To address this problem, let's carefully examine the chemical reaction and the changes occurring:
The provided reaction is:
[tex]$ Fe (s) + CuSO _4(a q) \rightarrow FeSO _4(a q) + Cu (s) $[/tex]
Here's a step-by-step analysis:
1. Identify the Reactants and Products:
- Reactants: Iron (Fe), Copper sulfate (CuSO₄)
- Products: Iron (II) sulfate (FeSO₄), Copper (Cu)
2. Determine the Oxidation States:
- In the reactant CuSO₄, copper exists as [tex]\(Cu^{2+}\)[/tex] and sulfate exists as [tex]\(SO_4^{2-}\)[/tex].
- Iron [tex]\(Fe\)[/tex] is in its elemental state, with an oxidation state of 0.
- In the product FeSO₄, iron exists as [tex]\(Fe^{2+}\)[/tex] and sulfate as [tex]\(SO_4^{2-}\)[/tex].
- Copper [tex]\(Cu\)[/tex] in the product is in its elemental state with an oxidation state of 0.
3. Observe the Change in Oxidation States:
- Iron ([tex]\(Fe\)[/tex]) changes from an oxidation state of 0 to [tex]\(Fe^{2+}\)[/tex] (oxidized, losing electrons).
- Copper ([tex]\(Cu^{2+}\)[/tex]) changes from an oxidation state of [tex]\(Cu^{2+}\)[/tex] to 0 (reduced, gaining electrons).
4. Identify the Type of Reaction for Copper (Cu):
- Copper ([tex]\(Cu^{2+}\)[/tex]) gains electrons to form neutral copper (Cu). This process of gaining electrons is known as reduction.
By understanding the loss and gain of electrons, we can conclude that the copper ion ([tex]\(Cu^{2+}\)[/tex]) is reduced.
Therefore, the correct statement about the reaction is:
C. Cu undergoes reduction.
