Chapter 18, Problem 18A

(a)

Interpretation Introduction

Interpretation: The element which is oxidized and which is reduced in the given reaction needs to be identified by assigning oxidation number.

  ${\text{Zn}}_{\text{(s)}}{\text{+ 2 HNO}}_3{}_{\text{(aq)}}\stackrel{}{\to }{\text{ Zn(NO}}_3{)}_2{}_{\text{(aq)}}{\text{+ H}}_{\text{2}}{}_{\text{(g)}}$

Concept Introduction: A Redox reaction leads to oxidation and reduction processes simultaneously.

Oxidation and reduction reactions are the processes involved in loss or gain of electron.

Oxidation is the process that is involved in the loss of electrons and oxidized to cation whereas reduction is the process that is involved in the gain of electrons and reduced to anion.

Answer to Problem 18A

• H= Reduce.
• $\text{Zn}$ = Oxidize.

Explanation of Solution

The substance, which is oxidized, is called as reducing agent whereas the substance, which gets reduce, is called as oxidizing agent.

  ${\text{Zn}}_{\text{(s)}}{\text{+ 2 HNO}}_3{}_{\text{(aq)}}\stackrel{}{\to }{\text{ Zn(NO}}_3{)}_2{}_{\text{(aq)}}{\text{+ H}}_{\text{2}}{}_{\text{(g)}}$

The oxidation number of elements are:

  $\begin{array}{l}{\text{Zn}}_{\text{(s)}}{\text{+ 2 HNO}}_3{}_{\text{(aq)}}\stackrel{}{\to }{\text{ Zn(NO}}_3{)}_2{}_{\text{(aq)}}{\text{+ H}}_{\text{2}}{}_{\text{(g)}}\\ 0\text{ +1+5-2 +2 +5 -2 0}\end{array}$

In the given reaction, the oxidation number of hydrogen in ${\text{HNO}}_3$ changes from +1 to 0 so it is a reduction process. Hence hydrogen in ${\text{HNO}}_3$ reduces during reaction.

Zinc in $\text{Zn}$ oxidizes from 0 to +2, thus, $\text{Zn}$ oxidized during the given reaction.

(b)

Interpretation Introduction

Interpretation: The element which is oxidized and which is reduced in the given reaction needs to be identified by assigning oxidation number.

  ${\text{H}}_{\text{2}}{}_{\text{(g)}}{\text{+ CuSO}}_{\text{4(aq)}}\stackrel{}{\to }{\text{Cu}}_{\text{(s)}}{\text{+ H}}_{\text{2}}{\text{SO}}_{\text{4}}{}_{\text{(aq)}}$

Concept Introduction: A Redox reaction leads to oxidation and reduction processes simultaneously.

Oxidation and reduction reactions are the processes involved in loss or gain of electron.

Oxidation is the process that is involved in the loss of electrons and oxidized to cation whereas reduction is the process that is involved in the gain of electrons and reduced to anion.

Answer to Problem 18A

• Cu = Reduce.
• $\text{H}$ = Oxidize.

Explanation of Solution

The substance, which is oxidized, is called as reducing agent whereas the substance, which gets reduce, is called as oxidizing agent.

  ${\text{H}}_{\text{2}}{}_{\text{(g)}}{\text{+ CuSO}}_{\text{4(aq)}}\stackrel{}{\to }{\text{Cu}}_{\text{(s)}}{\text{+ H}}_{\text{2}}{\text{SO}}_{\text{4}}{}_{\text{(aq)}}$

The oxidation number of elements are:

  $\begin{array}{l}{\text{H}}_{\text{2}}{}_{\text{(g)}}{\text{+ CuSO}}_{\text{4(aq)}}\stackrel{}{\to }{\text{ Cu}}_{\text{(s)}}{\text{+ H}}_{\text{2}}{\text{SO}}_{\text{4}}{}_{\text{(aq)}}\\ 0\text{ +2+6-2 0 +1+6-2}\end{array}$

In the given reaction, the oxidation number ofCu in ${\text{CuSO}}_{\text{4(aq)}}$ changes from +2 to 0 so it is a reduction process. Hence Cuin ${\text{CuSO}}_{\text{4(aq)}}$ reduces during reaction.

Hydrogen in ${\text{H}}_{\text{2}}$ oxidizes from 0 to +1, thus, ${\text{H}}_{\text{2}}$ oxidized during the given reaction.

(c)

Interpretation Introduction

Interpretation: The element which is oxidized and which is reduced in the given reaction needs to be identified by assigning oxidation number.

  ${\text{N}}_{\text{2}}{}_{\text{(g)}}{\text{+ 3 Br}}_{\text{2(l)}}\stackrel{}{\to }{\text{ 2 NBr}}_{\text{3}}{}_{\text{(g)}}$

Concept Introduction: A Redox reaction leads to oxidation and reduction processes simultaneously.

Oxidation and reduction reactions are the processes involved in loss or gain of electron.

Oxidation is the process that is involved in the loss of electrons and oxidized to cation whereas reduction is the process that is involved in the gain of electrons and reduced to anion.

Answer to Problem 18A

• Br= Reduce.
• $\text{N}$ = Oxidize.

Explanation of Solution

The substance, which is oxidized, is called as reducing agent whereas the substance, which gets reduce, is called as oxidizing agent.

  ${\text{N}}_{\text{2}}{}_{\text{(g)}}{\text{+ 3 Br}}_{\text{2(l)}}\stackrel{}{\to }{\text{ 2 NBr}}_{\text{3}}{}_{\text{(g)}}$

The oxidation number of elements are:

  $\begin{array}{l}{\text{N}}_{\text{2}}{}_{\text{(g)}}{\text{+ 3 Br}}_{\text{2(l)}}\stackrel{}{\to }{\text{ 2 NBr}}_{\text{3}}{}_{\text{(g)}}\\ 0\text{ 0 +3-1}\end{array}$

In the given reaction, the oxidation number ofBr in ${\text{Br}}_{\text{2(l)}}$ changes from 0 to -1 so it is a reduction process. Hence Brin ${\text{Br}}_{\text{2(l)}}$ reduces during reaction.

Nitrogen in ${\text{N}}_{\text{2}}$ oxidizes from 0 to +3, thus, ${\text{N}}_{\text{2}}$ oxidized during the given reaction.

(d)

Interpretation Introduction

Interpretation: The element which is oxidized and which is reduced in the given reaction needs to be identified by assigning oxidation number.

  ${\text{2 KBr}}_{\text{(aq)}}{\text{+ Cl}}_{\text{2(g)}}\stackrel{}{\to }{\text{ 2 KCl}}_{\text{(aq) }}{\text{+ Br}}_{\text{2(l)}}$

Concept Introduction: A Redox reaction leads to oxidation and reduction processes simultaneously.

Oxidation and reduction reactions are the processes involved in loss or gain of electron.

Oxidation is the process that is involved in the loss of electrons and oxidized to cation whereas reduction is the process that is involved in the gain of electrons and reduced to anion.

Answer to Problem 18A

• Br = oxidize.
• $\text{Cl}$ = Reduce.

Explanation of Solution

The substance, which is oxidized, is called as reducing agent whereas the substance, which gets reduce, is called as oxidizing agent.

  ${\text{2 KBr}}_{\text{(aq)}}{\text{+ Cl}}_{\text{2(g)}}\stackrel{}{\to }{\text{ 2 KCl}}_{\text{(aq) }}{\text{+ Br}}_{\text{2(l)}}$

The oxidation number of elements are:

  $\begin{array}{l}{\text{2 KBr}}_{\text{(aq)}}{\text{+ Cl}}_{\text{2(g)}}\stackrel{}{\to }{\text{ 2 KCl}}_{\text{(aq) }}{\text{+ Br}}_{\text{2(l)}}\\ \text{ +1 -1 }0\text{ +1-1 0}\end{array}$

In the given reaction, the oxidation number ofBr in $\text{KBr}$ changes from -1 to 0 so it is a oxidation process. Hence Br in $\text{KBr}$ oxidizes during reaction.

Chlorine in ${\text{Cl}}_{\text{2}}$ oxidizes from 0 to -1, thus, ${\text{Cl}}_{\text{2}}$ reduces during the given reaction.