The reaction $2H_2 S + SO_2 \to 3S + 2H_2 O$ is

To determine the nature of this reaction, we need to analyze the oxidation states of the elements involved. A redox reaction occurs when there is a change in the oxidation state of atoms during a chemical process.

1. Analyze the oxidation states

Let's look at the oxidation states of the elements in the reactants and products:

• In $H_2S$: Hydrogen is $+1$. Since the molecule is neutral, $2(+1) + S = 0$, so Sulfur ($S$) is $-2$.
• In $SO_2$: Oxygen is $-2$. Since the molecule is neutral, $S + 2(-2) = 0$, so Sulfur ($S$) is $+4$.
• In $3S$ (elemental sulfur): The oxidation state of an element in its standard state is $0$.
• In $H_2O$: Hydrogen is $+1$ and Oxygen is $-2$.

2. Identify the changes

• Sulfur in $H_2S$: Changes from $-2$ to $0$. This is an increase in oxidation number, which is oxidation. The substance that undergoes oxidation is the reductant (reducing agent). Therefore, $H_2S$ is the reductant.
• Sulfur in $SO_2$: Changes from $+4$ to $0$. This is a decrease in oxidation number, which is reduction. The substance that undergoes reduction is the oxidant (oxidizing agent). Therefore, $SO_2$ is the oxidant.

3. Conclusion

Since both oxidation and reduction have occurred, this is definitely a redox reaction. Specifically, $H_2S$ loses electrons (is oxidized) and $SO_2$ gains electrons (is reduced).

Comparing this to your options:

• Option D states: "a redox reaction in which $SO_2$ is the oxidant and $H_2S$ is the reductant." This matches our findings perfectly.

Exam Tip

Remember the mnemonic OIL RIG:

• Oxidation Is Loss (of electrons/increase in oxidation number).
• Reduction Is Gain (of electrons/decrease in oxidation number). The substance that gets oxidized is the reducing agent, and the substance that gets reduced is the oxidizing agent.