Chapter 18, Problem 4A

(a)

Interpretation Introduction

Interpretation: The element, which is oxidized and being reduced in the given reaction needs to be determined.

  ${\text{Ca}}_{\text{(s)}}{\text{+ 2 H}}_{\text{2}}{\text{O}}_{\text{(l)}}\stackrel{}{\to }{\text{Ca(OH)}}_{\text{2}}{}_{\text{(s, aq)}}{\text{ + H}}_{\text{2(g)}}$

Concept Introduction: A Redox reaction leads to oxidation and reduction processes simultaneously. Oxidation and reduction reactions are the processes involved loss or gain 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 4A

• $\text{H}$ = Reduce.
• $\text{Ca}$ = 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.

  ${\text{Ca}}_{\text{(s)}}{\text{+ 2 H}}_{\text{2}}{\text{O}}_{\text{(l)}}\stackrel{}{\to }{\text{Ca(OH)}}_{\text{2}}{}_{\text{(s, aq)}}{\text{ + H}}_{\text{2(g)}}$

In the given reaction, the oxidation number of hydrogen in ${\text{H}}_{\text{2}}\text{O}$ changes from +1 to 0 so it is a reduction process. Hence hydrogen in ${\text{H}}_{\text{2}}\text{O}$ reduces during reaction.

Iodine in $\text{Ca}$ oxidizes from 0 to +1, thus, $\text{Ca}$ oxidized during the given reaction.

(b)

Interpretation Introduction

Interpretation: The element, which is oxidized and being reduced in the given reaction needs to be determined.

  ${\text{H}}_{\text{2}}{}_{\text{(g)}}{\text{+ F}}_{\text{2}}{}_{\text{(g)}}\stackrel{}{\to }{\text{ 2HF}}_{\text{(g)}}$

Concept Introduction: A Redox reaction leads to oxidation and reduction processes simultaneously. Oxidation and reduction reactions are the processes involved loss or gain 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 4A

• $\text{F}$= 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.

  ${\text{H}}_{\text{2}}{}_{\text{(g)}}{\text{+ F}}_{\text{2}}{}_{\text{(g)}}\stackrel{}{\to }{\text{ 2HF}}_{\text{(g)}}$

In the given reaction, the oxidation number of hydrogen in ${\text{H}}_{\text{2}}$ changes from 0 to +1 so it is a oxidation process. Hence hydrogen in ${\text{H}}_{\text{2}}$ oxidized during reaction.

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

(c)

Interpretation Introduction

Interpretation: The element, which is oxidized and being reduced in the given reaction needs to be determined.

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

Concept Introduction: A Redox reaction leads to oxidation and reduction processes simultaneously. Oxidation and reduction reactions are the processes involved loss or gain 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 4A

• $\text{O}$ = Reduce.
• $\text{Fe}$ = 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.

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

In the given reaction, the oxidation number of hydrogen in $\text{Fe}$ changes from 0 to +3 so it is a oxidation process. Hence $\text{Fe}$ oxidized during reaction.

Oxygen in ${\text{O}}_{\text{2}}$ reduces from 0 to -2, thus, ${\text{O}}_{\text{2}}$ reduces during the given reaction.

(d)

Interpretation Introduction

Interpretation: The element, which is oxidized and being reduced in the given reaction needs to be determined.

  ${\text{4 Fe}}_{\text{(s)}}{\text{+ 3 Cl}}_{\text{2}}{}_{\text{(g)}}\stackrel{}{\to }\text{ 2 FeCl}{}_{\text{(s)}}$

Concept Introduction: A Redox reaction leads to oxidation and reduction processes simultaneously. Oxidation and reduction reactions are the processes involved loss or gain 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 4A

• $\text{Cl}$= Reduce.
• $\text{Fe}$ = 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.

  ${\text{4 Fe}}_{\text{(s)}}{\text{+ 3 Cl}}_{\text{2}}{}_{\text{(g)}}\stackrel{}{\to }\text{ 2 FeCl}{}_{\text{(s)}}$

In the given reaction, the oxidation number of hydrogen in $\text{Fe}$ changes from 0 to +3 so it is a oxidation process. Hence $\text{Fe}$ oxidized during reaction.

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