Chapter 18, Problem 18.33QE

(a)

Interpretation Introduction

Interpretation:

The balanced equation for the reaction has to be given.

  ${\text{Al(s) +ClO}}^{-}\left(\text{aq}\right)\to \text{Al}{\left(\text{OH}\right)}_4{}^{-}\text{(aq)}{\text{+ClO}}^{-}\left(\text{aq}\right)$

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:

  ${\text{Al(s) +ClO}}^{-}\left(\text{aq}\right)\to \text{Al}{\left(\text{OH}\right)}_4{}^{-}\text{(aq)}{\text{+Cl}}^{-}\left(\text{aq}\right)$

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

  $\begin{array}{l}\text{Al}\to \text{Al}{\left(\text{OH}\right)}_4{}^{-}\\ \\ {\text{ClO}}^{-}\to {\text{Cl}}^{-}\end{array}$

Steps for balancing half –reactions in BASIC solution:

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

   $\begin{array}{l}\text{Al}\to \text{Al}{\left(\text{OH}\right)}_4{}^{-}\\ \\ {\text{ClO}}^{-}\to {\text{Cl}}^{-}\end{array}$

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

   $\begin{array}{l}\text{Al}+4{\text{H}}_{\text{2}}\text{O}\to \text{Al}{\left(\text{OH}\right)}_4{}^{-}\\ \\ {\text{ClO}}^{-}\to {\text{Cl}}^{-}+{\text{H}}_{\text{2}}\text{O}\end{array}$

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

   $\begin{array}{l}\text{Al}+4{\text{H}}_{\text{2}}\text{O}\to \text{Al}{\left(\text{OH}\right)}_4{}^{-}+4{\text{H}}^{\text{+}}\\ \\ {\text{ClO}}^{-}+2{\text{H}}^{\text{+}}\to {\text{Cl}}^{-}+{\text{H}}_{\text{2}}\text{O}\end{array}$

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

$\begin{array}{l}\text{Al}+4{\text{H}}_{\text{2}}{\text{O+4OH}}^{-}\to \text{Al}{\left(\text{OH}\right)}_4{}^{-}+4{\text{H}}_2\text{O}\\ \\ {\text{ClO}}^{-}+2{\text{H}}_2\text{O}\to {\text{Cl}}^{-}+{\text{H}}_{\text{2}}{\text{O+2OH}}^{-}\end{array}$

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

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

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

   $\begin{array}{l}\text{2Al}+8{\text{H}}_{\text{2}}{\text{O+8OH}}^{-}\to 2\text{Al}{\left(\text{OH}\right)}_4{}^{-}+8{\text{H}}_2\text{O}{\text{+6e}}^{-}\\ \\ {\text{3ClO}}^{-}+6{\text{H}}_2{\text{O+6e}}^{-}\to 3{\text{Cl}}^{-}+3{\text{H}}_{\text{2}}{\text{O+6OH}}^{-}\end{array}$

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:

   $\text{2Al}{\text{+ 3ClO}}^{-}+3{\text{H}}_2{\text{O+2OH}}^{-}\rightleftarrows 2\text{Al}{\left(\text{OH}\right)}_4{}^{-}\text{+}3{\text{Cl}}^{-}$

Therefore, the balanced reaction is as follows.

$\text{2Al}{\text{+ 3ClO}}^{-}+3{\text{H}}_2{\text{O+2OH}}^{-}\rightleftarrows 2\text{Al}{\left(\text{OH}\right)}_4{}^{-}\text{+}3{\text{Cl}}^{-}$

(b)

Interpretation Introduction

Interpretation:

The balanced equation for the reaction has to be given.

  ${\text{MnO}}_4{}^{-}\left(\text{aq}\right){\text{ +SO}}_3{}^{2-}\left(\text{aq}\right)\to {\text{MnO}}_2\left(\text{s}\right){\text{+SO}}_4{}^{2-}\left(\text{aq}\right)$

Concept Introduction:

Refer to (a)

Explanation of Solution

The given reaction is as follows:

  ${\text{MnO}}_4{}^{-}\left(\text{aq}\right){\text{ +SO}}_3{}^{2-}\left(\text{aq}\right)\to {\text{MnO}}_2\left(\text{s}\right){\text{+SO}}_4{}^{2-}\left(\text{aq}\right)$

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

  $\begin{array}{l}{\text{MnO}}_4{}^{-}\to {\text{MnO}}_2\\ \\ {\text{SO}}_3{}^{2-}\to {\text{SO}}_4{}^{2-}\end{array}$

Steps for balancing half –reactions in BASIC solution:

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

$\begin{array}{l}{\text{MnO}}_4{}^{-}\to {\text{MnO}}_2\\ \\ {\text{SO}}_3{}^{2-}\to {\text{SO}}_4{}^{2-}\end{array}$

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

$\begin{array}{l}{\text{MnO}}_4{}^{-}\to {\text{MnO}}_2+2{\text{H}}_{\text{2}}\text{O}\\ \\ {\text{SO}}_3{}^{2-}+{\text{H}}_{\text{2}}\text{O}\to {\text{SO}}_4{}^{2-}\end{array}$

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

$\begin{array}{l}{\text{MnO}}_4{}^{-}+4{\text{H}}^{\text{+}}\to {\text{MnO}}_2+2{\text{H}}_{\text{2}}\text{O}\\ \\ {\text{SO}}_3{}^{2-}+{\text{H}}_{\text{2}}\text{O}\to {\text{SO}}_4{}^{2-}+2{\text{H}}^{\text{+}}\end{array}$

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

$\begin{array}{l}{\text{MnO}}_4{}^{-}+4{\text{H}}_{\text{2}}\text{O}\to {\text{MnO}}_2+2{\text{H}}_{\text{2}}{\text{O+4OH}}^{-}\\ \\ {\text{SO}}_3{}^{2-}+{\text{H}}_{\text{2}}\text{O}{\text{+2OH}}^{-}\to {\text{SO}}_4{}^{2-}+2{\text{H}}_{\text{2}}\text{O}\end{array}$

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

   $\begin{array}{l}{\text{MnO}}_4{}^{-}+4{\text{H}}_{\text{2}}\text{O+3}{\text{e}}^{-}\to {\text{MnO}}_2+2{\text{H}}_{\text{2}}{\text{O+4OH}}^{-}\\ \\ {\text{SO}}_3{}^{2-}+{\text{H}}_{\text{2}}\text{O}{\text{+2OH}}^{-}\to {\text{SO}}_4{}^{2-}+2{\text{H}}_{\text{2}}\text{O+2}{\text{e}}^{-}\end{array}$

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

$\begin{array}{l}{\text{2MnO}}_4{}^{-}+8{\text{H}}_{\text{2}}\text{O+6}{\text{e}}^{-}\to 2{\text{MnO}}_2+4{\text{H}}_{\text{2}}{\text{O+8OH}}^{-}\\ \\ {\text{3SO}}_3{}^{2-}+3{\text{H}}_{\text{2}}\text{O}{\text{+6OH}}^{-}\to 3{\text{SO}}_4{}^{2-}+6{\text{H}}_{\text{2}}\text{O+6}{\text{e}}^{-}\end{array}$

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:

   ${\text{2MnO}}_4{}^{-}{\text{+ 3SO}}_3{}^{2-}+{\text{H}}_2\text{O}\rightleftarrows 2{\text{MnO}}_2\text{+}3{\text{SO}}_4{}^{2-}{\text{+2OH}}^{-}$

Therefore, the balanced reaction is as follows.

  ${\text{2MnO}}_4{}^{-}{\text{+ 3SO}}_3{}^{2-}+{\text{H}}_2\text{O}\rightleftarrows 2{\text{MnO}}_2\text{+}3{\text{SO}}_4{}^{2-}{\text{+2OH}}^{-}$

(c)

Interpretation Introduction

Interpretation:

The balanced equation for the reaction has to be given.

  ${\text{Zn(s) +NO}}_3{}^{-}\left(\text{aq}\right)\to \text{Zn}{\left(\text{OH}\right)}_4{}^{2-}\text{(aq)}{\text{+NH}}_3\left(\text{aq}\right)$

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:

  ${\text{Zn(s) +NO}}_3{}^{-}\left(\text{aq}\right)\to \text{Zn}{\left(\text{OH}\right)}_4{}^{2-}\text{(aq)}{\text{+NH}}_3\left(\text{aq}\right)$

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

  $\begin{array}{l}\text{Zn}\to \text{Zn}{\left(\text{OH}\right)}_4{}^{2-}\\ \\ {\text{NO}}_3{}^{-}\to {\text{NH}}_3\end{array}$

Steps for balancing half –reactions in BASIC solution:

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

$\begin{array}{l}\text{Zn}\to \text{Zn}{\left(\text{OH}\right)}_4{}^{2-}\\ \\ {\text{NO}}_3{}^{-}\to {\text{NH}}_3\end{array}$

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

$\begin{array}{l}\text{Zn}+4{\text{H}}_{\text{2}}\text{O}\to \text{Zn}{\left(\text{OH}\right)}_4{}^{2-}\\ \\ {\text{NO}}_3{}^{-}\to {\text{NH}}_3+3{\text{H}}_{\text{2}}\text{O}\end{array}$

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

$\begin{array}{l}\text{Zn}+4{\text{H}}_{\text{2}}\text{O}\to \text{Zn}{\left(\text{OH}\right)}_4{}^{2-}+{\text{4H}}^{\text{+}}\\ \\ {\text{NO}}_3{}^{-}+9{\text{H}}^{\text{+}}\to {\text{NH}}_3+3{\text{H}}_{\text{2}}\text{O}\end{array}$

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

$\begin{array}{l}\text{Zn}+4{\text{H}}_{\text{2}}\text{O}+{\text{4OH}}^{-}\to \text{Zn}{\left(\text{OH}\right)}_4{}^{2-}+{\text{4H}}_{\text{2}}\text{O}\\ \\ {\text{NO}}_3{}^{-}+9{\text{H}}_{\text{2}}\text{O}\to {\text{NH}}_3+3{\text{H}}_{\text{2}}\text{O}+9{\text{OH}}^{-}\end{array}$

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

$\begin{array}{l}\text{Zn}+4{\text{H}}_{\text{2}}\text{O}+{\text{4OH}}^{-}\to \text{Zn}{\left(\text{OH}\right)}_4{}^{2-}+{\text{4H}}_{\text{2}}{\text{O+2e}}^{-}\\ \\ {\text{NO}}_3{}^{-}+9{\text{H}}_{\text{2}}\text{O}{\text{+8e}}^{-}\to {\text{NH}}_3+3{\text{H}}_{\text{2}}\text{O}+9{\text{OH}}^{-}\end{array}$

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

$\begin{array}{l}\text{4Zn}+16{\text{H}}_{\text{2}}\text{O}+16{\text{OH}}^{-}\to 4\text{Zn}{\left(\text{OH}\right)}_4{}^{2-}+16{\text{H}}_{\text{2}}{\text{O+8e}}^{-}\\ \\ {\text{NO}}_3{}^{-}+9{\text{H}}_{\text{2}}\text{O}{\text{+8e}}^{-}\to {\text{NH}}_3+3{\text{H}}_{\text{2}}\text{O}+9{\text{OH}}^{-}\end{array}$

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:

   $\text{4Zn}{\text{+ NO}}_3{}^{-}+6{\text{H}}_2{\text{O+7OH}}^{-}\rightleftarrows 4\text{Zn}{\left(\text{OH}\right)}_4{}^{2-}{\text{+NH}}_3$

Therefore, the balanced reaction is as follows.

  $\text{4Zn}{\text{+ NO}}_3{}^{-}+6{\text{H}}_2{\text{O+7OH}}^{-}\rightleftarrows 4\text{Zn}{\left(\text{OH}\right)}_4{}^{2-}{\text{+NH}}_3$