To answer questions about the reducing and oxidizing processes in the given reaction, it is essential to analyze the oxidation states of each element in the reactants and products.
The given reaction is:
[tex]\[ 2 \text{NO} (g) + 2 \text{CO} (g) \rightarrow \text{N}_2 (g) + 2 \text{CO}_2 (g) \][/tex]
First, let's assign and track the oxidation states of each element.
1. Nitrogen in NO:
- In NO, nitrogen is bonded to oxygen, which has an oxidation state of -2.
- To balance the -2 from oxygen, nitrogen must have an oxidation state of +2 (, since NO is a neutral molecule).
2. Carbon in CO:
- In CO, oxygen has an oxidation state of -2.
- Therefore, carbon must have an oxidation state of +2 to balance out the -2 from oxygen (since CO is neutral).
3. Products:
- [tex]\(\text{N}_2\)[/tex]: In diatomic nitrogen ([tex]\(\text{N}_2\)[/tex]), each nitrogen atom has an oxidation state of 0.
- [tex]\(\text{CO}_2\)[/tex]: In carbon dioxide ([tex]\(\text{CO}_2\)[/tex]), each oxygen has an oxidation state of -2. Since there are 2 oxygens, the total oxidation state for oxygen is -4. To balance this, carbon must have an oxidation state of +4.
Now, summarizing the changes:
- Nitrogen:
- Changes from +2 in NO to 0 in [tex]\(\text{N}_2\)[/tex]. This indicates a reduction since the oxidation number decreases.
- Carbon:
- Changes from +2 in CO to +4 in [tex]\(\text{CO}_2\)[/tex]. This indicates oxidation since the oxidation number increases.
Based on these changes, we can make the following observations:
- CO is oxidized (its oxidation state increases from +2 to +4), making CO the reducing agent.
- NO is reduced (its oxidation state decreases from +2 to 0), making NO the oxidizing agent.
The correct statement among the options provided is:
The oxidation state of nitrogen in NO changes from +2 to 0, and the oxidation state of carbon in CO changes from +2 to +4 as the reaction proceeds.
