Chapter 18, Problem 18.21QE

(a)

Interpretation Introduction

Interpretation:

The balanced equation for the reaction has to be given and also has to specify the species which are oxidised and reduced.

  ${\text{H}}_2+{\text{O}}_2\to {\text{H}}_2\text{O}$

Explanation of Solution

The given reaction is as follows:

  ${\text{H}}_2+{\text{O}}_2\to {\text{H}}_2\text{O}$

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

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

Steps for balancing half –reactions are given below:

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

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

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

   $\begin{array}{l}\begin{array}{l}{\text{H}}_2\to {\text{H}}_2\text{O}\\ \end{array}\hfill \\ {\text{O}}_2\to 2{\text{H}}_2\text{O}\hfill \end{array}$

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

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

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

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

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

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

2. 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:

   $2{\text{H}}_2+{\text{O}}_2\rightleftarrows 2{\text{H}}_2\text{O}$

Therefore, the balanced reaction is as follows.

$2{\text{H}}_2+{\text{O}}_2\rightleftarrows 2{\text{H}}_2\text{O}$

The change in oxidation state of reactants and products are given below:

$\stackrel{0}{2{\text{H}}_2}+\stackrel{0}{{\text{O}}_2}\rightleftarrows 2\stackrel{+1}{{\text{H}}_2}\stackrel{-2}{\text{O}}$

Oxidation state of hydrogen increases in the reaction and thus it is oxidised. Oxidation state of oxygen decreases in the reaction and thus it is reduced.

(b)

Interpretation Introduction

Interpretation:

The balanced equation for the reaction has to be given and also has to specify the species which are oxidised and reduced.

  $\text{Fe}+{\text{O}}_2\to {\text{Fe}}_2{\text{O}}_3$

Explanation of Solution

The given reaction is as follows:

  $\text{Fe}+{\text{O}}_2\to {\text{Fe}}_2{\text{O}}_3$

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

  $\begin{array}{l}\begin{array}{l}\text{Fe}\to {\text{Fe}}_2{\text{O}}_3\\ \end{array}\hfill \\ {\text{O}}_2\to {\text{H}}_2\text{O}\hfill \end{array}$

Steps for balancing half –reactions are given below:

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

   $\begin{array}{l}\begin{array}{l}2\text{Fe}\to {\text{Fe}}_2{\text{O}}_3\\ \end{array}\hfill \\ {\text{O}}_2\to 2{\text{H}}_2\text{O}\hfill \end{array}$

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

   $\begin{array}{l}\begin{array}{l}2\text{Fe}+3{\text{H}}_2\text{O}\to {\text{Fe}}_2{\text{O}}_3\\ \end{array}\hfill \\ {\text{O}}_2\to 2{\text{H}}_2\text{O}\hfill \end{array}$

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

   $\begin{array}{l}\begin{array}{l}2\text{Fe}+3{\text{H}}_2\text{O}\to {\text{Fe}}_2{\text{O}}_3+6{\text{H}}^{+}\\ \end{array}\hfill \\ {\text{O}}_2+4{\text{H}}^{+}\to 2{\text{H}}_2\text{O}\hfill \end{array}$

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

$\begin{array}{l}\begin{array}{l}2\text{Fe}+3{\text{H}}_2\text{O}\to {\text{Fe}}_2{\text{O}}_3+6{\text{H}}^{+}+6{\text{e}}^{-}\\ \end{array}\hfill \\ {\text{O}}_2+4{\text{H}}^{+}+4{\text{e}}^{-}\to 2{\text{H}}_2\text{O}\hfill \end{array}$

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

   $\begin{array}{l}\begin{array}{l}4\text{Fe}+6{\text{H}}_2\text{O}\to 2{\text{Fe}}_2{\text{O}}_3+12{\text{H}}^{+}+12{\text{e}}^{-}\\ \end{array}\hfill \\ 3{\text{O}}_2+12{\text{H}}^{+}+12{\text{e}}^{-}\to 6{\text{H}}_2\text{O}\hfill \end{array}$

2. 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:

   $4\text{Fe}+3{\text{O}}_2\rightleftarrows 2{\text{Fe}}_2{\text{O}}_3$

Therefore, the balanced reaction is as follows.

$4\text{Fe}+3{\text{O}}_2\rightleftarrows 2{\text{Fe}}_2{\text{O}}_3$

The change in oxidation state of reactants and products are given below:

$\stackrel{0}{4\text{Fe}}+\stackrel{0}{3{\text{O}}_2}\rightleftarrows 2{\stackrel{+3}{\text{Fe}}}_2\stackrel{-2}{{\text{O}}_3}$

Oxidation state of iron increases in the reaction and thus it is oxidised. Oxidation state of oxygen decreases in the reaction and thus it is reduced.

(c)

Interpretation Introduction

Interpretation:

The balanced equation for the reaction has to be given and also has to specify the species which are oxidised and reduced.

  ${\text{Al}}_2{\text{O}}_3+\text{C}\to \text{Al}+{\text{CO}}_2$

Explanation of Solution

The given reaction is as follows:

  ${\text{Al}}_2{\text{O}}_3+\text{C}\to \text{Al}+{\text{CO}}_2$

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

  $\begin{array}{l}\begin{array}{l}\text{C}\to {\text{CO}}_2\\ \end{array}\hfill \\ {\text{Al}}_{\text{2}}{\text{O}}_3\to \text{Al}\hfill \end{array}$

Steps for balancing half –reactions are given below:

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

   $\begin{array}{l}\begin{array}{l}\text{C}\to {\text{CO}}_2\\ \end{array}\hfill \\ {\text{Al}}_{\text{2}}{\text{O}}_3\to 2\text{Al}\hfill \end{array}$

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

   $\begin{array}{l}\begin{array}{l}\text{C}+2{\text{H}}_2\text{O}\to {\text{CO}}_2\\ \end{array}\hfill \\ {\text{Al}}_2{\text{O}}_3\to 2\text{Al}+3{\text{H}}_2\text{O}\hfill \end{array}$

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

   $\begin{array}{l}\begin{array}{l}\text{C}+2{\text{H}}_2\text{O}\to {\text{CO}}_2+4{\text{H}}^{+}\\ \end{array}\hfill \\ {\text{Al}}_2{\text{O}}_3+6{\text{H}}^{+}\to 2\text{Al}+3{\text{H}}_2\text{O}\hfill \end{array}$

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

$\begin{array}{l}\begin{array}{l}\text{C}+2{\text{H}}_2\text{O}\to {\text{CO}}_2+4{\text{H}}^{+}+4{\text{e}}^{-}\\ \end{array}\hfill \\ {\text{Al}}_2{\text{O}}_3+6{\text{H}}^{+}+6{\text{e}}^{-}\to 2\text{Al}+3{\text{H}}_2\text{O}\hfill \end{array}$

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

   $\begin{array}{l}\begin{array}{l}3\text{C}+6{\text{H}}_2\text{O}\to 3{\text{CO}}_2+12{\text{H}}^{+}+12{\text{e}}^{-}\\ \end{array}\hfill \\ 2{\text{Al}}_2{\text{O}}_3+12{\text{H}}^{+}+12{\text{e}}^{-}\to 4\text{Al}+6{\text{H}}_2\text{O}\hfill \end{array}$

2. 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:

   $3\text{C}+2{\text{Al}}_2{\text{O}}_3\rightleftarrows 3{\text{CO}}_2+4\text{Al}$

Therefore, the balanced reaction is as follows.

$3\text{C}+2{\text{Al}}_2{\text{O}}_3\rightleftarrows 3{\text{CO}}_2+4\text{Al}$

The change in oxidation state of reactants and products are given below:

$3\stackrel{0}{\text{C}}+2\stackrel{+3}{{\text{Al}}_2}{\stackrel{-2}{\text{O}}}_3\rightleftarrows 3\stackrel{+4}{\text{C}}{\stackrel{-2}{\text{O}}}_2+4\stackrel{0}{\text{Al}}$

Oxidation state of carbon increases in the reaction and thus it is oxidised. Oxidation state of aluminium decreases in the reaction and thus it is reduced.