When the gases sulphur dioxide and hydrogen sulphide mix in the presence of water, the reaction
`SO_(2) + 2H_(2) S to 2H_(2) O + 3S`
Here hydrogen sulphide is acting as

Options

A. an oxidising agent
B. a reducing agent
C. a dehydrating agent
D. a catalyst

Answer

To determine how hydrogen sulfide (H₂S) is acting in the reaction with sulfur dioxide (SO₂) in the presence of water, we can analyze the reaction step by step.

### Step 1: Write the Reaction
The reaction given is:
\[ \text{SO}_2 + 2\text{H}_2\text{S} \rightarrow 2\text{H}_2\text{O} + 3\text{S} \]

### Step 2: Identify the Oxidation States
Next, we need to determine the oxidation states of sulfur in the reactants and products:
- In SO₂, sulfur (S) has an oxidation state of +4.
- In H₂S, sulfur has an oxidation state of -2.
- In the products, elemental sulfur (S) has an oxidation state of 0 in its elemental form.

### Step 3: Analyze the Changes in Oxidation States
Now, let's analyze the changes:
- The sulfur in SO₂ goes from +4 to 0 (reduction).
- The sulfur in H₂S goes from -2 to 0 (oxidation).

### Step 4: Determine the Role of H₂S
Since H₂S is losing electrons (its oxidation state is increasing from -2 to 0), it is acting as a reducing agent. A reducing agent is a substance that donates electrons to another substance, causing the other substance to be reduced.

### Conclusion
In this reaction, hydrogen sulfide (H₂S) is acting as a reducing agent because it undergoes oxidation while causing the sulfur dioxide (SO₂) to be reduced.

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