Chapter 19, Problem 62A

Interpretation Introduction

(a)

Interpretation:

The given redox reaction is to be balanced using oxidation number method.

Concept introduction:

The reaction can be balanced be oxidation number method.

Oxidation number method involve:-

• Assign the oxidation number to each element.
• Check which element is oxidized and which element is reduced.
• Check the change in oxidation number for the elements which are oxidized and reduced.
• Balance the change in oxidation number by adjusting the coefficients in the equation.
• Then balance the other elements.

Answer to Problem 62A

$2PbS+3O_2\to 2PbO+2S{O}_2$

Explanation of Solution

The given redox reaction is:

$PbS+O_2\to PbO+S{O}_2$

Assigning the oxidation number-

$Pb\stackrel{-2}{S}+{\stackrel{0}{O}}_2\to Pb\stackrel{-2}{O}+\stackrel{+4}{S}{O}_2$

The oxidation numbers of atoms shows that

$S$ is oxidized and $O_2$ is reduced.

Change in oxidation number of $S=+6$

Change in oxidation number of $Operatom=-4$

Thus adjusting the coefficients:

$2PbS+3O_2\to PbO+S{O}_2$

After readjustment:

$2PbS+3O_2\to 2PbO+2S{O}_2$

Interpretation Introduction

(a)

Interpretation:

The given redox reaction is to be balanced using oxidation number method.

Concept introduction:

The reaction can be balanced be oxidation number method.

Oxidation number method involve:-

• Assign the oxidation number to each element.
• Check which element is oxidized and which element is reduced.
• Check the change in oxidation number for the elements which are oxidized and reduced.
• Balance the change in oxidation number by adjusting the coefficients in the equation.
• Then balance the other elements.

Answer to Problem 62A

$4NaW{O}_3+4NaOH+O_2\to 4N{a}_{2}W{O}_4+2H_{2}O$

Explanation of Solution

The given redox reaction is:

$NaW{O}_3+NaOH+O_2\to N{a}_{2}W{O}_4+H_{2}O$

Assigning the oxidation number-

$Na\stackrel{+5}{W}{O}_3+NaOH+\stackrel{0}{O_2}\to N{a}_2\stackrel{+6}{W}{O}_4+H_2\stackrel{-2}{O}$

The oxidation numbers of atoms shows that

$W$ is oxidized and $O$ is reduced.

Change in oxidation number of $W=+1$

Change in oxidation number of $Operatom=-4$

Thus adjusting the coefficients:

$4NaW{O}_3+NaOH+O_2\to N{a}_{2}W{O}_4+H_{2}O$

After readjustment:

$4NaW{O}_3+4NaOH+O_2\to 4N{a}_{2}W{O}_4+2H_{2}O$

Interpretation Introduction

(a)

Interpretation:

The given redox reaction is to be balanced using oxidation number method.

Concept introduction:

The reaction can be balanced be oxidation number method.

Oxidation number method involve:-

• Assign the oxidation number to each element.
• Check which element is oxidized and which element is reduced.
• Check the change in oxidation number for the elements which are oxidized and reduced.
• Balance the change in oxidation number by adjusting the coefficients in the equation.
• Then balance the other elements.

Answer to Problem 62A

$2N{H}_3+3CuO\to 3Cu+N_2+3H_{2}O$

Explanation of Solution

The given redox reaction is:

$N{H}_3+CuO\to Cu+N_2+H_{2}O$

Assigning the oxidation number-

$\stackrel{-3}{N}{H}_3+\stackrel{+2}{Cu}O\to \stackrel{0}{Cu}+\stackrel{0}{N_2}+H_{2}O$

The oxidation numbers of atoms shows that

$N$ is oxidized and $Cu$ is reduced.

Change in oxidation number of $N=+3$

Change in oxidation number of $Cu=-2$

Thus adjusting the coefficients:

$2N{H}_3+3CuO\to Cu+N_2+H_{2}O$

After readjustment:

$2N{H}_3+3CuO\to 3Cu+N_2+3H_{2}O$

Interpretation Introduction

(a)

Interpretation:

The given redox reaction is to be balanced using oxidation number method.

Concept introduction:

The reaction can be balanced be oxidation number method.

Oxidation number method involve:-

• Assign the oxidation number to each element.
• Check which element is oxidized and which element is reduced.
• Check the change in oxidation number for the elements which are oxidized and reduced.
• Balance the change in oxidation number by adjusting the coefficients in the equation.
• Then balance the other elements.

Answer to Problem 62A

$A{l}_2{O}_3+3C+3C{l}_2\to 2AlC{l}_3+3CO$

Explanation of Solution

The given redox reaction is:

$A{l}_2{O}_3+C+C{l}_2\to AlC{l}_3+CO$

Assigning the oxidation number-

$A{l}_2{O}_3+\stackrel{0}{C}+\stackrel{0}{C{l}_2}\to Al{\stackrel{-1}{Cl}}_3+\stackrel{+2}{C}O$

The oxidation numbers of atoms shows that

$C$ is oxidized and $C{l}_2$ is reduced.

Change in oxidation number of $C=+2$

Change in oxidation number of $Clperatom=-2$

Thus adjusting the coefficients:

$A{l}_2{O}_3+C+C{l}_2\to AlC{l}_3+CO$

After readjustment:

$A{l}_2{O}_3+3C+3C{l}_2\to 2AlC{l}_3+3CO$