To determine how the oxidation state of lithium (Li) changes in the given reaction, we can follow these steps:
[tex]\[
Li (s) + NaOH (aq) \rightarrow LiOH (aq) + Na (s)
\][/tex]
1. Identify the reactants and products involving lithium:
- Reactants: \( Li (s) \)
- Products: \( LiOH (aq) \)
2. Determine the oxidation state of lithium in the reactants:
- In \( Li (s) \), lithium is in its elemental form. The oxidation state of any element in its elemental form is always 0.
3. Determine the oxidation state of lithium in the products:
- In \( LiOH \), we need to consider the oxidation states of the other atoms to determine that of lithium.
- Oxygen typically has an oxidation state of -2.
- Hydrogen typically has an oxidation state of +1.
- The compound \( LiOH \) is neutral, so the sum of the oxidation states of all atoms in the molecule must equal 0.
- Let the oxidation state of lithium in \( LiOH \) be \( x \).
Hence, the equation can be written as:
[tex]\[
x + (-2) + (+1) = 0
\][/tex]
Simplifying this,
[tex]\[
x - 2 + 1 = 0 \implies x - 1 = 0 \implies x = +1
\][/tex]
4. Compare the oxidation states:
- Initially, in the reactant, the oxidation state of lithium is 0.
- In the product \( LiOH \), the oxidation state of lithium is +1.
5. Conclusion:
- The oxidation state of lithium changes from 0 in \( Li (s) \) to +1 in \( LiOH (aq) \).
Therefore, the correct answer is:
[tex]\[
\text{B. It goes from 0 to +1.}
\][/tex]
