Chapter 19, Problem 42A

Interpretation Introduction

(a)

Interpretation:

The oxidizing agent and reducing agent in given reaction is to be determined.

Concept introduction:

Oxidizing agent: Oxidizing agent is a species which helps for oxidation.

Reducing agent: Reducing agent is a species which helps for reduction.

The substance which undergoes decrease in oxidation number is oxidizing agent and which undergoes increase in oxidation number is reducing agent.

Answer to Problem 42A

$N_2$ is oxidizing agent and $H_2$ is reducing agent.

Explanation of Solution

The given equation is:

$N_2+3H_2\to 2N{H}_3$

Oxidation number of various atoms is:

$\stackrel{0}{N_2}+3\stackrel{0}{H_2}\to 2\stackrel{-3}{N}{\stackrel{+1}{H}}_3$

Oxidation no. of nitrogen is decreasing hence it is oxidizing agent.

Oxidation no. of oxygen is increasing hence it is reducing agent.

Interpretation Introduction

(b)

Interpretation:

The oxidizing agent and reducing agent in given reaction is to be determined.

Concept introduction:

Oxidizing agent: Oxidizing agent is a species which helps for oxidation.

Reducing agent: Reducing agent is a species which helps for reduction.

The substance which undergoes decrease in oxidation number is oxidizing agent and which undergoes increase in oxidation number is reducing agent.

Answer to Problem 42A

$I_2$ is oxidizing agent and $Na$ is reducing agent.

Explanation of Solution

The given equation is:

$2Na+I_2\to 2NaI$

Oxidation number of various atoms is:

$2\stackrel{0}{Na}+{\stackrel{0}{I}}_2\to 2\stackrel{+1}{Na}\stackrel{-1}{I}$

Oxidation no. of iodine is decreasing hence it is oxidizing agent.

Oxidation no. of sodium is increasing hence it is reducing agent.