H$_2$S$_{(g)}$  +  Cl$_2$$_{(g)}$ → 2HCl$_{(g)}$ + S$_{(s)}$ What is the change in oxidation state of sulphur from reactant to product?

To determine the change in the oxidation state of sulphur, we need to analyze the oxidation numbers of the atoms in the reactant and the product.

1. Analyze the Reactant: $H_2S$

In a compound, the sum of oxidation states must equal zero.

• Hydrogen ($H$) typically has an oxidation state of $+1$.
• Let the oxidation state of Sulphur ($S$) be $x$.
• The equation is: $2(+1) + x = 0$
• $2 + x = 0$
• $x = -2$ So, the oxidation state of sulphur in $H_2S$ is $-2$.

2. Analyze the Product: $S_{(s)}$

• According to the rules of oxidation states, any element in its standard elemental state (uncombined form) has an oxidation state of $0$.
• Therefore, the oxidation state of sulphur in $S_{(s)}$ is $0$.

3. Determine the Change

• The sulphur atom changes from an oxidation state of $-2$ (in $H_2S$) to $0$ (in $S$).
• This represents an increase in oxidation number, which confirms that $H_2S$ is being oxidized.

Conclusion

Comparing our result to the options provided, the change is from $-2$ to $0$. This matches option C.

Exam Tip: Always remember that in redox reactions, elements like $O_2, H_2, Cl_2,$ or $S_8$ (solid sulphur) always have an oxidation state of zero. This is a quick way to identify the product side of an oxidation process!