Chapter 20.2, Problem 13SP

(a)

Interpretation Introduction

Interpretation: The oxidizing and reducing agents present in the reaction, $2{\text{H}}_2\left(g\right)+{\text{O}}_{\text{2}}\left(g\right)\to 2{\text{H}}_{\text{2}}\text{O}\left(l\right)$ is to be determined.

Concept Introduction: In a redox reaction, the reagent that helps in the oxidation and loss of electrons of chemical species is referred to as a reducing agent. And the reagent that helps in the reduction of chemical species and gain of electrons is referred to as an oxidizing agent. So, oxidizing agents accept electrons while reducing agents lose electrons.

Answer to Problem 13SP

The reducing agent is ${\text{H}}_2$ and the oxidizing agent is ${\text{O}}_{\text{2}}$ in the given reaction.

Explanation of Solution

In the reaction,

  $2{\text{H}}_2\left(g\right)+{\text{O}}_{\text{2}}\left(g\right)\to 2{\text{H}}_{\text{2}}\text{O}\left(l\right)$

The oxidation number of ${\text{H}}_2$ is increased to +2 from zero by losing electrons. While the oxidation number of ${\text{O}}_{\text{2}}$ is decreased to -2 from zero by gaining electrons. So the chemical species that is oxidized is ${\text{H}}_2$ and the reduced is ${\text{O}}_{\text{2}}$ .

Therefore the reducing agent is ${\text{H}}_2$ and the oxidizing agent is ${\text{O}}_{\text{2}}$ .

(b)

Interpretation Introduction

Interpretation: The oxidizing and reducing agents present in the reaction, $2{\text{KNO}}_3\left(s\right)\to {\text{ 2KNO}}_{\text{2}}\left(s\right){\text{+O}}_{\text{2}}\left(g\right)$ is to be determined.

Concept Introduction: In a redox reaction, the reagent that helps in the oxidation and loss of electrons of chemical species is referred to as a reducing agent. And the reagent that helps in the reduction of chemical species and gain of electrons is referred to as an oxidizing agent. So, oxidizing agents accept electrons while reducing agents lose electrons.

Answer to Problem 13SP

The reducing agent is oxygen and the oxidizing agent is nitrogen in the given reaction.

Explanation of Solution

In the reaction,

  $2{\text{KNO}}_3\left(s\right)\to {\text{ 2KNO}}_{\text{2}}\left(s\right){\text{+O}}_{\text{2}}\left(g\right)$

The oxidation number of nitrogen in the reactant is,

  $\begin{array}{c}x-6+1=0\\ x=5\end{array}$

The oxidation number of nitrogen in the product is,

  $\begin{array}{c}x-4+1=0\\ x=3\end{array}$

The oxidation number of nitrogen is decreased from +5 to +3. So, nitrogen is said to be reduced.

The oxidation number of oxygen, ${\text{O}}_{\text{2}}$ is increased from $-6$ to 0. Since elemental compounds have an oxidation number of zero. So, oxygen is said to be oxidized.

Therefore the reducing agent is oxygen and the oxidizing agent is nitrogen.