To determine how the oxidation state of lithium (Li) changes in the reaction:
[tex]\[ \text{Li (s)} + \text{NaOH (aq)} \rightarrow \text{LiOH (aq)} + \text{Na (s)} \][/tex]
let's analyze it step-by-step.
### Step 1: Identify the reactants and products
- Reactants: Li (s) and NaOH (aq)
- Products: LiOH (aq) and Na (s)
### Step 2: Determine the oxidation state of Li in the reactants and products
1. Oxidation State of Li in the reactants:
- Lithium in its elemental form (Li (s)) has an oxidation state of 0 because it is not combined with any other element.
2. Oxidation State of Li in the products:
- In lithium hydroxide (LiOH), lithium typically has an oxidation state of +1, as it combines with hydroxide (OH⁻) where hydroxide always has an oxidation state of -1.
### Step 3: Calculate the change in oxidation state
- Initial oxidation state of Li: 0
- Final oxidation state of Li: +1
The change in oxidation state is calculated as follows:
[tex]\[ \text{Change in oxidation state} = \text{Final oxidation state} - \text{Initial oxidation state} \][/tex]
[tex]\[ \text{Change in oxidation state} = +1 - 0 \][/tex]
[tex]\[ \text{Change in oxidation state} = +1 \][/tex]
### Step 4: Interpret the result
Given the change in oxidation state of +1, we conclude that lithium's oxidation state goes from 0 to +1.
### Answer:
D. It goes from 0 to +1.
