Question

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

4Al+3O2→2Al2O3

(1 point)
Responses

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 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 two electrons, and each aluminum atom lost three electrons.

Answer 1

To determine the correct description of the electron transfer during the given oxidation-reduction reaction \(4Al + 3O_2 \rightarrow 2Al_2O_3\), we first need to examine the oxidation states of the elements involved before and after the reaction.

1. **Oxidation States:**
- In elemental aluminum (Al), the oxidation state is 0.
- In \(Al_2O_3\), aluminum has an oxidation state of +3.
- In molecular oxygen (\(O_2\)), the oxidation state is 0.
- In \(Al_2O_3\), oxygen has an oxidation state of -2.

2. **Changes in Oxidation States:**
- Aluminum (Al) goes from 0 to +3. Each aluminum atom loses 3 electrons.
- Oxygen (O) goes from 0 to -2. Each oxygen atom gains 2 electrons.

3. **Counting Electrons:**
- Since there are 4 aluminum atoms (each losing 3 electrons), a total of \(4 \times 3 = 12\) electrons are lost by aluminum.
- There are 3 oxygen molecules (each gaining 2 electrons), giving a total of \(3 \times 2 = 6\) electrons gained by oxygen.

Thus, each aluminum atom loses 3 electrons and each oxygen atom gains 2 electrons.

Based on this analysis, the correct answer is:

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