Oxygen is acting as an oxidizing agent in all of the following reactions EXCEPT — Electrochemistry Chemistry Question

STEPS:

1. Define Oxidizing Agent: An oxidizing agent is a substance that causes another species to be oxidized by accepting electrons. In this process, the oxidizing agent itself is reduced, meaning its oxidation number decreases.
2. Identify Initial Oxidation States: In all the provided reactions, oxygen starts in its elemental form as $O_2(g)$. The oxidation number of any element in its free, uncombined state is 0.
3. Evaluate the Role of Electronegativity: Oxygen is the second most electronegative element. In most reactions with other elements, oxygen will attract electrons and take on a negative oxidation state (usually -2), which makes it a very common oxidizing agent.
4. Identify the Exception (Fluorine): The only element on the periodic table with a higher electronegativity than oxygen is Fluorine. When these two elements react, Fluorine will "win" the electrons.
5. Analyze Choice C ($OF_2$): In the compound oxygen difluoride ($OF_2$), Fluorine must take its standard oxidation state of -1. To balance the two Fluorine atoms and maintain a neutral molecule, the oxygen atom must have an oxidation number of +2.
6. Determine the Electron Transfer in C: Because oxygen's oxidation number increases from 0 (in $O_2$) to +2 (in $OF_2$), it has lost electrons and has been oxidized.
7. Conclusion: Since oxygen is oxidized in reaction C, it is acting as a reducing agent in that specific case, while Fluorine acts as the oxidizing agent.

WHY_OTHERS_WRONG:

• A, B, D, and E: In these reactions, oxygen is combined with elements that are less electronegative than itself ($C, S, Na,$ and $Mg$). In the resulting products ($CO, SO_2, Na_2O_2,$ and $MgO$), oxygen takes on a negative oxidation state (-2 or -1). Because oxygen's oxidation number decreases from 0 to a negative value in all these cases, it is being reduced and is successfully acting as an oxidizing agent.