Assign oxidation states to all atoms in the reactants and products.

The given reaction is:

• In $\text{H}_2\text{S}$: Hydrogen (H) has an oxidation state of +1. Since there are two H atoms, the total positive charge is +2. For the compound to be neutral, Sulfur (S) must have an oxidation state of -2.
• In $\text{SO}_2$: Oxygen (O) has an oxidation state of -2. Since there are two O atoms, the total negative charge is -4. For the compound to be neutral, Sulfur (S) must have an oxidation state of +4.
• In $\text{H}_2\text{O}$: Hydrogen (H) has an oxidation state of +1, and Oxygen (O) has an oxidation state of -2.
• In $\text{S}$ (elemental sulfur): Elemental forms of elements have an oxidation state of 0.

• Sulfur in $\text{H}_2\text{S}$ changes from -2 to 0 (in elemental S). This is an increase in oxidation state.
• Sulfur in $\text{SO}_2$ changes from +4 to 0 (in elemental S). This is a decrease in oxidation state.
• Hydrogen and Oxygen do not change their oxidation states.

• A substance that undergoes oxidation (increase in oxidation state) is the reducing agent.
• A substance that undergoes reduction (decrease in oxidation state) is the oxidizing agent.

Based on the changes:

• $\text{H}_2\text{S}$ is oxidized because the oxidation state of sulfur increases from -2 to 0. Therefore, $\text{H}_2\text{S}$ is the reducing agent.

• $\text{SO}_2$ is reduced because the oxidation state of sulfur decreases from +4 to 0. Therefore, $\text{SO}_2$ is the oxidizing agent.

• Reducing Agent (R.A.): $\boxed{\text{H}_2\text{S}}$

  • Reason: The sulfur in $\text{H}_2\text{S}$ is oxidized from an oxidation state of -2 to 0, meaning it loses electrons and causes the reduction of $\text{SO}_2$.

• Oxidizing Agent (O.A.): $\boxed{\text{SO}_2}$

  • Reason: The sulfur in $\text{SO}_2$ is reduced from an oxidation state of +4 to 0, meaning it gains electrons and causes the oxidation of $\text{H}_2\text{S}$.