Which element is oxidized in this chemical reaction?
4Na(s) + O2(g) → 2Na2O(s)
To determine which element is oxidized in a chemical reaction, you need to identify the changes in oxidation numbers of the elements involved. Here's how you can do that for the given chemical equation:
Step 1: Assign oxidation numbers to each element in the reactants and products.
In the reactants:
- Sodium (Na) is a metal, and its oxidation number is always +1.
- Oxygen (O) is usually assigned an oxidation number of -2 when it is in compounds.
So, in Na₂O, each Na ion has an oxidation number of +1, and the O atom has an oxidation number of -2.
In the product:
- Sodium (Na) still has an oxidation number of +1.
- Oxygen (O) takes on an oxidation number of -2 in compounds.
Step 2: Compare the oxidation numbers before and after the reaction.
- The oxidation number of sodium remains +1 on both sides of the equation.
- The oxidation number of oxygen in O₂ is 0 since it is in its elemental form.
- In Na₂O, the oxidation number of oxygen is -2.
Step 3: Identify the element that had a change in oxidation number.
In this reaction, the oxidation number of oxygen changed from 0 in O₂ to -2 in Na₂O.
Step 4: Determine the element that is oxidized.
Since oxygen gained electrons (its oxidation number decreased from 0 to -2), we can conclude that oxygen is the element that is being reduced. The element that is oxidized is the one that loses electrons. In this case, since only oxygen is undergoing a change in oxidation number, it is the element that is being oxidized.
Therefore, in the given chemical reaction, oxygen is oxidized.