What is happening to [tex]$NO_3^{-}$[/tex] in the following reaction?

[tex]Fe + Cu(NO_3)_2 \rightarrow Cu + Fe(NO_3)_2[/tex]

A. It is getting oxidized
B. It is getting reduced
C. It is not getting oxidized or reduced



Answer :

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]