Chapter 19, Problem 69A

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 basic 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 hydroxyl 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 69A

$4N{H}_3+3N{O}_2\to 7N_2+6H_{2}O$

Explanation of Solution

The given redox reaction is:

$N{H}_3+N{O}_2\to N_2+H_{2}O$

Assigning the oxidation number-

$\stackrel{-3}{N}{H}_3+\stackrel{+4}{N}{O}_2\to \stackrel{0}{N_2}+H_{2}O$

The oxidation numbers of atoms shows that

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

Change in oxidation number of $N=+3$

Change in oxidation number of $N=-4$

$N{H}_3\to N_2+3e^{-}(Oxidation)$

$N{O}_2+4e^{-}\to N_2(\mathrm{Re}duction)$

Balancing the atoms and charges:

$N{H}_3+3O{H}^{-}\to N_2+3e^{-}+3H_{2}O(Oxidation)$

$N{O}_2+4e^{-}+2H_{2}O\to N_2+4O{H}^{-}(\mathrm{Re}duction)$

Now adding both reactions after multiplying oxidation half cell reaction with 4 and reduction half reaction with 3,

$4N{H}_3+3N{O}_2\to 7N_2+6H_{2}O$

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 basic 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 hydroxyl 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 69A

$6B{r}_2+12O{H}^{-}\to 10B{r}^{-}+2Br{O}_3{}^{-}+6H_{2}O$

Explanation of Solution

The given redox reaction is:

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

Assigning the oxidation number-

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

The oxidation numbers of atoms shows that

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

Change in oxidation number of $Brperatom=+10$

Change in oxidation number of $Brperatom=-2$

$B{r}_2\to Br{O}_3{}^{-}+10e^{-}(Oxidation)$

$B{r}_2+2e^{-}\to B{r}^{-}(\mathrm{Re}duction)$

Balancing the atoms and charges:

$B{r}_2+12O{H}^{-}\to 2Br{O}_3{}^{-}+10e^{-}+6H_{2}O(Oxidation)$

$B{r}_2+2e^{-}\to 2B{r}^{-}(\mathrm{Re}duction)$

Now adding both reactions after multiplying oxidation half cell reaction with 3

$6B{r}_2+12O{H}^{-}\to 10B{r}^{-}+2Br{O}_3{}^{-}+6H_{2}O$