In the reaction `H_(2)S+SO_(2) to S+H_(2)O, H_(2)S` is

Options

A. an oxidising agent
B. Reducing agent
C. Both oxidising and reducing agent
D. Evolved

Answer

To determine how \( H_2S \) is acting in the reaction \( H_2S + SO_2 \rightarrow S + H_2O \), we need to analyze the oxidation states of the elements involved in the reaction.

### Step-by-Step Solution:

1. **Identify the Reactants and Products:**
The reactants are \( H_2S \) and \( SO_2 \), and the products are sulfur (\( S \)) and water (\( H_2O \)).

2. **Assign Oxidation States:**
- In \( H_2S \):
- Hydrogen (\( H \)) has an oxidation state of +1.
- Let the oxidation state of sulfur (\( S \)) be \( x \).
- The equation for \( H_2S \) is: \( 2(+1) + x = 0 \) → \( x = -2 \).
- In \( SO_2 \):
- Oxygen (\( O \)) has an oxidation state of -2.
- Let the oxidation state of sulfur (\( S \)) be \( y \).
- The equation for \( SO_2 \) is: \( y + 2(-2) = 0 \) → \( y = +4 \).
- In the product \( S \):
- The oxidation state of sulfur is 0 (elemental form).
- In \( H_2O \):
- Hydrogen is +1 and oxygen is -2, so the oxidation state of oxygen is -2.

3. **Determine Changes in Oxidation States:**
- For sulfur in \( H_2S \): Changes from -2 to 0 (oxidation).
- For sulfur in \( SO_2 \): Changes from +4 to 0 (reduction).

4. **Identify the Role of \( H_2S \):**
- Since sulfur in \( H_2S \) is losing electrons (oxidation), \( H_2S \) is acting as a reducing agent.
- The substance that is reduced (gaining electrons) is \( SO_2 \), which is acting as an oxidizing agent.

5. **Conclusion:**
- Therefore, in the reaction \( H_2S + SO_2 \rightarrow S + H_2O \), \( H_2S \) is a **reducing agent**.