Chapter 17, Problem 48AE

(a)

Interpretation Introduction

Interpretation:

Atom or ion oxidized in reactions below has to be determined.

  $\begin{array}{l}1.{\text{ C}}_{\text{3}}{\text{H}}_{\text{8}}+{\text{O}}_2\to {\text{CO}}_{\text{2}}+{\text{H}}_{\text{2}}\text{O}\\ 2.{\text{ HNO}}_3+{\text{H}}_{\text{2}}\text{S}\to \text{NO}+\text{S}+{\text{H}}_{\text{2}}\text{O}\\ 3.\text{ CuO}+{\text{NH}}_{\text{3}}\to {\text{N}}_2+{\text{H}}_{\text{2}}\text{O}+\text{Cu}\\ 4.{\text{ H}}_{\text{2}}{\text{O}}_2+{\text{Na}}_{\text{2}}{\text{SO}}_3\to {\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}+{\text{H}}_{\text{2}}\text{O}\\ 5.{\text{ H}}_{\text{2}}{\text{O}}_2\to {\text{H}}_{\text{2}}\text{O}+{\text{O}}_{\text{2}}\end{array}$

Concept Introduction:

Oxidation number is integer value allotted to every element. It is formal charge occupied by atom if all of its bonds are dissociated heterolytically. Below mentioned are rules to assign oxidation numbers to various elements.

1. Elements present in their free state have zero oxidation number.

2. Oxidation number of hydrogen is generally $+1$, except for metal hydrides.

3. Oxidation number of oxygen is $-2$, except for peroxides.

4. Metals have positive oxidation numbers.

5. Negative oxidation numbers are assigned to most electronegative element in covalent compounds.

6. Sum of oxidation numbers of different elements in neutral atom is zero.

7. Sum of oxidation numbers of various elements in polyatomic ion is equal to charge present on ion.

Explanation of Solution

Given reaction (1) is as follows:

  ${\text{C}}_{\text{3}}{\text{H}}_{\text{8}}+{\text{O}}_2\to {\text{CO}}_{\text{2}}+{\text{H}}_{\text{2}}\text{O}$

Oxidation number of each atom in reaction (1) is as follows:

  $\stackrel{-\text{8/3}}{{\text{C}}_{\text{3}}}\stackrel{+1}{{\text{H}}_{\text{8}}}+\stackrel{0}{{\text{O}}_2}\to \stackrel{+4}{\text{C}}\stackrel{-2}{{\text{O}}_{\text{2}}}+\stackrel{+1}{{\text{H}}_{\text{2}}}\stackrel{-2}{\text{O}}$

Since oxidation state of carbon changes from $-\text{8/3}$ to $+4$ thus carbon is oxidized in this reaction.

Given reaction (2) is as follows:

  ${\text{HNO}}_3+{\text{H}}_{\text{2}}\text{S}\to \text{NO}+\text{S}+{\text{H}}_{\text{2}}\text{O}$

Oxidation number of each atom in reaction (2) is as follows:

  $\stackrel{+1}{\text{H}}\stackrel{+5}{\text{N}}\stackrel{-2}{{\text{O}}_3}+\stackrel{+1}{{\text{H}}_{\text{2}}}\stackrel{-2}{\text{S}}\to \stackrel{+2}{\text{N}}\stackrel{-2}{\text{O}}+\stackrel{0}{\text{S}}+\stackrel{+1}{{\text{H}}_{\text{2}}}\stackrel{-2}{\text{O}}$

Since oxidation state of sulfur changes from $-\text{2}$ to 0 thus sulfur is oxidized in this reaction.

Given reaction (3) is as follows:

  $\text{CuO}+{\text{NH}}_{\text{3}}\to {\text{N}}_2+{\text{H}}_{\text{2}}\text{O}+\text{Cu}$

Oxidation number of each atom in reaction (3) is as follows:

  $\stackrel{+2}{\text{Cu}}\stackrel{-2}{\text{O}}+\stackrel{-3}{\text{N}}\stackrel{+1}{{\text{H}}_{\text{3}}}\to \stackrel{0}{{\text{N}}_2}+\stackrel{+1}{{\text{H}}_{\text{2}}}\stackrel{-2}{\text{O}}+\stackrel{0}{\text{Cu}}$

Since oxidation state of nitrogen changes from $-\text{3}$ to 0 thus nitrogen oxidized in this reaction.

Given reaction (4) is as follows:

  ${\text{H}}_{\text{2}}{\text{O}}_2+{\text{Na}}_{\text{2}}{\text{SO}}_3\to {\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}+{\text{H}}_{\text{2}}\text{O}$

Oxidation number of each atom in reaction (4) is as follows:

  $\stackrel{+1}{{\text{H}}_{\text{2}}}\stackrel{-1}{{\text{O}}_2}+\stackrel{+1}{{\text{Na}}_{\text{2}}}\stackrel{+4}{\text{S}}\stackrel{-2}{{\text{O}}_3}\to \stackrel{+1}{{\text{Na}}_{\text{2}}}\stackrel{+6}{\text{S}}\stackrel{-2}{{\text{O}}_{\text{4}}}+\stackrel{+1}{{\text{H}}_{\text{2}}}\stackrel{-2}{\text{O}}$

Since oxidation state of sulfur changes from $+4$ to $+6$ thus sulfur oxidized in this reaction.

Given reaction (5) is as follows:

  ${\text{H}}_{\text{2}}{\text{O}}_2\to {\text{H}}_{\text{2}}\text{O}+{\text{O}}_{\text{2}}$

Oxidation number of each atom in reaction (5) is as follows:

  $\stackrel{+1}{{\text{H}}_{\text{2}}}\stackrel{-1}{{\text{O}}_2}\to \stackrel{+1}{{\text{H}}_{\text{2}}}\stackrel{-2}{\text{O}}+\stackrel{0}{{\text{O}}_{\text{2}}}$

Since oxidation state of oxygen changes from $-1$ to 0 thus oxygen oxidized in this reaction.

(b)

Interpretation Introduction

Interpretation:

Atom or ion reduced in reactions below has to be determined.

  $\begin{array}{l}1.{\text{ C}}_{\text{3}}{\text{H}}_{\text{8}}+{\text{O}}_2\to {\text{CO}}_{\text{2}}+{\text{H}}_{\text{2}}\text{O}\\ 2.{\text{ HNO}}_3+{\text{H}}_{\text{2}}\text{S}\to \text{NO}+\text{S}+{\text{H}}_{\text{2}}\text{O}\\ 3.\text{ CuO}+{\text{NH}}_{\text{3}}\to {\text{N}}_2+{\text{H}}_{\text{2}}\text{O}+\text{Cu}\\ 4.{\text{ H}}_{\text{2}}{\text{O}}_2+{\text{Na}}_{\text{2}}{\text{SO}}_3\to {\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}+{\text{H}}_{\text{2}}\text{O}\\ 5.{\text{ H}}_{\text{2}}{\text{O}}_2\to {\text{H}}_{\text{2}}\text{O}+{\text{O}}_{\text{2}}\end{array}$

Concept Introduction:

Refer to part (a).

Explanation of Solution

Given reaction (1) is as follows:

  ${\text{C}}_{\text{3}}{\text{H}}_{\text{8}}+{\text{O}}_2\to {\text{CO}}_{\text{2}}+{\text{H}}_{\text{2}}\text{O}$

Oxidation number of each atom in reaction (1) is as follows:

  $\stackrel{-\text{8/3}}{{\text{C}}_{\text{3}}}\stackrel{+1}{{\text{H}}_{\text{8}}}+\stackrel{0}{{\text{O}}_2}\to \stackrel{+4}{\text{C}}\stackrel{-2}{{\text{O}}_{\text{2}}}+\stackrel{+1}{{\text{H}}_{\text{2}}}\stackrel{-2}{\text{O}}$

Since oxidation state of oxygen changes from 0 to $-2$ thus oxygen reduced in this reaction.

Given reaction (2) is as follows:

  ${\text{HNO}}_3+{\text{H}}_{\text{2}}\text{S}\to \text{NO}+\text{S}+{\text{H}}_{\text{2}}\text{O}$

Oxidation number of each atom in reaction (2) is as follows:

  $\stackrel{+1}{\text{H}}\stackrel{+5}{\text{N}}\stackrel{-2}{{\text{O}}_3}+\stackrel{+1}{{\text{H}}_{\text{2}}}\stackrel{-2}{\text{S}}\to \stackrel{+2}{\text{N}}\stackrel{-2}{\text{O}}+\stackrel{0}{\text{S}}+\stackrel{+1}{{\text{H}}_{\text{2}}}\stackrel{-2}{\text{O}}$

Since oxidation state of nitrogen changes from $+5$ to $+2$ thus nitrogen reduced in this reaction.

Given reaction (3) is as follows:

  $\text{CuO}+{\text{NH}}_{\text{3}}\to {\text{N}}_2+{\text{H}}_{\text{2}}\text{O}+\text{Cu}$

Oxidation number of each atom in reaction (3) is as follows:

  $\stackrel{+2}{\text{Cu}}\stackrel{-2}{\text{O}}+\stackrel{-3}{\text{N}}\stackrel{+1}{{\text{H}}_{\text{3}}}\to \stackrel{0}{{\text{N}}_2}+\stackrel{+1}{{\text{H}}_{\text{2}}}\stackrel{-2}{\text{O}}+\stackrel{0}{\text{Cu}}$

Since oxidation state of copper changes from $+2$ to 0 thus copper reduced in this reaction.

Given reaction (4) is as follows:

  ${\text{H}}_{\text{2}}{\text{O}}_2+{\text{Na}}_{\text{2}}{\text{SO}}_3\to {\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}+{\text{H}}_{\text{2}}\text{O}$

Oxidation number of each atom in reaction (4) is as follows:

  $\stackrel{+1}{{\text{H}}_{\text{2}}}\stackrel{-1}{{\text{O}}_2}+\stackrel{+1}{{\text{Na}}_{\text{2}}}\stackrel{+4}{\text{S}}\stackrel{-2}{{\text{O}}_3}\to \stackrel{+1}{{\text{Na}}_{\text{2}}}\stackrel{+6}{\text{S}}\stackrel{-2}{{\text{O}}_{\text{4}}}+\stackrel{+1}{{\text{H}}_{\text{2}}}\stackrel{-2}{\text{O}}$

Since oxidation state of oxygen changes from $-1$ to $-2$ thus oxygen reduced in this reaction.

Given reaction (5) is as follows:

  ${\text{H}}_{\text{2}}{\text{O}}_2\to {\text{H}}_{\text{2}}\text{O}+{\text{O}}_{\text{2}}$

Oxidation number of each atom in reaction (5) is as follows:

  $\stackrel{+1}{{\text{H}}_{\text{2}}}\stackrel{-1}{{\text{O}}_2}\to \stackrel{+1}{{\text{H}}_{\text{2}}}\stackrel{-2}{\text{O}}+\stackrel{0}{{\text{O}}_{\text{2}}}$

Since oxidation state of oxygen changes from $-1$ to $-2$ thus oxygen reduced in this reaction.

(c)

Interpretation Introduction

Interpretation:

Oxidizing agent in reactions below has to be determined.

  $\begin{array}{l}1.{\text{ C}}_{\text{3}}{\text{H}}_{\text{8}}+{\text{O}}_2\to {\text{CO}}_{\text{2}}+{\text{H}}_{\text{2}}\text{O}\\ 2.{\text{ HNO}}_3+{\text{H}}_{\text{2}}\text{S}\to \text{NO}+\text{S}+{\text{H}}_{\text{2}}\text{O}\\ 3.\text{ CuO}+{\text{NH}}_{\text{3}}\to {\text{N}}_2+{\text{H}}_{\text{2}}\text{O}+\text{Cu}\\ 4.{\text{ H}}_{\text{2}}{\text{O}}_2+{\text{Na}}_{\text{2}}{\text{SO}}_3\to {\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}+{\text{H}}_{\text{2}}\text{O}\\ 5.{\text{ H}}_{\text{2}}{\text{O}}_2\to {\text{H}}_{\text{2}}\text{O}+{\text{O}}_{\text{2}}\end{array}$

Concept Introduction:

Refer to part (a).

Explanation of Solution

Given reaction (1) is as follows:

  ${\text{C}}_{\text{3}}{\text{H}}_{\text{8}}+{\text{O}}_2\to {\text{CO}}_{\text{2}}+{\text{H}}_{\text{2}}\text{O}$

Oxidation number of each atom in reaction (1) is as follows:

  $\stackrel{-\text{8/3}}{{\text{C}}_{\text{3}}}\stackrel{+1}{{\text{H}}_{\text{8}}}+\stackrel{0}{{\text{O}}_2}\to \stackrel{+4}{\text{C}}\stackrel{-2}{{\text{O}}_{\text{2}}}+\stackrel{+1}{{\text{H}}_{\text{2}}}\stackrel{-2}{\text{O}}$

Since oxidation state of oxygen changes from 0 to $-2$ thus oxygen reduced in this reaction. Hence oxygen gas is oxidizing agent in this reaction.

Given reaction (2) is as follows:

  ${\text{HNO}}_3+{\text{H}}_{\text{2}}\text{S}\to \text{NO}+\text{S}+{\text{H}}_{\text{2}}\text{O}$

Oxidation number of each atom in reaction (2) is as follows:

  $\stackrel{+1}{\text{H}}\stackrel{+5}{\text{N}}\stackrel{-2}{{\text{O}}_3}+\stackrel{+1}{{\text{H}}_{\text{2}}}\stackrel{-2}{\text{S}}\to \stackrel{+2}{\text{N}}\stackrel{-2}{\text{O}}+\stackrel{0}{\text{S}}+\stackrel{+1}{{\text{H}}_{\text{2}}}\stackrel{-2}{\text{O}}$

Since oxidation state of nitrogen changes from $+5$ to $+2$ thus nitrogen reduced in this reaction. Hence ${\text{HNO}}_3$ is oxidizing agent in this reaction.

Given reaction (3) is as follows:

  $\text{CuO}+{\text{NH}}_{\text{3}}\to {\text{N}}_2+{\text{H}}_{\text{2}}\text{O}+\text{Cu}$

Oxidation number of each atom in reaction (3) is as follows:

  $\stackrel{+2}{\text{Cu}}\stackrel{-2}{\text{O}}+\stackrel{-3}{\text{N}}\stackrel{+1}{{\text{H}}_{\text{3}}}\to \stackrel{0}{{\text{N}}_2}+\stackrel{+1}{{\text{H}}_{\text{2}}}\stackrel{-2}{\text{O}}+\stackrel{0}{\text{Cu}}$

Since oxidation state of copper changes from $+2$ to 0 thus copper reduced in this reaction. Hence $\text{CuO}$ is oxidizing agent in this reaction.

Given reaction (4) is as follows:

  ${\text{H}}_{\text{2}}{\text{O}}_2+{\text{Na}}_{\text{2}}{\text{SO}}_3\to {\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}+{\text{H}}_{\text{2}}\text{O}$

Oxidation number of each atom in reaction (4) is as follows:

  $\stackrel{+1}{{\text{H}}_{\text{2}}}\stackrel{-1}{{\text{O}}_2}+\stackrel{+1}{{\text{Na}}_{\text{2}}}\stackrel{+4}{\text{S}}\stackrel{-2}{{\text{O}}_3}\to \stackrel{+1}{{\text{Na}}_{\text{2}}}\stackrel{+6}{\text{S}}\stackrel{-2}{{\text{O}}_{\text{4}}}+\stackrel{+1}{{\text{H}}_{\text{2}}}\stackrel{-2}{\text{O}}$

Since oxidation state of oxygen changes from $-1$ to $-2$ thus oxygen reduced in this reaction. Hence ${\text{H}}_{\text{2}}{\text{O}}_2$ is oxidizing agent in this reaction.

Given reaction (5) is as follows:

  ${\text{H}}_{\text{2}}{\text{O}}_2\to {\text{H}}_{\text{2}}\text{O}+{\text{O}}_{\text{2}}$

Oxidation number of each atom in reaction (5) is as follows:

  $\stackrel{+1}{{\text{H}}_{\text{2}}}\stackrel{-1}{{\text{O}}_2}\to \stackrel{+1}{{\text{H}}_{\text{2}}}\stackrel{-2}{\text{O}}+\stackrel{0}{{\text{O}}_{\text{2}}}$

Since oxidation state of oxygen changes from $-1$ to $-2$ thus oxygen reduced in this reaction. Hence oxygen in ${\text{H}}_{\text{2}}{\text{O}}_2$ is oxidizing agent in this reaction.

(d)

Interpretation Introduction

Interpretation:

Reducing agent in reactions below has to be determined.

  $\begin{array}{l}1.{\text{ C}}_{\text{3}}{\text{H}}_{\text{8}}+{\text{O}}_2\to {\text{CO}}_{\text{2}}+{\text{H}}_{\text{2}}\text{O}\\ 2.{\text{ HNO}}_3+{\text{H}}_{\text{2}}\text{S}\to \text{NO}+\text{S}+{\text{H}}_{\text{2}}\text{O}\\ 3.\text{ CuO}+{\text{NH}}_{\text{3}}\to {\text{N}}_2+{\text{H}}_{\text{2}}\text{O}+\text{Cu}\\ 4.{\text{ H}}_{\text{2}}{\text{O}}_2+{\text{Na}}_{\text{2}}{\text{SO}}_3\to {\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}+{\text{H}}_{\text{2}}\text{O}\\ 5.{\text{ H}}_{\text{2}}{\text{O}}_2\to {\text{H}}_{\text{2}}\text{O}+{\text{O}}_{\text{2}}\end{array}$

Concept Introduction:

Refer to part (a).

Explanation of Solution

Given reaction (1) is as follows:

  ${\text{C}}_{\text{3}}{\text{H}}_{\text{8}}+{\text{O}}_2\to {\text{CO}}_{\text{2}}+{\text{H}}_{\text{2}}\text{O}$

Oxidation number of each atom in reaction (1) is as follows:

  $\stackrel{-\text{8/3}}{{\text{C}}_{\text{3}}}\stackrel{+1}{{\text{H}}_{\text{8}}}+\stackrel{0}{{\text{O}}_2}\to \stackrel{+4}{\text{C}}\stackrel{-2}{{\text{O}}_{\text{2}}}+\stackrel{+1}{{\text{H}}_{\text{2}}}\stackrel{-2}{\text{O}}$

Since oxidation state of carbon changes from $-\text{8/3}$ to $+4$ thus carbon oxidized in this reaction. Hence ${\text{C}}_{\text{3}}{\text{H}}_{\text{8}}$ is reducing agent in this reaction.

Given reaction (2) is as follows:

  ${\text{HNO}}_3+{\text{H}}_{\text{2}}\text{S}\to \text{NO}+\text{S}+{\text{H}}_{\text{2}}\text{O}$

Oxidation number of each atom in reaction (2) is as follows:

  $\stackrel{+1}{\text{H}}\stackrel{+5}{\text{N}}\stackrel{-2}{{\text{O}}_3}+\stackrel{+1}{{\text{H}}_{\text{2}}}\stackrel{-2}{\text{S}}\to \stackrel{+2}{\text{N}}\stackrel{-2}{\text{O}}+\stackrel{0}{\text{S}}+\stackrel{+1}{{\text{H}}_{\text{2}}}\stackrel{-2}{\text{O}}$

Since oxidation state of sulfur changes from $-\text{2}$ to 0 thus sulfur oxidized in this reaction. Hence ${\text{H}}_{\text{2}}\text{S}$ is reducing agent in this reaction.

Given reaction (3) is as follows:

  $\text{CuO}+{\text{NH}}_{\text{3}}\to {\text{N}}_2+{\text{H}}_{\text{2}}\text{O}+\text{Cu}$

Oxidation number of each atom in reaction (3) is as follows:

  $\stackrel{+2}{\text{Cu}}\stackrel{-2}{\text{O}}+\stackrel{-3}{\text{N}}\stackrel{+1}{{\text{H}}_{\text{3}}}\to \stackrel{0}{{\text{N}}_2}+\stackrel{+1}{{\text{H}}_{\text{2}}}\stackrel{-2}{\text{O}}+\stackrel{0}{\text{Cu}}$

Since oxidation state of nitrogen changes from $-\text{3}$ to 0 thus nitrogen oxidized in this reaction. Hence ${\text{NH}}_{\text{3}}$ is reducing agent in this reaction.

Given reaction (4) is as follows:

  ${\text{H}}_{\text{2}}{\text{O}}_2+{\text{Na}}_{\text{2}}{\text{SO}}_3\to {\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}+{\text{H}}_{\text{2}}\text{O}$

Oxidation number of each atom in reaction (4) is as follows:

  $\stackrel{+1}{{\text{H}}_{\text{2}}}\stackrel{-1}{{\text{O}}_2}+\stackrel{+1}{{\text{Na}}_{\text{2}}}\stackrel{+4}{\text{S}}\stackrel{-2}{{\text{O}}_3}\to \stackrel{+1}{{\text{Na}}_{\text{2}}}\stackrel{+6}{\text{S}}\stackrel{-2}{{\text{O}}_{\text{4}}}+\stackrel{+1}{{\text{H}}_{\text{2}}}\stackrel{-2}{\text{O}}$

Since oxidation state of sulfur changes from $+4$ to $+6$ thus sulfur oxidized in this reaction. Hence ${\text{Na}}_{\text{2}}{\text{SO}}_3$ is reducing agent in this reaction.

Given reaction (5) is as follows:

  ${\text{H}}_{\text{2}}{\text{O}}_2\to {\text{H}}_{\text{2}}\text{O}+{\text{O}}_{\text{2}}$

Oxidation number of each atom in reaction (5) is as follows:

  $\stackrel{+1}{{\text{H}}_{\text{2}}}\stackrel{-1}{{\text{O}}_2}\to \stackrel{+1}{{\text{H}}_{\text{2}}}\stackrel{-2}{\text{O}}+\stackrel{0}{{\text{O}}_{\text{2}}}$

Since oxidation state of oxygen changes from $-1$ to 0 thus oxygen oxidized in this reaction. Hence oxygen in ${\text{H}}_{\text{2}}{\text{O}}_2$ is reducing agent in this reaction.

(e)

Interpretation Introduction

Interpretation:

Oxidation state change in oxidizing agent in reactions below has to be determined.

  $\begin{array}{l}1.{\text{ C}}_{\text{3}}{\text{H}}_{\text{8}}+{\text{O}}_2\to {\text{CO}}_{\text{2}}+{\text{H}}_{\text{2}}\text{O}\\ 2.{\text{ HNO}}_3+{\text{H}}_{\text{2}}\text{S}\to \text{NO}+\text{S}+{\text{H}}_{\text{2}}\text{O}\\ 3.\text{ CuO}+{\text{NH}}_{\text{3}}\to {\text{N}}_2+{\text{H}}_{\text{2}}\text{O}+\text{Cu}\\ 4.{\text{ H}}_{\text{2}}{\text{O}}_2+{\text{Na}}_{\text{2}}{\text{SO}}_3\to {\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}+{\text{H}}_{\text{2}}\text{O}\\ 5.{\text{ H}}_{\text{2}}{\text{O}}_2\to {\text{H}}_{\text{2}}\text{O}+{\text{O}}_{\text{2}}\end{array}$

Concept Introduction:

Refer to part (a).

Explanation of Solution

Given reaction (1) is as follows:

  ${\text{C}}_{\text{3}}{\text{H}}_{\text{8}}+{\text{O}}_2\to {\text{CO}}_{\text{2}}+{\text{H}}_{\text{2}}\text{O}$

Oxidation number of each atom in reaction (1) is as follows:

  $\stackrel{-\text{8/3}}{{\text{C}}_{\text{3}}}\stackrel{+1}{{\text{H}}_{\text{8}}}+\stackrel{0}{{\text{O}}_2}\to \stackrel{+4}{\text{C}}\stackrel{-2}{{\text{O}}_{\text{2}}}+\stackrel{+1}{{\text{H}}_{\text{2}}}\stackrel{-2}{\text{O}}$

Since oxidation state of oxygen changes from 0 to $-2$ thus oxygen reduced in this reaction. Thus oxygen is oxidizing agent in this reaction.

Hence oxidation state of oxidizing agent in this reaction changes from 0 to $-2$.

Given reaction (2) is as follows:

  ${\text{HNO}}_3+{\text{H}}_{\text{2}}\text{S}\to \text{NO}+\text{S}+{\text{H}}_{\text{2}}\text{O}$

Oxidation number of each atom in reaction (2) is as follows:

  $\stackrel{+1}{\text{H}}\stackrel{+5}{\text{N}}\stackrel{-2}{{\text{O}}_3}+\stackrel{+1}{{\text{H}}_{\text{2}}}\stackrel{-2}{\text{S}}\to \stackrel{+2}{\text{N}}\stackrel{-2}{\text{O}}+\stackrel{0}{\text{S}}+\stackrel{+1}{{\text{H}}_{\text{2}}}\stackrel{-2}{\text{O}}$

Since oxidation state of nitrogen changes from $+5$ to $+2$ thus nitrogen reduced in this reaction. Thus nitrogen is oxidizing agent in this reaction.

Hence oxidation state of oxidizing agent in this reaction changes from $+5$ to $+2$.

Given reaction (3) is as follows:

  $\text{CuO}+{\text{NH}}_{\text{3}}\to {\text{N}}_2+{\text{H}}_{\text{2}}\text{O}+\text{Cu}$

Oxidation number of each atom in reaction (3) is as follows:

  $\stackrel{+2}{\text{Cu}}\stackrel{-2}{\text{O}}+\stackrel{-3}{\text{N}}\stackrel{+1}{{\text{H}}_{\text{3}}}\to \stackrel{0}{{\text{N}}_2}+\stackrel{+1}{{\text{H}}_{\text{2}}}\stackrel{-2}{\text{O}}+\stackrel{0}{\text{Cu}}$

Since oxidation state of copper changes from $+2$ to 0 thus copper reduced in this reaction. Thus copper is oxidizing agent in this reaction.

Hence oxidation state of oxidizing agent in this reaction changes from $+2$ to 0.

Given reaction (4) is as follows:

  ${\text{H}}_{\text{2}}{\text{O}}_2+{\text{Na}}_{\text{2}}{\text{SO}}_3\to {\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}+{\text{H}}_{\text{2}}\text{O}$

Oxidation number of each atom in reaction (4) is as follows:

  $\stackrel{+1}{{\text{H}}_{\text{2}}}\stackrel{-1}{{\text{O}}_2}+\stackrel{+1}{{\text{Na}}_{\text{2}}}\stackrel{+4}{\text{S}}\stackrel{-2}{{\text{O}}_3}\to \stackrel{+1}{{\text{Na}}_{\text{2}}}\stackrel{+6}{\text{S}}\stackrel{-2}{{\text{O}}_{\text{4}}}+\stackrel{+1}{{\text{H}}_{\text{2}}}\stackrel{-2}{\text{O}}$

Since oxidation state of oxygen changes from $-1$ to $-2$ thus oxygen reduced in this reaction. Thus oxygen is oxidizing agent in this reaction.

Hence oxidation state of oxidizing agent in this reaction changes from $-1$ to $-2$.

Given reaction (5) is as follows:

  ${\text{H}}_{\text{2}}{\text{O}}_2\to {\text{H}}_{\text{2}}\text{O}+{\text{O}}_{\text{2}}$

Oxidation number of each atom in reaction (5) is as follows:

  $\stackrel{+1}{{\text{H}}_{\text{2}}}\stackrel{-1}{{\text{O}}_2}\to \stackrel{+1}{{\text{H}}_{\text{2}}}\stackrel{-2}{\text{O}}+\stackrel{0}{{\text{O}}_{\text{2}}}$

Since oxidation state of oxygen changes from $-1$ to $-2$ thus oxygen reduced in this reaction. Thus oxygen is oxidizing agent in this reaction.

Hence oxidation state of oxidizing agent in this reaction changes from $-1$ to $-2$.

(f)

Interpretation Introduction

Interpretation:

Oxidation state change in reducing agent in reactions below has to be determined.

  $\begin{array}{l}1.{\text{ C}}_{\text{3}}{\text{H}}_{\text{8}}+{\text{O}}_2\to {\text{CO}}_{\text{2}}+{\text{H}}_{\text{2}}\text{O}\\ 2.{\text{ HNO}}_3+{\text{H}}_{\text{2}}\text{S}\to \text{NO}+\text{S}+{\text{H}}_{\text{2}}\text{O}\\ 3.\text{ CuO}+{\text{NH}}_{\text{3}}\to {\text{N}}_2+{\text{H}}_{\text{2}}\text{O}+\text{Cu}\\ 4.{\text{ H}}_{\text{2}}{\text{O}}_2+{\text{Na}}_{\text{2}}{\text{SO}}_3\to {\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}+{\text{H}}_{\text{2}}\text{O}\\ 5.{\text{ H}}_{\text{2}}{\text{O}}_2\to {\text{H}}_{\text{2}}\text{O}+{\text{O}}_{\text{2}}\end{array}$

Concept Introduction:

Refer to part (a).

Explanation of Solution

Given reaction (1) is as follows:

  ${\text{C}}_{\text{3}}{\text{H}}_{\text{8}}+{\text{O}}_2\to {\text{CO}}_{\text{2}}+{\text{H}}_{\text{2}}\text{O}$

Oxidation number of each atom in reaction (1) is as follows:

  $\stackrel{-\text{8/3}}{{\text{C}}_{\text{3}}}\stackrel{+1}{{\text{H}}_{\text{8}}}+\stackrel{0}{{\text{O}}_2}\to \stackrel{+4}{\text{C}}\stackrel{-2}{{\text{O}}_{\text{2}}}+\stackrel{+1}{{\text{H}}_{\text{2}}}\stackrel{-2}{\text{O}}$

Since oxidation state of carbon changes from $-\text{8/3}$ to $+4$ thus carbon oxidized in this reaction. Thus carbon is reducing agent in this reaction.

Hence oxidation state of reducing agent in this reaction changes from $-\text{8/3}$ to $+4$.

Given reaction (2) is as follows:

  ${\text{HNO}}_3+{\text{H}}_{\text{2}}\text{S}\to \text{NO}+\text{S}+{\text{H}}_{\text{2}}\text{O}$

Oxidation number of each atom in reaction (2) is as follows:

  $\stackrel{+1}{\text{H}}\stackrel{+5}{\text{N}}\stackrel{-2}{{\text{O}}_3}+\stackrel{+1}{{\text{H}}_{\text{2}}}\stackrel{-2}{\text{S}}\to \stackrel{+2}{\text{N}}\stackrel{-2}{\text{O}}+\stackrel{0}{\text{S}}+\stackrel{+1}{{\text{H}}_{\text{2}}}\stackrel{-2}{\text{O}}$

Since oxidation state of sulfur changes from $-\text{2}$ to 0 thus sulfur oxidized in this reaction. Thus sulfur is reducing agent in this reaction.

Hence oxidation state of reducing agent in this reaction changes from $-\text{2}$ to 0.

Given reaction (3) is as follows:

  $\text{CuO}+{\text{NH}}_{\text{3}}\to {\text{N}}_2+{\text{H}}_{\text{2}}\text{O}+\text{Cu}$

Oxidation number of each atom in reaction (3) is as follows:

  $\stackrel{+2}{\text{Cu}}\stackrel{-2}{\text{O}}+\stackrel{-3}{\text{N}}\stackrel{+1}{{\text{H}}_{\text{3}}}\to \stackrel{0}{{\text{N}}_2}+\stackrel{+1}{{\text{H}}_{\text{2}}}\stackrel{-2}{\text{O}}+\stackrel{0}{\text{Cu}}$

Since oxidation state of nitrogen changes from $-\text{3}$ to 0 thus nitrogen oxidized in this reaction. Thus nitrogen is reducing agent in this reaction.

Hence oxidation state of reducing agent in this reaction changes from $-\text{3}$ to 0.

Given reaction (4) is as follows:

  ${\text{H}}_{\text{2}}{\text{O}}_2+{\text{Na}}_{\text{2}}{\text{SO}}_3\to {\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}+{\text{H}}_{\text{2}}\text{O}$

Oxidation number of each atom in reaction (4) is as follows:

  $\stackrel{+1}{{\text{H}}_{\text{2}}}\stackrel{-1}{{\text{O}}_2}+\stackrel{+1}{{\text{Na}}_{\text{2}}}\stackrel{+4}{\text{S}}\stackrel{-2}{{\text{O}}_3}\to \stackrel{+1}{{\text{Na}}_{\text{2}}}\stackrel{+6}{\text{S}}\stackrel{-2}{{\text{O}}_{\text{4}}}+\stackrel{+1}{{\text{H}}_{\text{2}}}\stackrel{-2}{\text{O}}$

Since oxidation state of sulfur changes from $+4$ to $+6$ thus sulfur oxidized in this reaction. Thus sulfur is reducing agent in this reaction.

Hence oxidation state of reducing agent in this reaction changes from $+4$ to $+6$.

Given reaction (5) is as follows:

  ${\text{H}}_{\text{2}}{\text{O}}_2\to {\text{H}}_{\text{2}}\text{O}+{\text{O}}_{\text{2}}$

Oxidation number of each atom in reaction (5) is as follows:

  $\stackrel{+1}{{\text{H}}_{\text{2}}}\stackrel{-1}{{\text{O}}_2}\to \stackrel{+1}{{\text{H}}_{\text{2}}}\stackrel{-2}{\text{O}}+\stackrel{0}{{\text{O}}_{\text{2}}}$

Since oxidation state of oxygen changes from $-1$ to 0 thus oxygen oxidized in this reaction. Hence oxygen is reducing agent in this reaction.

Hence oxidation state of reducing agent in this reaction changes from $-1$ to 0.