H2 2 S(g) ( g ) + Cl2 2 (g) ( g ) → 2HCl(g) ( g ) + S(s) ( s ) What is the change in oxidation state of sulphur from reactant to product?

H2 ${}_2$S(g) ${}_{(g)}$  +  Cl2 ${}_2$(g) ${}_{(g)}$ → 2HCl(g) ${}_{(g)}$ + S(s) ${}_{(s)}$

What is the change in oxidation state of sulphur from reactant to product?

Answer Details

To determine the change in oxidation state of sulfur, follow these steps:

In the given reaction:

H₂S(g) + Cl₂(g) → 2HCl(g) + S(s)

We observe:

• **Hydrogen sulfide (H₂S)** is a molecule where hydrogen (H) with an oxidation state of **+1** bonds with sulfur (S). Since the molecule is neutral, the oxidation state of sulfur in H₂S must balance out the positive oxidation states of the two hydrogen atoms, which is **+1** each. Therefore, the oxidation state of sulfur in H₂S is **-2**.
• **Sulfur (S) in elemental form** at the products is just sulfur solid (S_(s)), and for elements in their natural state, the oxidation state is **0**.

Thus, the change in oxidation state of sulfur when moving from the reactants to the products is from **-2** to **0**. This indicates that sulfur is being oxidized.

The correct answer is that the oxidation state of sulfur changes from **-2 to 0**.