To determine how the oxidation state of lithium (Li) changes in the given reaction, we will follow these steps:
1. Identify the oxidation state of lithium in its elemental form (reactant side):
- In the elemental form, lithium (Li) exists as a solid metal. The oxidation state of any element in its pure, uncombined form is always 0.
Therefore, the initial oxidation state of lithium is 0.
2. Identify the oxidation state of lithium in the compound on the product side:
- In the product side of the reaction, lithium is part of the compound LiOH (lithium hydroxide). In compounds, the oxidation state of lithium is typically +1, as it tends to lose one electron to achieve a stable electronic configuration.
Hence, the final oxidation state of lithium in LiOH is +1.
3. Calculate the change in oxidation state:
- The change in oxidation state can be calculated by subtracting the initial oxidation state from the final oxidation state.
[tex]\[
\text{Change in oxidation state} = \text{Final oxidation state} - \text{Initial oxidation state}
\][/tex]
Substituting the known values:
[tex]\[
\text{Change in oxidation state} = +1 - 0 = +1
\][/tex]
4. Summarize the result:
- Lithium's oxidation state changes from 0 (in its elemental form) to +1 (when it becomes part of LiOH).
Based on this detailed analysis, the correct answer is:
A. It goes from 0 to +1.
