Let's analyze the reaction:
[tex]\[
2 Fe + 3 S \rightarrow Fe_2S_3
\][/tex]
### Determining Oxidation States
1. Iron (Fe):
- In the elemental form [tex]\( Fe \)[/tex], iron has an oxidation state of 0.
- In [tex]\( Fe_2S_3 \)[/tex], iron has a +3 oxidation state (each [tex]\( Fe \)[/tex] contributes +3).
2. Sulfur (S):
- In the elemental form [tex]\( S \)[/tex], sulfur has an oxidation state of 0.
- In [tex]\( Fe_2S_3 \)[/tex], sulfur has a -2 oxidation state (each [tex]\( S \)[/tex] contributes -2).
### Writing the Half-Reactions
#### Oxidation Half-Reaction
Oxidation involves the loss of electrons. Here, iron is losing electrons to go from an oxidation state of 0 to +3.
Thus, the balanced oxidation half-reaction is:
[tex]\[
2 Fe \rightarrow 2 Fe^{3+} + 6 e^-
\][/tex]
#### Reduction Half-Reaction
Reduction involves the gain of electrons. Here, sulfur is gaining electrons to go from an oxidation state of 0 to -2.
Thus, the balanced reduction half-reaction is:
[tex]\[
3 S + 6 e^- \rightarrow 3 S^{2-}
\][/tex]
### Summary
These balanced half-reactions describe the processes of oxidation and reduction in the given reaction:
\begin{tabular}{|l|l|}
\hline
oxidation: & [tex]\(2 Fe \rightarrow 2 Fe^{3+} + 6 e^-\)[/tex] \\
reduction: & [tex]\(3 S + 6 e^- \rightarrow 3 S^{2-}\)[/tex] \\
\hline
\end{tabular}
