Question

Which best describes the electron transfer that occurs during this oxidation-reduction reaction?

4Al+3O2→2Al2O3
4
Al
+
3
O
2
→
2
Al
2
O
3


(1 point)
Responses

Each oxygen atom gained four electrons, and each aluminum atom lost three electrons.
Each oxygen atom gained four electrons, and each aluminum atom lost three electrons.

Each oxygen atom gained three electrons, and each aluminum atom lost two electrons.
Each oxygen atom gained three electrons, and each aluminum atom lost two electrons.

Each oxygen atom gained three electrons, and each aluminum atom lost four electrons.
Each oxygen atom gained three electrons, and each aluminum atom lost four electrons.

Each oxygen atom gained two electrons, and each aluminum atom lost three electrons.

Answer 1

To analyze the oxidation-reduction reaction:

\[ 4Al + 3O_2 \rightarrow 2Al_2O_3 \]

we can determine the changes in oxidation states:

- Aluminum (Al) in its elemental form has an oxidation state of 0. In aluminum oxide (\(Al_2O_3\)), aluminum has an oxidation state of +3.
- Oxygen (O) in its elemental form (\(O_2\)) has an oxidation state of 0. In aluminum oxide, oxygen has an oxidation state of -2.

Thus, during the reaction:

- Each aluminum atom goes from 0 to +3, indicating that each aluminum atom **loses 3 electrons**.
- Each oxygen atom goes from 0 to -2, indicating that each oxygen atom **gains 2 electrons**.

Based on this information, the best description of the electron transfer in this reaction is:

**Each oxygen atom gained two electrons, and each aluminum atom lost three electrons.**

So the correct answer is:

**Each oxygen atom gained two electrons, and each aluminum atom lost three electrons.**