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

Step 1: Assign oxidation states to all atoms in the reactants and products. The given reaction is: 2H_2S + SO_2 → 2H_2O + 3S In H_2S: 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 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 H_2O: Hydrogen (H) has an oxidation state of +1, and Oxygen (O) has an oxidation state of -2. In S (elemental sulfur): Elemental forms of elements have an oxidation state of 0. Step 2: Identify the changes in oxidation states. Sulfur in H_2S changes from -2 to 0 (in elemental S). This is an increase* in oxidation state. Sulfur in 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. Step 3: Determine the Reducing Agent (R.A.) and Oxidizing Agent (O.A.). 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: H_2S is oxidized because the oxidation state of sulfur increases from -2 to 0. Therefore, H_2S is the reducing agent. SO_2 is reduced because the oxidation state of sulfur decreases from +4 to 0. Therefore, SO_2 is the oxidizing agent. Reducing Agent (R.A.): H_2S Reason: The sulfur in H_2S is oxidized from an oxidation state of -2 to 0, meaning it loses electrons and causes the reduction of SO_2. Oxidizing Agent (O.A.): SO_2 Reason: The sulfur in SO_2 is reduced from an oxidation state of +4 to 0, meaning it gains electrons and causes the oxidation of H_2S. That's 2 down. 3 left today — send the next one.