Chapter 18, Problem 18.35QE

(a)

Interpretation Introduction

Interpretation:

The balanced equation for the reaction has to be given.

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

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:

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

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

  $\begin{array}{l}\begin{array}{l}{\text{Cl}}_2\to {\text{ClO}}_3^{-}\\ \end{array}\hfill \\ {\text{Cl}}_2\to {\text{Cl}}^{-}\hfill \end{array}$

Steps for balancing half –reactions in BASIC solution:

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

   $\begin{array}{l}\begin{array}{l}{\text{Cl}}_2\to 2{\text{ClO}}_3^{-}\\ \end{array}\hfill \\ {\text{Cl}}_2\to 2{\text{Cl}}^{-}\hfill \end{array}$

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

   $\begin{array}{l}\begin{array}{l}{\text{Cl}}_2+6{\text{H}}_2\text{O}\to 2{\text{ClO}}_3^{-}\\ \end{array}\hfill \\ {\text{Cl}}_2\to 2{\text{Cl}}^{-}\hfill \end{array}$

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

   $\begin{array}{l}\begin{array}{l}{\text{Cl}}_2+6{\text{H}}_2\text{O}\to 2{\text{ClO}}_3^{-}+12{\text{H}}^{+}\\ \end{array}\hfill \\ {\text{Cl}}_2\to 2{\text{Cl}}^{-}\hfill \end{array}$

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

$\begin{array}{l}\begin{array}{l}{\text{Cl}}_2+6{\text{H}}_2\text{O}+12{\text{OH}}^{-}\to 2{\text{ClO}}_3^{-}+12{\text{H}}_2\text{O}\\ \end{array}\hfill \\ {\text{Cl}}_2\to 2{\text{Cl}}^{-}\hfill \end{array}$

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

$\begin{array}{l}\begin{array}{l}{\text{Cl}}_2+6{\text{H}}_2\text{O}+12{\text{OH}}^{-}\to 2{\text{ClO}}_3^{-}+12{\text{H}}_2\text{O}+10{\text{e}}^{-}\\ \end{array}\hfill \\ {\text{Cl}}_2+2{\text{e}}^{-}\to 2{\text{Cl}}^{-}\hfill \end{array}$

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

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

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:

   $3{\text{Cl}}_2+6{\text{OH}}^{-}\rightleftarrows {\text{ClO}}_3^{-}+5{\text{Cl}}^{-}+3{\text{H}}_2\text{O}$

Therefore, the balanced reaction is as follows.

$3{\text{Cl}}_2+6{\text{OH}}^{-}\rightleftarrows {\text{ClO}}_3^{-}+5{\text{Cl}}^{-}+3{\text{H}}_2\text{O}$

(b)

Interpretation Introduction

Interpretation:

The balanced equation for the reaction has to be given.

  ${\text{MnO}}_{\text{4}}{}^{\text{-}}\left(\text{aq}\right){\text{+I}}^{-}\left(\text{aq}\right)\to {\text{IO}}_3{}^{-}\left(\text{aq}\right){\text{+MnO}}_{\text{2}}\left(\text{s}\right)$

Concept Introduction:

Refer to part (a).

Explanation of Solution

The given reaction is as follows:

  ${\text{MnO}}_{\text{4}}{}^{\text{-}}\left(\text{aq}\right){\text{+I}}^{-}\left(\text{aq}\right)\to {\text{IO}}_3{}^{-}\left(\text{aq}\right){\text{+MnO}}_{\text{2}}\left(\text{s}\right)$

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

  $\begin{array}{l}\begin{array}{l}{\text{I}}^{-}\to {\text{IO}}_3^{-}\\ \end{array}\hfill \\ {\text{MnO}}_4^{-}\to {\text{MnO}}_2\hfill \end{array}$

Steps for balancing half –reactions in BASIC solution:

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

$\begin{array}{l}\begin{array}{l}{\text{I}}^{-}\to {\text{IO}}_3^{-}\\ \end{array}\hfill \\ {\text{MnO}}_4^{-}\to {\text{MnO}}_2\hfill \end{array}$

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

$\begin{array}{l}\begin{array}{l}{\text{I}}^{-}+3{\text{H}}_2\text{O}\to {\text{IO}}_3^{-}\\ \end{array}\hfill \\ {\text{MnO}}_4^{-}\to {\text{MnO}}_2+2{\text{H}}_2\text{O}\hfill \end{array}$

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

$\begin{array}{l}\begin{array}{l}{\text{I}}^{-}+3{\text{H}}_2\text{O}\to {\text{IO}}_3^{-}+6{\text{H}}^{+}\\ \end{array}\hfill \\ {\text{MnO}}_4^{-}+4{\text{H}}^{+}\to {\text{MnO}}_2+2{\text{H}}_2\text{O}\hfill \end{array}$

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

$\begin{array}{l}\begin{array}{l}{\text{I}}^{-}+3{\text{H}}_2\text{O}+6{\text{OH}}^{-}\to {\text{IO}}_3^{-}+6{\text{H}}_2\text{O}\\ \end{array}\hfill \\ {\text{MnO}}_4^{-}+4{\text{H}}_2\text{O}\to {\text{MnO}}_2+2{\text{H}}_2\text{O}+4{\text{OH}}^{-}\hfill \end{array}$

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

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

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

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

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:

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

Therefore, the balanced reaction is as follows.

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

(c)

Interpretation Introduction

Interpretation:

The balanced equation for the reaction has to be given.

  ${\text{ClO}}_3{}^{-}\left(\text{aq}\right){\text{ +CN}}^{-}\left(\text{aq}\right)\to {\text{Cl}}^{-}\text{(aq)}{\text{+CNO}}^{-}\left(\text{aq}\right)$

Concept introduction:

Refer to part (a).

Explanation of Solution

The given reaction is as follows:

  ${\text{ClO}}_3{}^{-}\left(\text{aq}\right){\text{ +CN}}^{-}\left(\text{aq}\right)\to {\text{Cl}}^{-}\text{(aq)}{\text{+CNO}}^{-}\left(\text{aq}\right)$

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

  $\begin{array}{l}\begin{array}{l}{\text{CN}}^{-}\to {\text{CNO}}^{-}\\ \end{array}\hfill \\ {\text{ClO}}_3^{-}\to {\text{Cl}}^{-}\hfill \end{array}$

Steps for balancing half –reactions in BASIC solution:

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

$\begin{array}{l}\begin{array}{l}{\text{CN}}^{-}\to {\text{CNO}}^{-}\\ \end{array}\hfill \\ {\text{ClO}}_3^{-}\to {\text{Cl}}^{-}\hfill \end{array}$

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

$\begin{array}{l}\begin{array}{l}{\text{CN}}^{-}+{\text{H}}_2\text{O}\to {\text{CNO}}^{-}\\ \end{array}\hfill \\ {\text{ClO}}_3^{-}\to {\text{Cl}}^{-}+3{\text{H}}_2\text{O}\hfill \end{array}$

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

   $\begin{array}{l}\begin{array}{l}{\text{CN}}^{-}+{\text{H}}_2\text{O}\to {\text{CNO}}^{-}+2{\text{H}}^{+}\\ \end{array}\hfill \\ {\text{ClO}}_3^{-}+6{\text{H}}^{+}\to {\text{Cl}}^{-}+3{\text{H}}_2\text{O}\hfill \end{array}$

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

$\begin{array}{l}\begin{array}{l}{\text{CN}}^{-}+{\text{H}}_2\text{O}+2{\text{OH}}^{-}\to {\text{CNO}}^{-}+2{\text{H}}_2\text{O}\\ \end{array}\hfill \\ {\text{ClO}}_3^{-}+6{\text{H}}_2\text{O}\to {\text{Cl}}^{-}+3{\text{H}}_2\text{O}+6{\text{OH}}^{-}\hfill \end{array}$

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

$\begin{array}{l}\begin{array}{l}{\text{CN}}^{-}+{\text{H}}_2\text{O}+2{\text{OH}}^{-}\to {\text{CNO}}^{-}+2{\text{H}}_2\text{O}+2{\text{e}}^{-}\\ \end{array}\hfill \\ {\text{ClO}}_3^{-}+6{\text{H}}_2\text{O}+6{\text{e}}^{-}\to {\text{Cl}}^{-}+3{\text{H}}_2\text{O}+6{\text{OH}}^{-}\hfill \end{array}$

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

$\begin{array}{l}\begin{array}{l}3{\text{CN}}^{-}+3{\text{H}}_2\text{O}+6{\text{OH}}^{-}\to 3{\text{CNO}}^{-}+6{\text{H}}_2\text{O}+6{\text{e}}^{-}\\ \end{array}\hfill \\ {\text{ClO}}_3^{-}+6{\text{H}}_2\text{O}+6{\text{e}}^{-}\to {\text{Cl}}^{-}+3{\text{H}}_2\text{O}+6{\text{OH}}^{-}\hfill \end{array}$

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:

   $3{\text{CN}}^{-}+{\text{ClO}}_3^{-}\rightleftarrows 3{\text{CNO}}^{-}+{\text{Cl}}^{-}$

Therefore, the balanced reaction is as follows.

  $3{\text{CN}}^{-}+{\text{ClO}}_3^{-}\rightleftarrows 3{\text{CNO}}^{-}+{\text{Cl}}^{-}$