Chapter 19.2, Problem 32SSC

(a)

Interpretation Introduction

Interpretation:

The given redox reaction is to be balanced using half reaction method.

Concept introduction:

Half reaction 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. Write oxidation and reduction half reactions.
• Check the change in oxidation number for the elements which are oxidized and reduced.
• Then add enough hydrogen ions and water molecules to the equation to balance hydrogen atoms on both sides.
• Adjust the coefficients so that number of electrons lost in oxidation is equal to electrons gained in reduction.
• Add the balanced half reactions.

Answer to Problem 32SSC

$3HCl{O}_3\to 2Cl{O}_2+H_{2}O+HCl{O}_4$

Explanation of Solution

The given redox reaction is:

$HCl{O}_3\to Cl{O}_2+HCl{O}_4+H_{2}O$

Reaction in ionic form is:

$Cl{O}_3{}^{-}\to Cl{O}_2+Cl{O}_4{}^{-}$

Assigning the oxidation number-

$\stackrel{+5}{Cl}{O}_3{}^{-}\to \stackrel{+4}{Cl}{O}_2+\stackrel{+7}{Cl}{O}_4{}^{-}$

The oxidation numbers of atoms shows that

$Cl$ is oxidized and $Cl$ is reduced.

Change in oxidation number of $Cl=+2$

Change in oxidation number of $Cl=-1$

$Cl{O}_3{}^{-}\to Cl{O}_4{}^{-}+2e^{-}$

$Cl{O}_3{}^{-}+e^{-}\to Cl{O}_2$

Balancing the atoms and charges:

$Cl{O}_3{}^{-}+H_{2}O\to Cl{O}_4{}^{-}+2e^{-}+2H^{+}$

$Cl{O}_3{}^{-}+e^{-}+2H^{+}\to Cl{O}_2+H_{2}O$

Now adding both reactions after multiplying reduction half reaction with 2

$3Cl{O}_3{}^{-}+2H^{+}\to 2Cl{O}_2+H_{2}O+Cl{O}_4{}^{-}$

Or

$3HCl{O}_3\to 2Cl{O}_2+H_{2}O+HCl{O}_4$

(b)

Interpretation Introduction

Interpretation:

The given redox reaction is to be balanced using half reaction method.

Concept introduction:

Half reaction 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. Write oxidation and reduction half reactions.
• Check the change in oxidation number for the elements which are oxidized and reduced.
• Then add enough hydrogen ions and water molecules to the equation to balance hydrogen atoms on both sides.
• Adjust the coefficients so that number of electrons lost in oxidation is equal to electrons gained in reduction.
• Add the balanced half reactions.

Answer to Problem 32SSC

$2HCl{O}_3+5H_{2}Se{O}_3\to C{l}_2+H_{2}O+5H_{2}Se{O}_4$

Explanation of Solution

The given redox reaction is:

$H_{2}Se{O}_3+HCl{O}_3\to HCl{O}_3+C{l}_2+H_{2}O$

Reaction in ionic form is:

$Cl{O}_3{}^{-}+Se{O}_3{}^{2-}\to C{l}_2+Se{O}_4{}^{2-}$

Assigning the oxidation number-

$\stackrel{+5}{Cl}{O}_3{}^{-}+\stackrel{+4}{Se}{O}_3{}^{2-}\to \stackrel{0}{C{l}_2}+\stackrel{+6}{Se}{O}_4{}^{2-}$

The oxidation numbers of atoms shows that

$Se$ is oxidized and $Cl$ is reduced.

Change in oxidation number of $Se=+2$

Change in oxidation number of $Cl=-5$

$Se{O}_3{}^{2-}\to Se{O}_4{}^{2-}+2e^{-}$

$Cl{O}_3{}^{-}+5e^{-}\to C{l}_2$

Balancing the atoms and charges:

$Se{O}_3{}^{2-}+H_{2}O\to Se{O}_4{}^{2-}+2e^{-}+2H^{+}$

$2Cl{O}_3{}^{-}+10e^{-}+12H^{+}\to C{l}_2+6H_{2}O$

Now adding both reactions after multiplying oxidation half reaction with 5

$2Cl{O}_3{}^{-}+2H^{+}+5Se{O}_3{}^{2-}\to C{l}_2+H_{2}O+5Se{O}_4{}^{2-}$

Or

$2HCl{O}_3+5H_{2}Se{O}_3\to C{l}_2+H_{2}O+5H_{2}Se{O}_4$

(c)

Interpretation Introduction

Interpretation:

The given redox reaction is to be balanced using half reaction method.

Concept introduction:

Half reaction 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. Write oxidation and reduction half reactions.
• Check the change in oxidation number for the elements which are oxidized and reduced.
• Then add enough hydrogen ions and water molecules to the equation to balance hydrogen atoms on both sides.
• Adjust the coefficients so that number of electrons lost in oxidation is equal to electrons gained in reduction.
• Add the balanced half reactions.

Answer to Problem 32SSC

$6F{e}^{2+}+C{r}_2{O}_7{}^{2-}+14H^{+}\to 2C{r}^{3+}+7H_{2}O+6F{e}^{3+}$

Explanation of Solution

The given redox reaction is:

$C{r}_2{O}_7{}^{2-}+F{e}^{2+}\to C{r}^{3+}+F{e}^{3+}$

Assigning the oxidation number-

${\stackrel{+6}{Cr}}_2{O}_7{}^{2-}+\stackrel{+2}{F{e}^{2+}}\to \stackrel{+3}{C{r}^{3+}}+\stackrel{+3}{F{e}^{3+}}$

The oxidation numbers of atoms shows that

$Fe$ is oxidized and $Cr$ is reduced.

Change in oxidation number of $Fe=+1$

Change in oxidation number of $Crperatom=-6$

$F{e}^{2+}\to F{e}^{3+}+e^{-}$

$C{r}_2{O}_7{}^{2-}+6e^{-}\to 2C{r}^{3+}$

Balancing the atoms and charges:

$F{e}^{2+}\to F{e}^{3+}+e^{-}$

$C{r}_2{O}_7{}^{2-}+6e^{-}+14H^{+}\to 2C{r}^{3+}+7H_{2}O$

Now adding both reactions after multiplying oxidation half reaction with 6

$6F{e}^{2+}+C{r}_2{O}_7{}^{2-}+14H^{+}\to 2C{r}^{3+}+7H_{2}O+6F{e}^{3+}$