Let's analyze the electron-transfer reaction given:
[tex]\[ 3 \text{Cd}^{2+}(aq) + 2 \text{Al}(s) \rightarrow 3 \text{Cd}(s) + 2 \text{Al}^{3+}(aq) \][/tex]
First, let’s determine which species is oxidized and which species is reduced. We need to identify the changes in oxidation states for each species:
1. Cadmium (Cd):
- In the reactants: Cadmium starts as [tex]\(\text{Cd}^{2+}(aq)\)[/tex], which means cadmium has an oxidation state of +2.
- In the products: Cadmium ends as [tex]\(\text{Cd}(s)\)[/tex], which has an oxidation state of 0.
Cadmium goes from +2 to 0, which means it gains 2 electrons. The gain of electrons is called reduction. Therefore, Cadmium ([tex]\(\text{Cd}^{2+}\)[/tex]) is the species that is reduced.
2. Aluminum (Al):
- In the reactants: Aluminum starts as [tex]\(\text{Al}(s)\)[/tex], which means aluminum has an oxidation state of 0.
- In the products: Aluminum ends as [tex]\(\text{Al}^{3+}(aq)\)[/tex], which has an oxidation state of +3.
Aluminum goes from 0 to +3, which means it loses 3 electrons. The loss of electrons is called oxidation. Therefore, Aluminum ([tex]\(\text{Al}\)[/tex]) is the species that is oxidized.
Summarizing this:
- Species oxidized: [tex]\(\text{Al}\)[/tex]
- Species reduced: [tex]\(\text{Cd}^{2+}\)[/tex]
Next, we need to determine the direction of the electron transfer. As the reaction proceeds:
- Electrons are transferred from the species that is oxidized to the species that is reduced.
- This means electrons are transferred from Aluminum ([tex]\(\text{Al}\)[/tex]) to Cadmium ions ([tex]\(\text{Cd}^{2+}\)[/tex]).
Thus:
- Electrons are transferred from: [tex]\(\text{Al}\)[/tex]
- Electrons are transferred to: [tex]\(\text{Cd}^{2+}\)[/tex]
Final answers:
- Species oxidized: [tex]\(\text{Al}\)[/tex]
- Species reduced: [tex]\(\text{Cd}^{2+}\)[/tex]
- Electrons are transferred from [tex]\(\text{Al}\)[/tex] to [tex]\(\text{Cd}^{2+}\)[/tex]
