For the following reaction, identify the substances oxidized, reduced and the oxidizing and reducing agents?

2SO2 + O2 --> 2SO3

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To determine the substances oxidized and reduced, as well as the oxidizing and reducing agents in a chemical reaction, we need to analyze the changes in oxidation numbers.

Let's start by assigning oxidation numbers to the elements in the equation. In this case, the oxidation numbers of sulfur (S) will be of particular interest.

In SO2, the oxidation number of sulfur is +4. Oxygen is usually assigned an oxidation number of -2, so the total oxidation number of sulfur would be balanced accordingly.

In SO3, the oxidation number of sulfur is +6, assuming the same oxidation number for oxygen.

Comparing the oxidation numbers of sulfur in SO2 (+4) and SO3 (+6), we can see that sulfur has been oxidized. It has gained two electrons, going from +4 to +6.

Therefore, sulfur is being oxidized in this reaction.

The substance that is reduced is the one gaining the electrons, in this case, oxygen. It goes from an oxidation number of 0 in O2 to -2 in SO3, gaining two electrons.

Therefore, oxygen is being reduced in this reaction.

Now, let's determine the oxidizing and reducing agents.

The oxidizing agent is the species that causes another substance to undergo oxidation. In this reaction, O2 is responsible for oxidizing sulfur, so O2 is the oxidizing agent.

The reducing agent is the species that causes another substance to undergo reduction. In this reaction, SO2 is responsible for reducing oxygen, so SO2 is the reducing agent.

To summarize:
- Sulfur (S) is oxidized in the reaction.
- Oxygen (O) is reduced in the reaction.
- The oxidizing agent is O2.
- The reducing agent is SO2.