Final answer:
In the given chemical reaction, Cu is reduced from an oxidation state of +1 in Cu2O to 0 in metallic Cu, while oxygen's state changes from a formal state of +1 in Cu2O to its elemental state of 0 in O2. The reaction showcases the concepts of oxidation and reduction which are integral parts of a redox reaction.
Step-by-step explanation:
The question relates to the change in oxidation number of the metal atom in the chemical reaction 2Cu2O → 4Cu + O2. When analyzing this reaction, we can establish that copper (Cu) is starting with an oxidation state of +1 in Cu2O, and in the product, it is present as Cu0 and O2.
This means Cu is reduced from +1 to 0, because oxidation is the loss of electrons and reduction is the gain of electrons. The element oxygen (O) is in a compound with copper, and oxygen normally has an oxidation state of -2, but in this compound, it has a part in forming Cu2O, so it effectively has a +1 state there.
When it becomes O2 in the products, it goes back to its elemental state with an oxidation number of 0. Therefore, we see that oxygen is neither reduced nor oxidized in this reaction.
Changes in Oxidation Number in Redox Reactions
The changes in oxidation states during the reaction show a disproportionation reaction, where the same element is simultaneously oxidized and reduced. This is a key concept in redox reactions, which involve a transfer of electrons leading to changes in oxidation states, indicating whether the substances are being oxidized (increase in oxidation number) or reduced (decrease in oxidation number).