To determine the reducing agent in the redox reaction, let's first understand what a reducing agent is. The reducing agent is the substance that donates electrons (is oxidized) in the reaction.
The given reaction is:
[tex]\[ 2 \text{Li} + \text{F}_2 \rightarrow 2 \text{LiF} \][/tex]
To identify the reducing agent:
1. Oxidation states:
- Lithium (`Li`) is in its elemental form, so its oxidation state is 0.
- Fluorine (`F_2`) is also in its elemental form, so its oxidation state is 0.
2. Oxidation states in the product:
- In `LiF`, lithium (`Li`) forms an ionic bond by losing an electron to fluorine (`F`).
- Therefore, the oxidation state of lithium (`Li`) becomes +1 in `LiF`.
- The oxidation state of fluorine (`F`) in `LiF` is -1 as it gains an electron from lithium.
3. Identifying electron transfers:
- Lithium (`Li`) goes from an oxidation state of 0 to +1, indicating it loses one electron per lithium atom.
- Fluorine (`F`) goes from 0 to -1, indicating each fluorine atom gains one electron.
The substance that donates electrons (is oxidized) in this reaction is lithium (`Li`). Therefore, lithium (`Li`) is the reducing agent.
Correct answer:
B. Li
