Chapter 5, Problem 1PE

When sodium reacts with molecular oxygen, ${\text{O}}_{\text{2}}$, the product is sodium peroxide, ${\text{Na}}_{\text{2}}\text{O2}$, which contains the peroxide ion, ${\text{O}}_{\text{2}}{}^{\text{2-}}$. In this reaction, is ${\text{O}}_{\text{2}}$ oxidized or reduced?(Hint: Which reactant gains electrons, and which loses electrons?)

Interpretation Introduction

Interpretation:

Whether ${\text{O}}_{\text{2}}$ is oxidized or reduced in the given reaction, is to be determined.

Concept Introduction:

Oxidation is termed as a chemical reaction during the course of which a number of electrons are lost.

Reduction is termed as a chemical reaction during the course of which a number of electrons are gained.

Answer to Problem 1PE

Solution: ${\text{O}}_{\text{2}}$ is reduced.

Explanation of Solution

Sodium reacts with molecular oxygen $\left({\text{O}}_{\text{2}}\right)$ and forms the product, sodium peroxide $\left({\text{Na}}_{\text{2}}{\text{O}}_{\text{2}}\right)$ which contains the peroxide ion $\left({\text{O}}_{\text{2}}{}^{2-}\right)$.

The reaction of sodium is as follows:

$\text{2Na}(s){\text{ + O}}_2(g)\to {\text{Na}}_2{\text{O}}_2(s)$

Oxygen is reduced in this reaction because it experiences gain of electrons and Sodium is oxidized because it experiences loss of electrons.

Conclusion

In this reaction, ${\text{O}}_{\text{2}}$ is reduced.