Chapter 18, Problem 18.23QE

(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.

  $\text{Na}+{\text{FeCl}}_3\to \text{Fe}+\text{NaCl}$

Explanation of Solution

The given reaction is as follows:

  $\text{Na}+{\text{FeCl}}_3\to \text{Fe}+\text{NaCl}$

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

  $\begin{array}{l}\begin{array}{l}\text{Na}\to \text{NaCl}\\ \end{array}\hfill \\ {\text{FeCl}}_3\to \text{Fe}\hfill \end{array}$

Steps for balancing half –reactions are given below:

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

   $\begin{array}{l}\begin{array}{l}3\text{Na}+{\text{FeCl}}_3\to 3\text{NaCl}+\text{Fe}\\ \end{array}\hfill \\ {\text{FeCl}}_3+3\text{Na}\to \text{Fe}+3\text{NaCl}\hfill \end{array}$

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

   $\begin{array}{l}\begin{array}{l}3\text{Na}+{\text{FeCl}}_3\to 3\text{NaCl}+\text{Fe}\\ \end{array}\hfill \\ {\text{FeCl}}_3+3\text{Na}\to \text{Fe}+3\text{NaCl}\hfill \end{array}$

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

   $\begin{array}{l}\begin{array}{l}3\text{Na}+{\text{FeCl}}_3\to 3\text{NaCl}+\text{Fe}\\ \end{array}\hfill \\ {\text{FeCl}}_3+3\text{Na}\to \text{Fe}+3\text{NaCl}\hfill \end{array}$

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

$\begin{array}{l}\begin{array}{l}3\text{Na}+{\text{FeCl}}_3\to 3\text{NaCl}+\text{Fe}\\ \end{array}\hfill \\ {\text{FeCl}}_3+3\text{Na}\to \text{Fe}+3\text{NaCl}\hfill \end{array}$

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

   $\begin{array}{l}\begin{array}{l}3\text{Na}+{\text{FeCl}}_3\to 3\text{NaCl}+\text{Fe}\\ \end{array}\hfill \\ {\text{FeCl}}_3+3\text{Na}\to \text{Fe}+3\text{NaCl}\hfill \end{array}$

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:

   $2{\text{FeCl}}_3+6\text{Na}\to 2\text{Fe}+6\text{NaCl}$

Therefore, the balanced reaction is as follows.

${\text{FeCl}}_3+3\text{Na}\to \text{Fe}+3\text{NaCl}$

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

$\stackrel{0}{3\text{Na}}+\stackrel{}{\stackrel{+3}{\text{Fe}}{\text{Cl}}_3}\to \stackrel{0}{\text{Fe}}+3\stackrel{}{\stackrel{+1}{\text{Na}}\text{Cl}}$

Oxidation state of sodium increases in the reaction and thus it is oxidised. Oxidation state of iron 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.

  ${\text{SnCl}}_2+{\text{FeCl}}_3\to {\text{SnCl}}_4+{\text{FeCl}}_2$

Explanation of Solution

The given reaction is as follows:

  ${\text{SnCl}}_2+{\text{FeCl}}_3\to {\text{SnCl}}_4+{\text{FeCl}}_2$

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

  $\begin{array}{l}{\text{SnCl}}_2\to {\text{SnCl}}_4\\ \\ {\text{FeCl}}_3\to {\text{FeCl}}_2\end{array}$

Steps for balancing half –reactions are given below:

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

   $\begin{array}{l}\begin{array}{l}{\text{SnCl}}_2+2{\text{FeCl}}_3\to {\text{SnCl}}_4+2{\text{FeCl}}_2\\ \end{array}\hfill \\ 2{\text{FeCl}}_3+{\text{SnCl}}_2\to 2{\text{FeCl}}_2+{\text{SnCl}}_4\hfill \end{array}$

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

   $\begin{array}{l}\begin{array}{l}{\text{SnCl}}_2+2{\text{FeCl}}_3\to {\text{SnCl}}_4+2{\text{FeCl}}_2\\ \end{array}\hfill \\ 2{\text{FeCl}}_3+{\text{SnCl}}_2\to 2{\text{FeCl}}_2+{\text{SnCl}}_4\hfill \end{array}$

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

   $\begin{array}{l}\begin{array}{l}{\text{SnCl}}_2+2{\text{FeCl}}_3\to {\text{SnCl}}_4+2{\text{FeCl}}_2\\ \end{array}\hfill \\ 2{\text{FeCl}}_3+{\text{SnCl}}_2\to 2{\text{FeCl}}_2+{\text{SnCl}}_4\hfill \end{array}$

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

$\begin{array}{l}\begin{array}{l}{\text{SnCl}}_2+2{\text{FeCl}}_3\to {\text{SnCl}}_4+2{\text{FeCl}}_2\\ \end{array}\hfill \\ 2{\text{FeCl}}_3+{\text{SnCl}}_2\to 2{\text{FeCl}}_2+{\text{SnCl}}_4\hfill \end{array}$

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

   $\begin{array}{l}\begin{array}{l}{\text{SnCl}}_2+2{\text{FeCl}}_3\to {\text{SnCl}}_4+2{\text{FeCl}}_2\\ \end{array}\hfill \\ 2{\text{FeCl}}_3+{\text{SnCl}}_2\to 2{\text{FeCl}}_2+{\text{SnCl}}_4\hfill \end{array}$

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:

   $4{\text{FeCl}}_3+2{\text{SnCl}}_2\to 4{\text{FeCl}}_2+2{\text{SnCl}}_4$

Therefore, the balanced reaction is as follows.

$2{\text{FeCl}}_3+{\text{SnCl}}_2\to 2{\text{FeCl}}_2+{\text{SnCl}}_4$

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

$2\stackrel{+3}{\text{Fe}}{\text{Cl}}_3+\stackrel{+2}{\text{Sn}}{\text{Cl}}_2\to 2\stackrel{+2}{\text{Fe}}{\text{Cl}}_2+\stackrel{+4}{\text{Sn}}{\text{Cl}}_4$

Oxidation state of tin increases in the reaction and thus it is oxidised. Oxidation state of iron 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.

  $\text{CO}+{\text{Cr}}_2{\text{O}}_3\to \text{Cr}+{\text{CO}}_2$

Explanation of Solution

The given reaction is as follows:

  $\text{CO}+{\text{Cr}}_2{\text{O}}_3\to \text{Cr}+{\text{CO}}_2$

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

  $\begin{array}{l}\begin{array}{l}\text{CO}\to {\text{CO}}_2\\ \end{array}\hfill \\ {\text{Cr}}_2{\text{O}}_3\to \text{Cr}\hfill \end{array}$

Steps for balancing half –reactions are given below:

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

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

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

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

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

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

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

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

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

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

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:

   $\text{3CO}+{\text{Cr}}_2{\text{O}}_3\to 2\text{Cr}+3{\text{CO}}_2$

Therefore, the balanced reaction is as follows.

$\text{3CO}+{\text{Cr}}_2{\text{O}}_3\to 2\text{Cr}+3{\text{CO}}_2$

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

$\text{3}\stackrel{+2}{\text{C}}\text{O}+{\stackrel{+3}{\text{Cr}}}_2{\text{O}}_3\to 2\stackrel{0}{\text{Cr}}+3\stackrel{+4}{\text{C}}{\text{O}}_2$

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