Chapter 18, Problem 18.32QE

(a)

Interpretation Introduction

Interpretation:

The balanced equation for the reaction has to be given.

  ${\text{Zn(s) +NO}}_3{}^{-}\left(\text{aq}\right)\to {\text{Zn}}^{\text{+2}}\text{(aq)}{\text{+N}}_2\text{(g)}$

Concept Introduction:

Steps for balancing half –reactions in ACIDIC 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.

Explanation of Solution

The given reaction is as follows:

  ${\text{Zn(s) +NO}}_3{}^{-}\left(\text{aq}\right)\to {\text{Zn}}^{\text{+2}}\text{(aq)}{\text{+N}}_2\text{(g)}$

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

  $\begin{array}{l}\text{Zn}\to {\text{Zn}}^{\text{+2}}\\ \\ {\text{NO}}_3{}^{-}\to {\text{N}}_2\end{array}$

Steps for balancing half –reactions in ACIDIC solution:

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

   $\begin{array}{l}\text{Zn}\to {\text{Zn}}^{\text{+2}}\\ \\ \text{2}{\text{NO}}_3{}^{-}\to {\text{N}}_2\end{array}$

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

   $\begin{array}{l}\text{Zn}\to {\text{Zn}}^{\text{+2}}\\ \\ \text{2}{\text{NO}}_3{}^{-}\to {\text{N}}_2+\text{6}{\text{H}}_{\text{2}}\text{O}\end{array}$

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

   $\begin{array}{l}\text{Zn}\to {\text{Zn}}^{\text{+2}}\\ \\ \text{2}{\text{NO}}_3{}^{-}+12{\text{H}}^{\text{+}}\to {\text{N}}_2+\text{6}{\text{H}}_{\text{2}}\text{O}\end{array}$

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

   $\begin{array}{l}\text{Zn}\to {\text{Zn}}^{\text{+2}}+{\text{2e}}^{-}\\ \\ \text{2}{\text{NO}}_3{}^{-}+12{\text{H}}^{\text{+}}+10{\text{e}}^{-}\to {\text{N}}_2+\text{6}{\text{H}}_{\text{2}}\text{O}\end{array}$

5. 5 Make the number of electrons the same in both half reactions by multiplication, while avoiding fractional number of electrons.

   $\begin{array}{l}\text{5}\text{Zn}\to 5{\text{Zn}}^{\text{+2}}+10{\text{e}}^{-}\\ \\ \text{2}{\text{NO}}_3{}^{-}+12{\text{H}}^{\text{+}}+10{\text{e}}^{-}\to {\text{N}}_2+\text{6}{\text{H}}_{\text{2}}\text{O}\end{array}$

6. 6 The half reactions can be added together and then same species on opposite side of the arrow can be cancelled for simplifying the equation:

   $\text{5}\text{Zn}\text{+ 2}{\text{NO}}_3{}^{-}+12{\text{H}}^{\text{+}}\rightleftarrows 5{\text{Zn}}^{\text{+2}}{\text{+N}}_2+\text{6}{\text{H}}_{\text{2}}\text{O}$

Therefore, the balanced reaction is as follows.

$\text{5}\text{Zn}\text{+ 2}{\text{NO}}_3{}^{-}+12{\text{H}}^{\text{+}}\rightleftarrows 5{\text{Zn}}^{\text{+2}}{\text{+N}}_2+\text{6}{\text{H}}_{\text{2}}\text{O}$

(b)

Interpretation Introduction

Interpretation:

The balanced equation for the reaction has to be given.

  ${\text{IO}}_3{}^{-}{\text{(aq) +I}}^{-}\text{(aq)}\to {\text{I}}_2\text{(aq)}$

Concept Introduction:

Refer to (a)

Explanation of Solution

The given reaction is as follows:

  ${\text{IO}}_3{}^{-}{\text{(aq) +I}}^{-}\text{(aq)}\to {\text{I}}_2\text{(aq)}$

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

  $\begin{array}{l}{\text{I}}^{-}\to {\text{I}}_2\\ \\ {\text{IO}}_3{}^{-}\to {\text{I}}_2\end{array}$

Steps for balancing half –reactions in ACIDIC solution:

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

$\begin{array}{l}\text{2}{\text{I}}^{-}\to {\text{I}}_2\\ \\ \text{2}{\text{IO}}_3{}^{-}\to {\text{I}}_2\end{array}$

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

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

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

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

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

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

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

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

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

$\begin{array}{l}\text{10}{\text{I}}^{-}\text{+2}{\text{IO}}_3{}^{-}+12{\text{H}}^{\text{+}}\to 6{\text{I}}_2+6{\text{H}}_{\text{2}}\text{O}\\ \\ 5{\text{I}}^{-}\text{+}{\text{IO}}_3{}^{-}+6{\text{H}}^{\text{+}}\to 3{\text{I}}_2+3{\text{H}}_{\text{2}}\text{O}\end{array}$

Therefore, the balanced reaction is as follows.

$5{\text{I}}^{-}\text{+}{\text{IO}}_3{}^{-}+6{\text{H}}^{\text{+}}\to 3{\text{I}}_2+3{\text{H}}_{\text{2}}\text{O}$

(c)

Interpretation Introduction

Interpretation:

The balanced equation for the reaction has to be given.

  ${\text{Ce}}^{\text{+4}}{\text{(aq) +Cl}}^{-}\left(\text{aq}\right)\to {\text{Cl}}_2\text{(aq)}{\text{+Ce}}^{\text{+3}}\text{(aq)}$

Concept Introduction:

Refer to (a)

Explanation of Solution

The given reaction is as follows:

  ${\text{Ce}}^{\text{+4}}{\text{(aq) +Cl}}^{-}\left(\text{aq}\right)\to {\text{Cl}}_2\text{(aq)}{\text{+Ce}}^{\text{+3}}\text{(aq)}$

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

  $\begin{array}{l}{\text{Ce}}^{\text{+4}}\to {\text{Ce}}^{\text{+3}}\\ \\ {\text{Cl}}^{-}\to {\text{Cl}}_2\end{array}$

Steps for balancing half –reactions in ACIDIC solution:

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

   $\begin{array}{l}\text{2}{\text{Cl}}^{-}\to {\text{Cl}}_2\\ \\ {\text{Ce}}^{\text{+4}}\to {\text{Ce}}^{\text{+3}}\end{array}$

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

   $\begin{array}{l}\text{2}{\text{Cl}}^{-}\to {\text{Cl}}_2\\ \\ {\text{Ce}}^{\text{+4}}\to {\text{Ce}}^{\text{+3}}\end{array}$

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

   $\begin{array}{l}\text{2}{\text{Cl}}^{-}\to {\text{Cl}}_2\\ \\ {\text{Ce}}^{\text{+4}}\to {\text{Ce}}^{\text{+3}}\end{array}$

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

   $\begin{array}{l}\text{2}{\text{Cl}}^{-}\to {\text{Cl}}_2+{\text{2e}}^{-}\\ \\ {\text{Ce}}^{\text{+4}}\text{ +}{\text{e}}^{-}\to {\text{Ce}}^{\text{+3}}\end{array}$

5. 5 Make the number of electrons the same in both half reactions by multiplication, while avoiding fractional number of electrons.

  $\begin{array}{l}\text{2}{\text{Cl}}^{-}\to {\text{Cl}}_2+{\text{2e}}^{-}\\ \\ {\text{2Ce}}^{\text{+4}}\text{ +}2{\text{e}}^{-}\to 2{\text{Ce}}^{\text{+3}}\end{array}$

1. 6 The half reactions can be added together and then same species on opposite side of the arrow can be cancelled for simplifying the equation:

   $\text{2}{\text{Cl}}^{-}{\text{+ 2Ce}}^{\text{+4}}\rightleftarrows {\text{Cl}}_2+2{\text{Ce}}^{\text{+3}}$

   Therefore, the balanced reaction is as follows.

  $\text{2}{\text{Cl}}^{-}{\text{+ 2Ce}}^{\text{+4}}\rightleftarrows {\text{Cl}}_2+2{\text{Ce}}^{\text{+3}}$