Sure, let's analyze the given reaction step by step to identify the reducing agent. The reaction is:
[tex]\[ \text{Cu} (s) + 4 \text{HNO}_3 (aq) \rightarrow \text{Cu}\left(\text{NO}_3\right)_2 (aq) + 2 \text{NO}_2 (g) + 2 \text{H}_2\text{O} (l) \][/tex]
To identify the reducing agent, we need to understand which species is being oxidized and which one is being reduced. Let's start by determining the oxidation states of the elements involved in the reaction.
- Copper (Cu) in its elemental form (solid state) has an oxidation state of 0.
- In [tex]\(\text{HNO}_3\)[/tex], nitrogen is typically in its +5 oxidation state (H is +1 and O is -2).
- In [tex]\(\text{Cu}\left(\text{NO}_3\right)_2\)[/tex]:
- The nitrate ion ([tex]\(\left(\text{NO}_3\right)^{-}\)[/tex]) has an overall charge of -1, with nitrogen typically being in the +5 oxidation state.
- The copper ion in [tex]\(\text{Cu}\left(\text{NO}_3\right)_2\)[/tex] has an oxidation state of +2 to balance the two nitrate ions (each -1 charge).
- In [tex]\(\text{NO}_2\)[/tex], nitrogen has an oxidation state of +4 (O is -2).
Now let’s track the changes in oxidation states:
1. Copper (Cu):
- Reactant: [tex]\(\text{Cu}\)[/tex] (s) has an oxidation state of 0.
- Product: [tex]\(\text{Cu}^{2+}\)[/tex] in [tex]\(\text{Cu}\left(\text{NO}_3\right)_2\)[/tex] has an oxidation state of +2.
2. Nitrogen (N) in HNO[tex]\(_3\)[/tex]:
- Reactant: Nitrogen in [tex]\(\text{HNO}_3\)[/tex] has an oxidation state of +5.
- Product: Nitrogen in [tex]\(\text{NO}_2\)[/tex] has an oxidation state of +4.
From these changes, we can see:
- Copper (Cu) is losing 2 electrons, going from an oxidation state of 0 to +2. Therefore, copper is being oxidized.
- Nitrogen in [tex]\(\text{NO}_3^-\)[/tex] gains 1 electron per nitrogen atom, going from an oxidation state of +5 to +4 (in [tex]\(\text{NO}_2\)[/tex]). Therefore, the nitrogen is being reduced.
The reducing agent is defined as the species that donates electrons and gets oxidized in the process. Here, copper (Cu) is donating electrons and getting oxidized.
Thus, the reducing agent in this reaction is [tex]\(\text{Cu}\)[/tex].
