To determine which element is getting reduced in the reaction, we need to look at the oxidation states of the elements involved in the reactants and products:
[tex]\[
Fe + Cu\left(NO_3\right)_2 \rightarrow Cu + Fe\left(NO_3\right)_2
\][/tex]
1. Oxidation States in Reactants:
- [tex]\(Fe\)[/tex]: In its elemental form, it has an oxidation state of [tex]\(0\)[/tex].
- [tex]\(Cu\left(NO_3\right)_2\)[/tex]: The nitrate ion [tex]\(\left(NO_3^{-}\right)\)[/tex] has a charge of [tex]\(-1\)[/tex]. Since there are two nitrates, their total charge is [tex]\(-2\)[/tex]. For the compound to be neutral, the copper ion must have a charge of [tex]\(+2\)[/tex]. Hence, the oxidation state of copper in [tex]\(Cu\left(NO_3\right)_2\)[/tex] is [tex]\(+2\)[/tex].
2. Oxidation States in Products:
- [tex]\(Cu\)[/tex]: In its elemental form, it has an oxidation state of [tex]\(0\)[/tex].
- [tex]\(Fe\left(NO_3\right)_2\)[/tex]: Similar to the reactants, the nitrate ion still has a [tex]\(-1\)[/tex] charge, and since there are two nitrates, their total charge is [tex]\(-2\)[/tex]. Therefore, the iron ion must have a [tex]\(+2\)[/tex] charge for the compound to be neutral. Hence, the oxidation state of iron in [tex]\(Fe\left(NO_3\right)_2\)[/tex] is [tex]\(+2\)[/tex].
3. Identify the Changes in Oxidation States:
- Iron (Fe): Goes from [tex]\(0\)[/tex] to [tex]\(+2\)[/tex], indicating it is losing electrons and getting oxidized.
- Copper (Cu): Goes from [tex]\(+2\)[/tex] to [tex]\(0\)[/tex], indicating it is gaining electrons and getting reduced.
4. Conclusion:
The element that is getting reduced is the one gaining electrons, which is copper in this case. The copper ion [tex]\(\mathbf{Cu^{+2}}\)[/tex], supplied by [tex]\(Cu\left(NO_3\right)_2\)[/tex], is the species getting reduced as it gains 2 electrons to become elemental copper ([tex]\(\mathbf{Cu}\)[/tex]).
Thus, the correct answer is:
[tex]\[\boxed{Cu^{+2} \text{ (supplied by the } Cu\left(NO_3\right)_2\text{ )}}\][/tex]
