Chapter 19, Problem 72A

Interpretation Introduction

(a)

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 72A

$3C{l}^{-}+2N{O}_3{}^{-}+2H^{+}\to 2NO+3Cl{O}^{-}+H_{2}O$

Explanation of Solution

The given redox reaction is:

$C{l}^{-}+N{O}_3{}^{-}\to NO+Cl{O}^{-}$

Assigning the oxidation number-

$\stackrel{-1}{C{l}^{-}}+\stackrel{+5}{N}{O}_3{}^{-}\to \stackrel{+2}{N}O+\stackrel{+1}{Cl}{O}^{-}$

The oxidation numbers of atoms shows that

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

Change in oxidation number of $Cl=+2$

Change in oxidation number of $N=-3$

$C{l}^{-}\to Cl{O}^{-}+2e^{-}$

$N{O}_3{}^{-}+3e^{-}\to NO$

Balancing the atoms and charges:

$C{l}^{-}+H_{2}O\to Cl{O}^{-}+2e^{-}+2H^{+}$

$N{O}_3{}^{-}+3e^{-}+4H^{+}\to NO+2H_{2}O$

Now adding both reactions after multiplying oxidation half cell reaction with 3 and reduction half reaction with 2,

$3C{l}^{-}+2N{O}_3{}^{-}+2H^{+}\to 2NO+3Cl{O}^{-}+H_{2}O$

Interpretation Introduction

(b)

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 72A

$5B{r}^{-}+I{O}_3{}^{-}+6H^{+}\to 2B{r}_2+IBr+3H_{2}O$

Explanation of Solution

The given redox reaction is:

$B{r}^{-}+I{O}_3{}^{-}\to B{r}_2+IBr$

Assigning the oxidation number-

$\stackrel{-1}{B{r}^{-}}+\stackrel{+5}{I}{O}_3{}^{-}\to \stackrel{0}{B{r}_2}+\stackrel{+1}{I}Br$

The oxidation numbers of atoms shows that

$Br$ is oxidized and $I$ is reduced.

Change in oxidation number of $Br=+1$

Change in oxidation number of $I=-4$

$B{r}^{-}\to B{r}_2+e^{-}$

$B{r}^{-}+I{O}_3{}^{-}+4e^{-}\to IBr$

Balancing the atoms and charges:

$2B{r}^{-}\to B{r}_2+2e^{-}$

$B{r}^{-}+I{O}_3{}^{-}+4e^{-}+6H^{+}\to IBr+3H_{2}O$

Now adding both reactions after multiplying oxidation half cell reaction with 2

$5B{r}^{-}+I{O}_3{}^{-}+6H^{+}\to 2B{r}_2+IBr+3H_{2}O$

Interpretation Introduction

(c)

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 72A

$I_2+S_2{O}_3{}^{2-}+H_{2}O\to S_2{O}_4{}^{2-}+2I^{-}+2H^{+}$

Explanation of Solution

The given redox reaction is:

$I_2+N{a}_2{S}_2{O}_3\to N{a}_2{S}_2{O}_4+NaI$

Reaction in ionic form:

$I_2+S_2{O}_3{}^{2-}\to S_2{O}_4{}^{2-}+I^{-}$

Assigning the oxidation number-

$\stackrel{0}{I_2}+{\stackrel{+2}{S}}_2{O}_3{}^{2-}\to {\stackrel{+3}{S}}_2{O}_4{}^{2-}+\stackrel{-1}{I^{-}}$

The oxidation numbers of atoms shows that

$S$ is oxidized and $I$ is reduced.

Change in oxidation number of $Speratom=+2$

Change in oxidation number of $Iperatom=-2$

$S_2{O}_3{}^{2-}\to S_2{O}_4{}^{2-}+2e^{-}$

$I_2+2e^{-}\to I^{-}$

Balancing the atoms and charges:

$S_2{O}_3{}^{2-}+H_{2}O\to S_2{O}_4{}^{2-}+2e^{-}+2H^{+}$

$I_2+2e^{-}\to 2I^{-}$

Now adding both reactions

$I_2+S_2{O}_3{}^{2-}+H_{2}O\to S_2{O}_4{}^{2-}+2I^{-}+2H^{+}$