H$_2$ S$_{(g)}$ + Cl$_{2(g)}$ → 2HCl$_{(g)}$ + S$_{(g)}$ In the reaction above, the substance that is reduced is

To determine which substance is reduced in the reaction, we need to analyze the changes in the oxidation states of the elements involved.

1. Analyze the concept

In chemistry, reduction is defined as:

• A decrease in oxidation state.
• The gain of electrons.
• The removal of oxygen or the addition of hydrogen.

We use the acronym OIL RIG: Oxidation Is Loss (of electrons), Reduction Is Gain (of electrons).

2. Step-by-step solution

Let's assign oxidation numbers to each element in the reaction: H$_2$S$_{(g)}$ + Cl$_{2(g)}$ → 2HCl$_{(g)}$ + S$_{(s)}$

• In H$_2$S: Hydrogen (H) is +1, so Sulfur (S) must be -2 (since the total charge of the molecule is 0).
• In Cl$_2$: This is an element in its standard state, so its oxidation number is 0.
• In HCl: Hydrogen (H) is +1, so Chlorine (Cl) must be -1.
• In S: This is an element in its standard state, so its oxidation number is 0.

Now, let's track the changes:

• Sulfur (S): Changes from -2 (in H$_2$S) to 0 (in S). An increase in oxidation number indicates oxidation.
• Chlorine (Cl): Changes from 0 (in Cl$_2$) to -1 (in HCl). A decrease in oxidation number indicates reduction.

3. Conclusion

Since the oxidation state of Chlorine decreased from 0 to -1, Chlorine (Cl$_2$) is the substance that gained electrons and was reduced.

Exam Tip

Always remember that in a reaction between a non-metal element (like Cl$_2$) and a compound containing a non-metal (like H$_2$S), the element that is "free" on the reactant side (Cl$_2$) is almost always the one being reduced, as it is gaining electrons to form an ion.