Chapter 18, Problem 18.34QE

(a)

Interpretation Introduction

Interpretation:

The balanced equation for the reaction has to be given.

  ${\text{ClO}}^{-}\left(\text{aq}\right){\text{ +CrO}}_2{}^{-}\left(\text{aq}\right)\to {\text{Cl}}^{-}\text{(aq)}{\text{+CrO}}_4{}^{2-}\left(\text{aq}\right)$

Concept introduction:

Steps for balancing half –reactions in BASIC solution:

• Balance all atoms except H and O in half reaction.
• Balance O atoms by adding water to the side missing O atoms.
• Balance the H atoms by adding ${\text{H}}^{\text{+}}$ to the side missing H atoms.
• Balance the charge by adding electrons to side with more total positive charge.
• Make the number of electrons the same in both half reactions by multiplication, while avoiding fractional number of electrons.
• Add the same numbers of ${\text{OH}}^{\text{-}}$ groups as there are ${\text{H}}^{\text{+}}$ present to both sides of the equation.

Explanation of Solution

The given reaction is as follows:

  ${\text{ClO}}^{-}\left(\text{aq}\right){\text{ +CrO}}_2{}^{-}\left(\text{aq}\right)\to {\text{Cl}}^{-}\text{(aq)}{\text{+CrO}}_4{}^{2-}\left(\text{aq}\right)$

The two half-cell reactions can be written as below:

  $\begin{array}{l}{\text{CrO}}_2{}^{-}\to {\text{CrO}}_4{}^{2-}\\ \\ {\text{ClO}}^{-}\to {\text{Cl}}^{-}\end{array}$

Steps for balancing half –reactions in BASIC solution:

1. 1. Balance all atoms except H and O in half reaction.

   $\begin{array}{l}{\text{CrO}}_2{}^{-}\to {\text{CrO}}_4{}^{2-}\\ \\ {\text{ClO}}^{-}\to {\text{Cl}}^{-}\end{array}$

2. 2. Balance O atoms by adding water to the side missing O atoms.

   $\begin{array}{l}{\text{CrO}}_2{}^{-}+2{\text{H}}_{\text{2}}\text{O}\to {\text{CrO}}_4{}^{2-}\\ \\ {\text{ClO}}^{-}\to {\text{Cl}}^{-}+{\text{H}}_{\text{2}}\text{O}\end{array}$

3. 3. Balance the H atoms by adding ${\text{H}}^{\text{+}}$ to the side missing H atoms.

   $\begin{array}{l}{\text{CrO}}_2{}^{-}+2{\text{H}}_{\text{2}}\text{O}\to {\text{CrO}}_4{}^{2-}+4{\text{H}}^{\text{+}}\\ \\ {\text{ClO}}^{-}+2{\text{H}}^{\text{+}}\to {\text{Cl}}^{-}+{\text{H}}_{\text{2}}\text{O}\end{array}$

4. 4. . Add the same numbers of ${\text{OH}}^{\text{-}}$ groups as there are ${\text{H}}^{\text{+}}$ present to both sides of the equation

$\begin{array}{l}{\text{CrO}}_2{}^{-}+2{\text{H}}_{\text{2}}\text{O}{\text{+4OH}}^{-}\to {\text{CrO}}_4{}^{2-}+4{\text{H}}_{\text{2}}\text{O}\\ \\ {\text{ClO}}^{-}+2{\text{H}}_{\text{2}}\text{O}\to {\text{Cl}}^{-}+{\text{H}}_{\text{2}}{\text{O+2OH}}^{-}\end{array}$

1. 5 Balance the charge by adding electrons to side with more total positive charge.

$\begin{array}{l}{\text{CrO}}_2{}^{-}+2{\text{H}}_{\text{2}}\text{O}{\text{+4OH}}^{-}\to {\text{CrO}}_4{}^{2-}+4{\text{H}}_{\text{2}}{\text{O+3e}}^{-}\\ \\ {\text{ClO}}^{-}+2{\text{H}}_{\text{2}}{\text{O+2e}}^{-}\to {\text{Cl}}^{-}+{\text{H}}_{\text{2}}{\text{O+2OH}}^{-}\end{array}$

1. 6 Make the number of electrons the same in both half- reactions by multiplication, while avoiding fractional number of electrons.

   $\begin{array}{l}{\text{2CrO}}_2{}^{-}+4{\text{H}}_{\text{2}}\text{O}{\text{+8OH}}^{-}\to 2{\text{CrO}}_4{}^{2-}+8{\text{H}}_{\text{2}}{\text{O+6e}}^{-}\\ \\ {\text{3ClO}}^{-}+6{\text{H}}_{\text{2}}{\text{O+6e}}^{-}\to 3{\text{Cl}}^{-}+3{\text{H}}_{\text{2}}{\text{O+6OH}}^{-}\end{array}$

2. 7 The half reactions can be added together and then same species on opposite side of the arrow can be cancelled for simplifying the equation:

   ${\text{2CrO}}_2{}^{-}{\text{+ 3ClO}}^{-}+{\text{2OH}}^{-}\rightleftarrows 2{\text{CrO}}_4{}^{2-}\text{+}3{\text{Cl}}^{-}+{\text{H}}_{\text{2}}\text{O}$

Therefore, the balanced reaction is as follows.

${\text{2CrO}}_2{}^{-}{\text{+ 3ClO}}^{-}+{\text{2OH}}^{-}\rightleftarrows 2{\text{CrO}}_4{}^{2-}\text{+}3{\text{Cl}}^{-}+{\text{H}}_{\text{2}}\text{O}$

(b)

Interpretation Introduction

Interpretation:

The balanced equation for the reaction has to be given.

  ${\text{Br}}_2\left(\text{aq}\right)\to {\text{Br}}^{-}\left(\text{aq}\right){\text{+BrO}}_3{}^{2-}\left(\text{aq}\right)$

Concept Introduction:

Refer to (a)

Explanation of Solution

The given reaction is as follows:

  ${\text{Br}}_2\left(\text{aq}\right)\to {\text{Br}}^{-}\left(\text{aq}\right){\text{+BrO}}_3{}^{-}\left(\text{aq}\right)$

The two half-cell reactions can be written as below:

  $\begin{array}{l}{\text{Br}}_2\to {\text{BrO}}_3{}^{-}\\ \\ {\text{Br}}_2\to {\text{Br}}^{-}\end{array}$

Steps for balancing half –reactions in BASIC solution:

1. 1. Balance all atoms except H and O in half reaction.

$\begin{array}{l}{\text{Br}}_2\to 2{\text{BrO}}_3{}^{-}\\ \\ {\text{Br}}_2\to 2{\text{Br}}^{-}\end{array}$

1. 2. Balance O atoms by adding water to the side missing O atoms.

$\begin{array}{l}{\text{Br}}_2+6{\text{H}}_{\text{2}}\text{O}\to 2{\text{BrO}}_3{}^{-}\\ \\ {\text{Br}}_2\to 2{\text{Br}}^{-}\end{array}$

1. 3. Balance the H atoms by adding ${\text{H}}^{\text{+}}$ to the side missing H atoms.

$\begin{array}{l}{\text{Br}}_2+6{\text{H}}_{\text{2}}\text{O}\to 2{\text{BrO}}_3{}^{-}+12{\text{H}}^{\text{+}}\\ \\ {\text{Br}}_2\to 2{\text{Br}}^{-}\end{array}$

1. 4. . Add the same numbers of ${\text{OH}}^{\text{-}}$ groups as there are ${\text{H}}^{\text{+}}$ present to both sides of the equation

$\begin{array}{l}{\text{Br}}_2+6{\text{H}}_{\text{2}}{\text{O+12OH}}^{-}\to 2{\text{BrO}}_3{}^{-}+12{\text{H}}_{\text{2}}\text{O}\\ \\ {\text{Br}}_2\to 2{\text{Br}}^{-}\end{array}$

1. 5. Balance the charge by adding electrons to side with more total positive charge.

   $\begin{array}{l}{\text{Br}}_2+6{\text{H}}_{\text{2}}{\text{O+12OH}}^{-}\to 2{\text{BrO}}_3{}^{-}+12{\text{H}}_{\text{2}}{\text{O+10e}}^{-}\\ \\ {\text{Br}}_2+2{\text{e}}^{-}\to 2{\text{Br}}^{-}\end{array}$

2. 6. Make the number of electrons the same in both half- reactions by multiplication, while avoiding fractional number of electrons.

$\begin{array}{l}{\text{Br}}_2+6{\text{H}}_{\text{2}}{\text{O+12OH}}^{-}\to 2{\text{BrO}}_3{}^{-}+12{\text{H}}_{\text{2}}{\text{O+10e}}^{-}\\ \\ {\text{5Br}}_2+10{\text{e}}^{-}\to 10{\text{Br}}^{-}\end{array}$

1. 7. The half reactions can be added together and then same species on opposite side of the arrow can be cancelled for simplifying the equation:

   ${\text{6Br}}_2{\text{+ 12OH}}^{-}\rightleftarrows 2{\text{BrO}}_3{}^{-}\text{+}10{\text{Br}}^{-}\text{+}6{\text{H}}_{\text{2}}\text{O}$

Therefore, the balanced reaction is as follows.

  ${\text{3Br}}_2{\text{+ 6OH}}^{-}\rightleftarrows {\text{BrO}}_3{}^{-}\text{+}5{\text{Br}}^{-}\text{+}3{\text{H}}_{\text{2}}\text{O}$

(c)

Interpretation Introduction

Interpretation:

The balanced equation for the reaction has to be given.

  ${\text{H}}_2{\text{O}}_2\left(\text{aq}\right){\text{ +N}}_2{\text{O}}_4\left(\text{aq}\right)\to {\text{N}}_2\text{(g)}{\text{+H}}_2\text{O}\left(l\right)$

Concept introduction:

Steps for balancing half –reactions in BASIC solution:

• Balance all atoms except H and O in half reaction.
• Balance O atoms by adding water to the side missing O atoms.
• Balance the H atoms by adding ${\text{H}}^{\text{+}}$ to the side missing H atoms.
• Balance the charge by adding electrons to side with more total positive charge.
• Make the number of electrons the same in both half reactions by multiplication, while avoiding fractional number of electrons.
• Add the same numbers of ${\text{OH}}^{\text{-}}$ groups as there are ${\text{H}}^{\text{+}}$ present to both sides of the equation.

Explanation of Solution

The given reaction is as follows:

  ${\text{H}}_2{\text{O}}_2\left(\text{aq}\right){\text{ +N}}_2{\text{O}}_4\left(\text{aq}\right)\to {\text{N}}_2\text{(g)}{\text{+H}}_2\text{O}\left(l\right)$

The two half-cell reactions can be written as below:

  $\begin{array}{l}{\text{H}}_2{\text{O}}_2\to {\text{H}}_2\text{O}\\ \\ {\text{N}}_2{\text{O}}_4\to {\text{N}}_2\end{array}$

Steps for balancing half –reactions in BASIC solution:

1. 1. Balance all atoms except H and O in half reaction.

$\begin{array}{l}{\text{H}}_2{\text{O}}_2\to {\text{H}}_2\text{O}\\ \\ {\text{N}}_2{\text{O}}_4\to {\text{N}}_2\end{array}$

1. 2. Balance O atoms by adding water to the side missing O atoms.

$\begin{array}{l}{\text{H}}_2{\text{O}}_2\to 2{\text{H}}_2\text{O}\\ \\ {\text{N}}_2{\text{O}}_4\to {\text{N}}_2+4{\text{H}}_{\text{2}}\text{O}\end{array}$

1. 3. Balance the H atoms by adding ${\text{H}}^{\text{+}}$ to the side missing H atoms.

   $\begin{array}{l}\begin{array}{l}{\text{N}}_2{\text{H}}_4\to {\text{N}}_2+4{\text{H}}^{+}\\ \end{array}\hfill \\ {\text{H}}_2{\text{O}}_2+2{\text{H}}^{+}\to 2{\text{H}}_2\text{O}\hfill \end{array}$

2. 4. . Add the same numbers of ${\text{OH}}^{\text{-}}$ groups as there are ${\text{H}}^{\text{+}}$ present to both sides of the equation

$\begin{array}{l}\begin{array}{l}{\text{N}}_2{\text{H}}_4+4{\text{OH}}^{-}\to {\text{N}}_2+4{\text{H}}_2\text{O}\\ \end{array}\hfill \\ {\text{H}}_2{\text{O}}_2+2{\text{H}}_2\text{O}\to 2{\text{H}}_2\text{O}+2{\text{OH}}^{-}\hfill \end{array}$

1. 5. Balance the charge by adding electrons to side with more total positive charge.

$\begin{array}{l}\begin{array}{l}{\text{N}}_2{\text{H}}_4+4{\text{OH}}^{-}\to {\text{N}}_2+4{\text{H}}_2\text{O}+4{\text{e}}^{-}\\ \end{array}\hfill \\ {\text{H}}_2{\text{O}}_2+2{\text{H}}_2\text{O}+2{\text{e}}^{-}\to 2{\text{H}}_2\text{O}+2{\text{OH}}^{-}\hfill \end{array}$

1. 6. Make the number of electrons the same in both half- reactions by multiplication, while avoiding fractional number of electrons.

$\begin{array}{l}\begin{array}{l}{\text{N}}_2{\text{H}}_4+4{\text{OH}}^{-}\to {\text{N}}_2+4{\text{H}}_2\text{O}+4{\text{e}}^{-}\\ \end{array}\hfill \\ 2{\text{H}}_2{\text{O}}_2+4{\text{H}}_2\text{O}+4{\text{e}}^{-}\to 4{\text{H}}_2\text{O}+4{\text{OH}}^{-}\hfill \end{array}$

1. 7. The half reactions can be added together and then same species on opposite side of the arrow can be cancelled for simplifying the equation:

   ${\text{N}}_2{\text{H}}_4+2{\text{H}}_2{\text{O}}_2\rightleftarrows {\text{N}}_2+4{\text{H}}_2\text{O}$

Therefore, the balanced reaction is as follows.

  ${\text{N}}_2{\text{H}}_4+2{\text{H}}_2{\text{O}}_2\rightleftarrows {\text{N}}_2+4{\text{H}}_2\text{O}$