Chapter 20, Problem 38A

(a)

Interpretation Introduction

Interpretation: The oxidizing and reducing agents in the reaction, $H_2\left(g\right)+S\left(s\right)\to H_{2}S\left(g\right)$ are to be determined.

Concept Introduction: The chemical species that undergo oxidation by the removal of electrons is the reducing agent while the one that undergoes reduction by the gain of electrons is the oxidizing agent.

Answer to Problem 38A

Oxidizing agent is S and the reducing agent is $H_2$ .

Explanation of Solution

In the reaction,

  $H_2\left(g\right)+S\left(s\right)\to H_{2}S\left(g\right)$

The chemical species that undergo oxidation is $H_2$ while that undergoes reduction is S. So the oxidizing agent is S and the reducing agent is $H_2$ .

(b)

Interpretation Introduction

Interpretation: The oxidizing and reducing agents in the reaction, $N_2\left(g\right)+3H_2\left(g\right)\to 2N{H}_3\left(g\right)$ is to be determined.

Concept Introduction: The chemical species that undergo oxidation by the removal of electrons is the reducing agent while the one that undergoes reduction by the gain of electrons is the oxidizing agent.

Answer to Problem 38A

Oxidizing agent is $N_2$ and the reducing agent is $H_2$ .

Explanation of Solution

In the reaction,

  $N_2\left(g\right)+3H_2\left(g\right)\to 2N{H}_3\left(g\right)$

The chemical species that undergo oxidation is $H_2$ while that undergoes reduction is $N_2$ . So, the oxidizing agent is $N_2$ and the reducing agent is $H_2$ .

(c)

Interpretation Introduction

Interpretation: The oxidizing and reducing agents in the reaction, $O_2\left(g\right)+S\left(s\right)\to S{O}_2\left(g\right)$ are to be determined.

Concept Introduction: The chemical species that undergo oxidation by the removal of electrons is the reducing agent while the one that undergoes reduction by the gain of electrons is the oxidizing agent.

Answer to Problem 38A

Oxidizing agent is $O_2$ and the reducing agent is $S$ .

Explanation of Solution

In the reaction,

  $O_2\left(g\right)+S\left(s\right)\to S{O}_2\left(g\right)$

The chemical reaction that involves oxidation is S and that involves reduction is $O_2$ . So the oxidizing agent is $O_2$ and the reducing agent is S.

(d)

Interpretation Introduction

Interpretation: The oxidizing and reducing agents in the reaction, $H_2\left(g\right)+O_2\left(g\right)\to H_{2}O\left(l\right)$ are to be determined.

Concept Introduction: The chemical species that undergo oxidation by the removal of electrons is the reducing agent while the one that undergoes reduction by the gain of electrons is the oxidizing agent.

Answer to Problem 38A

Oxidizing agent is $O_2$ and the reducing agent is $H_2$ .

Explanation of Solution

In the reaction,

  $H_2\left(g\right)+O_2\left(g\right)\to H_{2}O\left(l\right)$

The chemical species that undergo oxidation is $H_2$ while that undergoes reduction is $O_2$ . So, the oxidizing agent is $O_2$ and the reducing agent is $H_2$ .