Certainly! To analyze the given chemical reaction [tex]\( Fe + I_2 \rightarrow FeI_2 \)[/tex], we need to identify the oxidation and reduction half-reactions that occur. Here’s a detailed step-by-step breakdown of how to do this:
1. Identify the Reactants and Products:
- Reactants: Iron (Fe) and Iodine (I[tex]\(_2\)[/tex])
- Product: Iron(II) iodide (FeI[tex]\(_2\)[/tex])
2. Determine Oxidation States:
- Iron (Fe) starts with an oxidation state of 0.
- In the product [tex]\(FeI_2\)[/tex], iron (Fe) has an oxidation state of [tex]\(+2\)[/tex].
- Iodine (I[tex]\(_2\)[/tex]) in elemental form has an oxidation state of 0.
- In [tex]\(FeI_2\)[/tex], each iodine ([tex]\(I^-\)[/tex]) has an oxidation state of [tex]\(-1\)[/tex].
3. Write the Oxidation Half-Reaction:
- Iron is getting oxidized. The oxidation state of iron goes from 0 to +2.
- To balance the charges, iron loses 2 electrons.
- The oxidation half-reaction is:
[tex]\[
Fe \rightarrow Fe^{2+} + 2e^-
\][/tex]
4. Write the Reduction Half-Reaction:
- Iodine is getting reduced. The oxidation state of iodine goes from 0 to -1.
- Since [tex]\(I_2\)[/tex] is a diatomic molecule, it will gain a total of 2 electrons (1 electron per iodine atom).
- The reduction half-reaction is:
[tex]\[
I_2 + 2e^- \rightarrow 2I^-
\][/tex]
5. Combining the Half-Reactions:
- After identifying and balancing the half-reactions, we can see how the electrons are transferred in the overall reaction.
- Pairing the oxidation and reduction half-reactions together, we see that the 2 electrons lost by iron are gained by the iodine molecules, ensuring the reaction is balanced.
In conclusion, the balanced half-reactions describing the oxidation and reduction in this reaction are:
1. Oxidation Half-Reaction:
[tex]\[
Fe \rightarrow Fe^{2+} + 2e^-
\][/tex]
2. Reduction Half-Reaction:
[tex]\[
I_2 + 2e^- \rightarrow 2I^-
\][/tex]
These step-by-step details illustrate how the oxidation and reduction processes balance out in the given chemical reaction.
