Which element or compound has lost electrons in this oxidation-reduction reaction?

[tex]\[
4 \mathrm{I} + 2 \mathrm{CoO} \rightarrow 2 \mathrm{Co} + 2 \mathrm{I}_2
\][/tex]

A. CoO
B. [tex]$\mathrm{I}_2$[/tex]
C. I
D. Co
E. O



Answer :

Sure, let's analyze the given oxidation-reduction reaction to determine which element has lost electrons.

Given reaction:
[tex]\[ \text{Co} + 2\text{Cr}^{3+} \rightarrow \text{Co}^{2+} + 2\text{Cr}^{2+} \][/tex]

1. Determine the oxidation states:
- For Cobalt (\text{Co}):
- In elemental form: [tex]\(\text{Co}\)[/tex] has an oxidation state of 0.
- In the product [tex]\(\text{Co}^{2+}\)[/tex]: Cobalt has an oxidation state of +2.

- For Chromium (\text{Cr}):
- Initially in [tex]\(\text{Cr}^{3+}\)[/tex]: Chromium has an oxidation state of +3.
- In the product [tex]\(\text{Cr}^{2+}\)[/tex]: Chromium has an oxidation state of +2.

2. Identify the changes in oxidation states:
- Cobalt (\text{Co}) changes from 0 to +2. This indicates that cobalt has lost electrons (oxidation).
- Chromium (\text{Cr}) changes from +3 to +2. This indicates that chromium has gained electrons (reduction).

3. Oxidation is the loss of electrons:
- Since Cobalt (\text{Co}) changes from 0 to +2, it has lost 2 electrons.

Therefore, the element that has lost electrons in this reaction is Cobalt (\text{Co}).

- [tex]\(\boxed{\text{D. Co}}\)[/tex]