Chapter 19, Problem 76A

(a)

Interpretation Introduction

Interpretation: The reducing agent in given reaction needs to be identified.

  $4N{H}_3+5O_2\to 4NO+6H_{2}O$

Concept Introduction :

Reducing agent: Reducing agent is a species which reduces other and itself gets oxidized.

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

Answer to Problem 76A

  ${\text{NH}}_{\text{3}}$ is reducing agent.

Explanation of Solution

The given equation is:

  $4N{H}_3+5O_2\to 4NO+6H_{2}O$

Oxidation number of various atoms is:

  $4\stackrel{-3}{N}{H}_3+5\stackrel{0}{O_2}\to 4\stackrel{+2}{N}O+6H_2\stackrel{-2}{O}$

Oxidation number of oxygen decreases thus, it undergoes reduction and that of N increases thus, it undergoes oxidation and act as a reducing agent.

Thus, ${\text{NH}}_{\text{3}}$ is reducing agent.

(b)

Interpretation Introduction

Interpretation: The reducing agent in given reaction needs to be identified.

  $N{a}_{2}S{O}_4+4C\to N{a}_{2}S+4CO$

Concept Introduction :

Reducing agent: Reducing agent is a species which reduces other and itself gets oxidized.

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

Answer to Problem 76A

  $C$ is reducing agent.

Explanation of Solution

The given equation is:

  $N{a}_{2}S{O}_4+4C\to N{a}_{2}S+4CO$

Oxidation number of various atoms is:

  $N{a}_2\stackrel{+6}{S}{O}_4+4\stackrel{0}{C}\to N{a}_2\stackrel{-2}{S}+4\stackrel{+2}{C}O$

Oxidation number of carbon increases thus, it undergoes oxidation and act as a reducing agent.

(c)

Interpretation Introduction

Interpretation: The reducing agent in given reaction needs to be identified.

  $N{a}_{2}S{O}_4+4C\to N{a}_{2}S+4CO$

Concept Introduction :

Reducing agent: Reducing agent is a species which reduces other and itself gets oxidized.

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

Answer to Problem 76A

  $Ir$ is reducing agent.

Explanation of Solution

The given equation is:

  $4Ir{F}_5+Ir\to 5Ir{F}_4$

Oxidation number of various atoms is:

  $4\stackrel{+5}{Ir}{F}_5+\stackrel{0}{Ir}\to 5\stackrel{+4}{Ir}{F}_4$

Here, oxidation state of Ir increases from 0 to +4, it also decreases from +5 to +4 but that of F remains the same. Thus, Ir is getting oxidized and reduced as well.

From two species Ir is act as a reducing agent.