Chapter 18.2, Problem 2RQ

(a)

Interpretation Introduction

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

  ${\text{Cu}}_{\text{(s)}}{\text{+ 2 AgNO}}_{\text{3(aq)}}\stackrel{}{\to }{\text{Cu(NO}}_{\text{3}}{\text{)}}_{\text{2(aq)}}{\text{ + 2 Ag}}_{\text{(s)}}$

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

Answer to Problem 2RQ

• ${\text{Cu}}_{\text{(s)}}$= reducing agent.
• ${\text{AgNO}}_{\text{3(aq)}}$ = oxidizing agent.

Explanation of Solution

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

  $\begin{array}{l}{\text{Cu}}_{\text{(s)}}{\text{+ 2 AgNO}}_{\text{3(aq)}}\stackrel{}{\to }{\text{Cu(NO}}_{\text{3}}{\text{)}}_{\text{2(aq)}}{\text{ + 2 Ag}}_{\text{(s)}}\\ \text{0 +1 +2 0}\end{array}$

In the given reaction, the oxidation number of ${\text{Cu}}_{\text{(s)}}$ changes from 0 to +1 so it is an oxidation process. Hence ${\text{Cu}}_{\text{(s)}}$ must be reducing agent.

Silver in ${\text{AgNO}}_{\text{3(aq)}}$ reduces from +1 to 0 in solid silver ${\text{Ag}}_{\text{(s)}}$ . Thus, ${\text{AgNO}}_{\text{3(aq)}}$ must be oxidizing agent.

(b)

Interpretation Introduction

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

  ${\text{4 Al}}_{\text{(s)}}{\text{+ 3 O}}_{\text{2(g)}}\stackrel{}{\to }{\text{ 2 Al}}_{\text{2}}{\text{O}}_{\text{3}}{}_{\text{(s)}}$

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

Answer to Problem 2RQ

• ${\text{Al}}_{\text{(s)}}$ = reducing agent.
• ${\text{O}}_{\text{2(g)}}$ = oxidizing agent.

Explanation of Solution

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

  $\begin{array}{l}{\text{4 Al}}_{\text{(s)}}{\text{+ 3 O}}_{\text{2(g)}}\stackrel{}{\to }{\text{ 2 Al}}_{\text{2}}{\text{O}}_{\text{3}}{}_{\text{(s)}}\\ 0\text{ 0 +3 -2}\end{array}$

In the given reaction, the oxidation number of ${\text{Al}}_{\text{(s)}}$ changes from 0 to +3 so it is an oxidation process. Hence ${\text{Al}}_{\text{(s)}}$ must be reducing agent.

Silver in ${\text{O}}_{\text{2(g)}}$ reduces from 0 to -2 in ${\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}{}_{\text{(s)}}$ . Thus, ${\text{O}}_{\text{2(g)}}$ must be oxidizing agent.

(c)

Interpretation Introduction

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

  ${\text{C}}_{\text{(s)}}{\text{+ O}}_{\text{2(g)}}\stackrel{}{\to }{\text{CO}}_{\text{2}}{}_{\text{(g)}}$

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

Answer to Problem 2RQ

• ${\text{C}}_{\text{(s)}}$ = reducing agent.
• ${\text{O}}_{\text{2(g)}}$ = oxidizing agent.

Explanation of Solution

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

  $\begin{array}{l}{\text{C}}_{\text{(s)}}{\text{+ O}}_{\text{2(g)}}\stackrel{}{\to }{\text{CO}}_{\text{2}}{}_{\text{(g)}}\\ 0\text{ 0 +4 -2}\end{array}$

In the given reaction, the oxidation number of ${\text{C}}_{\text{(s)}}$ changes from 0 to +4 so it is an oxidation process. Hence ${\text{C}}_{\text{(s)}}$ must be reducing agent.

Silver in ${\text{O}}_{\text{2(g)}}$ reduces from 0 to -2 in ${\text{CO}}_{\text{2}}{}_{\text{(g)}}$ . Thus, ${\text{O}}_{\text{2(g)}}$ must be oxidizing agent.