Final answer:
In the reaction Fe (s) + CuSO4(aq) → FeSO4(aq) + Cu (s), CuSO4 is the oxidizing agent as it gets reduced from Cu with a +2 oxidation state to elemental Cu with an oxidation state of 0, while Fe is the reducing agent as it gets oxidized from elemental Fe with an oxidation state of 0 to Fe in FeSO4 with a +2 oxidation state.
Step-by-step explanation:
In the oxidation-reduction reaction Fe (s) + CuSO4(aq) → FeSO4(aq) + Cu (s), the element that is the oxidizing agent is CuSO4, as it contains copper (Cu) which goes from a +2 oxidation state in CuSO4 to 0 in elemental copper (Cu). This decrease in oxidation state indicates that copper (Cu) is reduced. The reducing agent is Fe (s), or iron, which is oxidized as its oxidation state increases from 0 in the elemental state to +2 in FeSO4. Consequently, we see the iron (Fe) losing electrons, and the copper (Cu) gaining electrons in this redox process.
The changes in oxidation states are as follows: Fe goes from 0 in the elemental state to +2 in FeSO4 (indicating oxidation), and Cu goes from +2 in CuSO4 to 0 in its elemental state (indicating reduction).