Chapter 19, Problem 79A

Interpretation Introduction

(a)

Interpretation:

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

Concept introduction:

Oxidation number method to balance ionic equation in acidic solution 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 add enough hydrogen ions and water molecules to the equation to balance hydrogen atoms on both sides.

Answer to Problem 79A

$5S{b}^{+3}+2Mn{O}_4{}^{-1}+12H_{2}O\to 5Sb{O}_4{}^{-3}+2M{n}^{+2}+24H^{+}$

Explanation of Solution

The given redox reaction is:

$S{b}^{+3}+Mn{O}_4{}^{-1}\to Sb{O}_4{}^{-3}+M{n}^{+2}$

Assigning the oxidation number-

$\stackrel{+3}{S{b}^{+3}}+\stackrel{+7}{Mn}{O}_4{}^{-1}\to \stackrel{+5}{Sb}{O}_4{}^{-3}+\stackrel{+2}{M{n}^{+2}}$

The oxidation numbers of atoms shows that

$Sb$ is oxidized and $Mn$ is reduced.

Change in oxidation number of $Sb=+2$

Change in oxidation number of $Mn=-5$

Thus adjusting the coefficients:

$5S{b}^{+3}+2Mn{O}_4{}^{-1}\to 5Sb{O}_4{}^{-3}+2M{n}^{+2}$

Now adding hydrogen ions on right side of equation to balance negative charge

$5S{b}^{+3}+2Mn{O}_4{}^{-1}\to 5Sb{O}_4{}^{-3}+2M{n}^{+2}+24H^{+}$

Adding water to left side:

$5S{b}^{+3}+2Mn{O}_4{}^{-1}+12H_{2}O\to 5Sb{O}_4{}^{-3}+2M{n}^{+2}+24H^{+}$

Interpretation Introduction

(b)

Interpretation:

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

Concept introduction:

Oxidation number method to balance ionic equation in basic solution 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 add enough hydroxyl ions to balance the charge and water molecules to the equation to balance hydrogen atoms on both sides.

Answer to Problem 79A

$N_{2}O+2Cl{O}^{-1}+2O{H}^{-}\to 2C{l}^{-}+2N{O}_2{}^{-}+H_{2}O$

Explanation of Solution

The given redox reaction is:

$N_{2}O+Cl{O}^{-1}\to C{l}^{-}+N{O}_2{}^{-}$

Assigning the oxidation number-

${\stackrel{+1}{N}}_{2}O+\stackrel{+1}{Cl}{O}^{-1}\to \stackrel{-1}{C{l}^{-}}+\stackrel{+3}{N}{O}_2{}^{-}$

The oxidation numbers of atoms shows that

$I$ is oxidized and $Mn$ is reduced.

Change in oxidation number of $Nperatom=+4$

Change in oxidation number of $Cl=-2$

Thus adjusting the coefficients:

$N_{2}O+2Cl{O}^{-1}\to 2C{l}^{-}+N{O}_2{}^{-}$

Balancing the atoms:

$N_{2}O+2Cl{O}^{-1}\to 2C{l}^{-}+2N{O}_2{}^{-}$

Now adding hydroxyl ions on left side of equation to balance negative charge

$N_{2}O+2Cl{O}^{-1}+2O{H}^{-}\to 2C{l}^{-}+2N{O}_2{}^{-}$

Adding water to water side:

$N_{2}O+2Cl{O}^{-1}+2O{H}^{-}\to 2C{l}^{-}+2N{O}_2{}^{-}+H_{2}O$