For the following reaction, identify the substance oxidized, reduced and the oxidizing and reducing agents.

2SO2 + O2 ----> 2SO3

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Apply the following definitions.
Oxidation is the loss of electrons.
Reduction is the gain of electrons.
The substance oxidized is the reducing agent.
The substance reduced is the oxidizing agent.

To get you started. S in SO2 has an oxidation state of +4. In SO3 S is +6
O in O2 i zero, in SO3 it is -2.

To identify the substance oxidized, reduced, and the oxidizing and reducing agents in the given reaction:

Step 1: Assign oxidation numbers to all the elements in the reaction.

In SO2, sulfur (S) has an oxidation number of +4, and in SO3, sulfur has an oxidation number of +6. Therefore, the oxidation state of sulfur changes from +4 to +6, indicating that sulfur is oxidized.

In O2, the oxidation state of oxygen is 0, and in SO3, the oxidation state of oxygen is -2. Therefore, the oxidation state of oxygen changes from 0 to -2, indicating that oxygen is reduced.

Step 2: Determine the oxidizing agent and reducing agent based on the changes in oxidation numbers.

The substance that causes another substance to be oxidized is the oxidizing agent, and the substance that causes another substance to be reduced is the reducing agent.

In this reaction, O2 causes sulfur (in SO2) to be oxidized from +4 to +6, so O2 is the oxidizing agent.

Similarly, sulfur in SO2 causes oxygen (in O2) to be reduced from 0 to -2, so sulfur is the reducing agent.

To summarize:
- The substance oxidized is sulfur (S).
- The substance reduced is oxygen (O2).
- The oxidizing agent is oxygen (O2).
- The reducing agent is sulfur (S).

To identify the substance that is oxidized, reduced, and the oxidizing and reducing agents in a reaction, we need to assign oxidation numbers to the elements involved.

Let's start by assigning oxidation numbers to the atoms in the reaction:

In SO2, sulfur has an oxidation number of +4, and each oxygen has an oxidation number of -2.
In O2, each oxygen atom has an oxidation number of 0.
In SO3, sulfur has an oxidation number of +6, and each oxygen has an oxidation number of -2.

Now, let's look at the changes in oxidation numbers for each element:

Sulfur goes from +4 in SO2 to +6 in SO3. It is an increase in oxidation number, which means sulfur has been oxidized.

Oxygen goes from -2 in SO2 to -2 in SO3. There is no change in oxidation number, which means oxygen has not been oxidized or reduced.

To identify the oxidizing and reducing agents, we need to look at the elements or compounds that caused the changes in oxidation number.

In this reaction, O2 caused sulfur to go from +4 to +6, which means it gained electrons and caused the oxidation of sulfur. Therefore, O2 is the oxidizing agent.

On the other hand, sulfur caused oxygen to stay at oxidation number -2, which means it lost electrons and caused the reduction of oxygen. Therefore, sulfur is the reducing agent.

In summary:
Substance oxidized: Sulfur (from +4 to +6)
Substance reduced: Oxygen (remains at -2)
Oxidizing agent: O2
Reducing agent: Sulfur