Let's analyze the reaction to determine what is happening to the nitrate ion, [tex]$NO_3^{-}$[/tex]:
Given reaction:
[tex]\[ Fe + Cu\left(NO_3\right)_2 \rightarrow Cu + Fe\left(NO_3\right)_2 \][/tex]
Step-by-step analysis:
1. Identify the reactants and products:
- Reactants: [tex]\( Fe \)[/tex] and [tex]\( Cu\left(NO_3\right)_2 \)[/tex]
- Products: [tex]\( Cu \)[/tex] and [tex]\( Fe\left(NO_3\right)_2 \)[/tex]
2. Separate the reaction into half-reactions to determine the oxidation and reduction processes:
- For iron ([tex]\( Fe \)[/tex]):
[tex]\[ Fe \rightarrow Fe^{2+} + 2e^{-} \][/tex]
Iron loses 2 electrons, hence it is oxidized.
- For copper in copper(II) nitrate ([tex]\( Cu(NO_3)_2 \)[/tex]):
[tex]\[ Cu^{2+} + 2e^{-} \rightarrow Cu \][/tex]
Copper(II) ion gains 2 electrons, hence it is reduced.
3. Look at the nitrate ion ([tex]$NO_3^-$[/tex]):
- In [tex]\( Cu(NO_3)_2 \)[/tex], nitrate ion is [tex]\( NO_3^- \)[/tex].
- In [tex]\( Fe(NO_3)_2 \)[/tex], nitrate ion is still [tex]\( NO_3^- \)[/tex].
4. Compare the states of the nitrate ion in reactants and products:
- In both reactants and products, the nitrate ion remains as [tex]\( NO_3^- \)[/tex].
- There is no change in the oxidation state of the nitrate ion.
Consequently, the nitrate ion ([tex]\( NO_3^- \)[/tex]) is not undergoing any oxidation or reduction; it remains unchanged throughout the reaction.
Thus, the correct answer is:
[tex]\[ \text{It is not getting oxidized or reduced} \][/tex]
