Answer :
To determine which element or compound has lost electrons in the given oxidation-reduction reaction, we need to analyze the oxidation states and the transfer of electrons:
The given oxidation-reduction reaction is:
[tex]\[ 4 \text{Li} + 2 \text{CoO} \rightarrow 2 \text{Co}^{2+} + 2 \text{Li}_2\text{O} \][/tex]
1. Identify the reactants and products:
- Reactants: Lithium (Li) and Cobalt(II) oxide (CoO)
- Products: Cobalt ions ([tex]\(\text{Co}^{2+}\)[/tex]) and Lithium oxide ([tex]\(\text{Li}_2\text{O}\)[/tex])
2. Determine the oxidation states of each element in the reactants and products:
- Lithium (Li) in its elemental form (reactants): Oxidation state = 0
- Cobalt (Co) in CoO: Oxidation state = +2 (since oxygen has a -2 oxidation state in CoO)
- Cobalt (Co) in [tex]\(\text{Co}^{2+}\)[/tex] (products): Oxidation state = +2
- Lithium (Li) in [tex]\(\text{Li}_2\text{O}\)[/tex]: Each Li has oxidation state = +1 (since oxygen has a -2 oxidation state and [tex]\(\text{Li}_2\text{O}\)[/tex] is a neutral compound)
3. Look at the changes in oxidation states to identify oxidation and reduction:
- Lithium (Li) goes from 0 (in reactants) to +1 (in products), indicating a loss of electrons (oxidation).
- Cobalt (Co) maintains the oxidation state of +2, indicating no change.
4. Oxidation involves the loss of electrons. Therefore, the element that has lost electrons in this reaction is Lithium (Li).
5. Verify the correct option by matching it to the given answer choices:
- A. [tex]\(\text{CoO}\)[/tex]
- B. [tex]\(\text{Li}_2\text{O}\)[/tex]
- C. [tex]\(\text{Li}\)[/tex]
- D. Co
- E. 0
Lithium (Li) corresponds to option C.
Therefore, the element that has lost electrons in this oxidation-reduction reaction is:
[tex]\[ \boxed{C. \text{Li}} \][/tex]
The given oxidation-reduction reaction is:
[tex]\[ 4 \text{Li} + 2 \text{CoO} \rightarrow 2 \text{Co}^{2+} + 2 \text{Li}_2\text{O} \][/tex]
1. Identify the reactants and products:
- Reactants: Lithium (Li) and Cobalt(II) oxide (CoO)
- Products: Cobalt ions ([tex]\(\text{Co}^{2+}\)[/tex]) and Lithium oxide ([tex]\(\text{Li}_2\text{O}\)[/tex])
2. Determine the oxidation states of each element in the reactants and products:
- Lithium (Li) in its elemental form (reactants): Oxidation state = 0
- Cobalt (Co) in CoO: Oxidation state = +2 (since oxygen has a -2 oxidation state in CoO)
- Cobalt (Co) in [tex]\(\text{Co}^{2+}\)[/tex] (products): Oxidation state = +2
- Lithium (Li) in [tex]\(\text{Li}_2\text{O}\)[/tex]: Each Li has oxidation state = +1 (since oxygen has a -2 oxidation state and [tex]\(\text{Li}_2\text{O}\)[/tex] is a neutral compound)
3. Look at the changes in oxidation states to identify oxidation and reduction:
- Lithium (Li) goes from 0 (in reactants) to +1 (in products), indicating a loss of electrons (oxidation).
- Cobalt (Co) maintains the oxidation state of +2, indicating no change.
4. Oxidation involves the loss of electrons. Therefore, the element that has lost electrons in this reaction is Lithium (Li).
5. Verify the correct option by matching it to the given answer choices:
- A. [tex]\(\text{CoO}\)[/tex]
- B. [tex]\(\text{Li}_2\text{O}\)[/tex]
- C. [tex]\(\text{Li}\)[/tex]
- D. Co
- E. 0
Lithium (Li) corresponds to option C.
Therefore, the element that has lost electrons in this oxidation-reduction reaction is:
[tex]\[ \boxed{C. \text{Li}} \][/tex]
