To determine the oxidizing and reducing agents in the chemical equation [tex]\(3 \text{FeS} + 8 \text{HNO}_3 \rightarrow 3 \text{FeSO}_4 + 8 \text{NO} + 4 \text{H}_2\text{O}\)[/tex], we need to analyze the changes in oxidation states of the elements involved.
1. Identify the initial oxidation states:
- Fe in FeS: The oxidation state of Fe in FeS is +2 (since sulfur is usually -2 in compounds).
- S in FeS: The oxidation state of sulfur in FeS is -2.
- N in HNO[tex]\(_3\)[/tex]: Nitrogen in HNO[tex]\(_3\)[/tex] has an oxidation state of +5 (since oxygen is -2 and hydrogen is +1).
2. Identify the final oxidation states:
- Fe in FeSO[tex]\(_4\)[/tex]: Iron (Fe) in FeSO[tex]\(_4\)[/tex] is +2.
- S in FeSO[tex]\(_4\)[/tex]: Sulfur (S) in FeSO[tex]\(_4\)[/tex] is +6 (since sulfate, SO[tex]\(_4^{2-}\)[/tex], has a -2 charge and each oxygen is -2).
- N in NO: Nitrogen in NO has an oxidation state of +2.
3. Determine the changes in oxidation states:
- Fe remains at +2, indicating no change for Fe.
- S goes from -2 in FeS to +6 in FeSO[tex]\(_4\)[/tex]. This is an increase of 8, meaning sulfur is oxidized.
- N goes from +5 in HNO[tex]\(_3\)[/tex] to +2 in NO. This is a decrease of 3, meaning nitrogen is reduced.
4. Determine the agents:
- The substance being oxidized (in losing electrons) is FeS. Hence, FeS is the reducing agent because it provides electrons to another species.
- The substance gaining electrons/reducing (increasing in oxidation state) is HNO[tex]\(_3\)[/tex], which is therefore the oxidizing agent as it accepts electrons from another species.
Thus, the correct answer is:
C. FeS is the reducing agent, and [tex]\( \text{HNO}_3 \)[/tex] is the oxidizing agent.
So, the true answer is option C.
