Now, we will analyze the changes in oxidation states:

- **Sulfur (S)** changes from -2 in PbS to +6 in PbSO4. This is an increase in oxidation state, indicating that sulfur is oxidized.

- **Oxygen (O)** in H2O2 changes from -1 to -2 in H2O. This is a decrease in oxidation state, indicating that oxygen is reduced.

### Step 3: Identify the Oxidizing and Reducing Agents
- The substance that is oxidized (loses electrons) acts as the **reducing agent**. In this case, PbS is oxidized, so it is the reducing agent.

- The substance that is reduced (gains electrons) acts as the **oxidizing agent**. Here, H2O2 is reduced, so it is the oxidizing agent.

### Conclusion
- **Oxidizing Agent**: H2O2
- **Reducing Agent**: PbS