SO42- + HAsO2+ 2H+H3AsO4 + SO2

In the above redox reaction, use oxidation numbers to identify the element oxidized, the element reduced, the oxidizing agent and the reducing agent.

1. name of the element oxidized:
2. name of the element reduced:
3. formula of the oxidizing agent:
4. formula of the reducing agent:

See your post above.

To identify the element oxidized and the element reduced, as well as the oxidizing agent and the reducing agent, we need to determine the changes in oxidation numbers for the elements involved in the redox reaction.

1. The element oxidized is the element that undergoes an increase in oxidation number. In this reaction, the sulfur atom in SO42- has an oxidation number of +6, but in the product H2SO4, it has an oxidation number of +4. Therefore, sulfur (S) is the element oxidized.

2. The element reduced is the element that undergoes a decrease in oxidation number. In this reaction, the arsenic atom in HAsO2+ has an oxidation number of +3, but in the product H3AsO4, it has an oxidation number of +5. Therefore, arsenic (As) is the element reduced.

3. The oxidizing agent is the species that causes the oxidation of another species by accepting electrons. In this reaction, HAsO2+ is reduced, suggesting that it acts as the oxidizing agent. The formula of the oxidizing agent is HAsO2+.

4. The reducing agent is the species that causes the reduction of another species by donating electrons. In this reaction, the SO42- ion is oxidized, implying that it acts as the reducing agent. The formula of the reducing agent is SO42-.

To summarize:
1. Element oxidized: Sulfur (S)
2. Element reduced: Arsenic (As)
3. Oxidizing agent: HAsO2+
4. Reducing agent: SO42-