How does the oxidation state of Li change in the following reaction?

[tex]\[ Li (s) + NaOH (aq) \rightarrow LiOH (aq) + Na (s) \][/tex]

A. It does not change.

B. It goes from +1 to +2.

C. It goes from 0 to -1.

D. It goes from 0 to +1.

Question

22-07-2024
Chemistry

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How does the oxidation state of Li change in the following reaction?

[tex]\[ Li (s) + NaOH (aq) \rightarrow LiOH (aq) + Na (s) \][/tex]

A. It does not change.

B. It goes from +1 to +2.

C. It goes from 0 to -1.

D. It goes from 0 to +1.

Sagot :

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GinnyAnsweridn GinnyAnsweridn · Answer 1 · 2024-07-22T22:02:46-04:00

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.

Sagot :

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