Chapter 18, Problem 18.73QE

a)

Interpretation Introduction

Interpretation:

Oxidation and reduction half-reactions have to be given for the reaction shown below:

  ${\text{C}}_3{\text{H}}_8\text{(g) +5}{\text{O}}_{\text{2}}\left(\text{g}\right)\to 3{\text{CO}}_{\text{2}}\text{(g)}\text{+4}{\text{H}}_2\text{O(l)}$

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:

  ${\text{C}}_3{\text{H}}_8\text{(g) +5}{\text{O}}_{\text{2}}\left(\text{g}\right)\to 3{\text{CO}}_{\text{2}}\text{(g)}\text{+4}{\text{H}}_2\text{O(l)}$

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

  $\begin{array}{l}{\text{C}}_3{\text{H}}_8\to {\text{CO}}_{\text{2}}\\ \\ {\text{O}}_{\text{2}}\to {\text{H}}_2\text{O}\end{array}$

Steps for balancing half –reactions in BASIC solution:

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

   $\begin{array}{l}{\text{C}}_3{\text{H}}_8\to 3{\text{CO}}_{\text{2}}\\ \\ {\text{O}}_{\text{2}}\to {\text{H}}_2\text{O}\end{array}$

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

   $\begin{array}{l}{\text{C}}_3{\text{H}}_8+6{\text{H}}_{\text{2}}\text{O}\to 3{\text{CO}}_{\text{2}}\\ \\ {\text{O}}_{\text{2}}\to 2{\text{H}}_2\text{O}\end{array}$

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

   $\begin{array}{l}{\text{C}}_3{\text{H}}_8+6{\text{H}}_{\text{2}}\text{O}\to 3{\text{CO}}_{\text{2}}+20{\text{H}}^{\text{+}}\\ \\ {\text{O}}_{\text{2}}+4{\text{H}}^{\text{+}}\to 2{\text{H}}_2\text{O}\end{array}$

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

$\begin{array}{l}{\text{C}}_3{\text{H}}_8+6{\text{H}}_{\text{2}}\text{O+20}{\text{OH}}^{-}\to 3{\text{CO}}_{\text{2}}+20{\text{H}}_{\text{2}}\text{O}\\ \\ {\text{O}}_{\text{2}}+4{\text{H}}_2\text{O}\to 2{\text{H}}_2\text{O+4}{\text{OH}}^{-}\end{array}$

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

$\begin{array}{l}{\text{C}}_3{\text{H}}_8+6{\text{H}}_{\text{2}}\text{O+20}{\text{OH}}^{-}\to 3{\text{CO}}_{\text{2}}+20{\text{H}}_{\text{2}}\text{O+20}{\text{e}}^{-}\\ \\ {\text{O}}_{\text{2}}+4{\text{H}}_2\text{O}\text{+4}{\text{e}}^{-}\to 2{\text{H}}_2\text{O+4}{\text{OH}}^{-}\end{array}$

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

$\begin{array}{l}{\text{C}}_3{\text{H}}_8+6{\text{H}}_{\text{2}}\text{O+20}{\text{OH}}^{-}\to 3{\text{CO}}_{\text{2}}+20{\text{H}}_{\text{2}}\text{O+20}{\text{e}}^{-}\\ \\ {\text{5O}}_{\text{2}}+20{\text{H}}_2\text{O}\text{+20}{\text{e}}^{-}\to 10{\text{H}}_2\text{O+20}{\text{OH}}^{-}\end{array}$

The half-cell reaction can be written as below:

  $\begin{array}{l}{\text{Oxidation: C}}_3{\text{H}}_8\text{+20}{\text{OH}}^{-}\to 3{\text{CO}}_{\text{2}}+14{\text{H}}_{\text{2}}\text{O+20}{\text{e}}^{-}\\ \\ \text{Reduction: }{\text{5O}}_{\text{2}}+20{\text{H}}_2\text{O}\text{+20}{\text{e}}^{-}\to \text{20}{\text{OH}}^{-}\end{array}$

b)

Interpretation Introduction

Interpretation:

The potential of the cell has to be determined from standard free energy change.

Explanation of Solution

The half-cell reaction can be written as below:

  $\begin{array}{l}{\text{Oxidation: C}}_3{\text{H}}_8\text{+20}{\text{OH}}^{-}\to 3{\text{CO}}_{\text{2}}+14{\text{H}}_{\text{2}}\text{O+20}{\text{e}}^{-}\\ \\ \text{Reduction: }{\text{5O}}_{\text{2}}+20{\text{H}}_2\text{O}\text{+20}{\text{e}}^{-}\to \text{20}{\text{OH}}^{-}\end{array}$

The overall reaction is given below:

  ${\text{C}}_3{\text{H}}_8\text{(g) +5}{\text{O}}_{\text{2}}\left(\text{g}\right)\to 3{\text{CO}}_{\text{2}}\text{(g)}\text{+4}{\text{H}}_2\text{O(l)}$

The standard free energy change for the reaction is,

$\begin{array}{c}{\text{C}}_{\text{3}}{\text{H}}_{\text{8}}\text{(g) +5}{\text{O}}_{\text{2}}\left(\text{g}\right)\to \text{3}{\text{CO}}_{\text{2}}\text{(g)}\text{+4}{\text{H}}_{\text{2}}\text{O(l)}\\ \\ {\text{ΔG}}_{\text{r}}{}^{\text{°}}\text{ = }\sum {\text{ΔG}}^{\text{°}}{}_{\text{products}}-\sum {\text{G}}^{\text{°}}{}_{\text{reactants}}\\ \text{=}\left({\text{3ΔG}}^{\text{°}}{}_{\text{f}}{\text{CO}}_{\text{2}}\text{(g)}\text{+4}{\text{ΔG}}^{\text{°}}{}_{\text{f}}{\text{H}}_{\text{2}}\text{O(l)}\right)\text{-}\left({\text{ΔG}}^{\text{°}}{}_{\text{f}}{\text{C}}_{\text{3}}{\text{H}}_{\text{8}}\text{(g)}\text{+5}{\text{ΔG}}^{\text{°}}{}_{\text{f}}{\text{O}}_{\text{2}}\left(\text{g}\right)\right)\\ \text{=}\left(\text{3×-394}\text{.36kJ/mol}\text{+4×-237}\text{.18}\text{kJ/mol}\right)\text{-}\left(\text{-23}\text{.49+5×0kJ/mol}\right)\\ \text{=}-\text{2108}\text{.31}\text{kJ}\end{array}$

${\text{E}}_{\text{cell}}^{\text{0}}$ can be calculated:

  $\begin{array}{c}{\text{ΔG}}^{\text{0}}\text{= }-{\text{nFE}}^{\text{0}}\\ {\text{E}}^{\text{0}}=\frac{{\text{ΔG}}^{\text{0}}}{-\text{nF}}\\ \\ =\frac{-2108.31\times {10}^3\text{J}}{-20\text{mol}{\text{e}}^{\text{-}}\times \text{96500C/mol}{\text{e}}^{\text{-}}}\\ \text{=}\\ \text{=}1.09\text{V}\end{array}$

${\text{E}}_{\text{cell}}^{\text{0}}$ is calculated to be $1.09\text{V}$.

c)

Interpretation Introduction

Interpretation:

Number of kilojoules of electrical energy produced when one gram of propane is consumed has to be determined.

Explanation of Solution

The standard free energy change for the reaction is,

$\begin{array}{c}{\text{C}}_{\text{3}}{\text{H}}_{\text{8}}\text{(g) +5}{\text{O}}_{\text{2}}\left(\text{g}\right)\to \text{3}{\text{CO}}_{\text{2}}\text{(g)}\text{+4}{\text{H}}_{\text{2}}\text{O(l)}\\ \\ {\text{ΔG}}_{\text{r}}{}^{\text{°}}\text{ = }\sum {\text{ΔG}}^{\text{°}}{}_{\text{products}}-\sum {\text{G}}^{\text{°}}{}_{\text{reactants}}\\ \text{=}\left({\text{3ΔG}}^{\text{°}}{}_{\text{f}}{\text{CO}}_{\text{2}}\text{(g)}\text{+4}{\text{ΔG}}^{\text{°}}{}_{\text{f}}{\text{H}}_{\text{2}}\text{O(l)}\right)\text{-}\left({\text{ΔG}}^{\text{°}}{}_{\text{f}}{\text{C}}_{\text{3}}{\text{H}}_{\text{8}}\text{(g)}\text{+5}{\text{ΔG}}^{\text{°}}{}_{\text{f}}{\text{O}}_{\text{2}}\left(\text{g}\right)\right)\\ \text{=}\left(\text{3×-394}\text{.36kJ/mol}\text{+4×-237}\text{.18}\text{kJ/mol}\right)\text{-}\left(\text{-23}\text{.49+5×0kJ/mol}\right)\\ \text{=}-\text{2108}\text{.31}\text{kJ}\end{array}$

Number of kilojoules of electrical energy produced when one gram of propane is consumed can be calculated as follows:

  $\text{1}\text{g}{\text{C}}_{\text{3}}{\text{H}}_{\text{8}}\text{×}\frac{\text{1}\text{mol}{\text{C}}_{\text{3}}{\text{H}}_{\text{8}}}{\text{44}\text{g}{\text{C}}_{\text{3}}{\text{H}}_{\text{8}}}\text{×}\frac{-\text{2108}\text{.31}\text{kJ}}{\text{1}\text{mol}{\text{C}}_{\text{3}}{\text{H}}_{\text{8}}}\text{=}-\text{47}\text{.9}\text{kJ=}-{\text{5×10}}^{\text{1}}\text{kJ}$

Number of kilojoules of electrical energy produced when one gram of propane is consumed is $-{\text{5×10}}^{\text{1}}\text{kJ}$.