Chapter 19, Problem 24QP

Calculate $△G°$ and $K_c$ for the following reactions at $25°\text{C:}$

$\begin{array}{l}(a)\\ \text{Mg}(s)+p{b}^{2+}(aq)\rightleftarrows {\text{Mg}}^{\text{2+}}(aq)+pb(s)\end{array}$

$\begin{array}{l}(b)\\ O_2(g)+4H^{+}(aq)+4F{e}^{2+}(aq)\rightleftarrows 2H_{2}O(I)+4F{e}^{3+}(aq)\end{array}$

$\begin{array}{l}(c)\\ 2A1(s)+{31}_2(s)\rightleftarrows 2A{1}^{3+}(aq)+{61}^{-}(aq)\end{array}$

Interpretation Introduction

Interpretation:

The equilibrium constant $\left(K\right)$ and the Gibbs free energy $\left(\Delta G^o\right)$ for each of the given chemical reactions areto be calculated.

Concept introduction:

The standard cell potential of a particular cell is given as:

$E^0{}_{cell}=E^0{}_{cathode}-E^0{}_{anode}$

Here, ${\text{E}}^{\text{0}}{}_{\text{cathode}}{\text{and E}}^{\text{0}}{}_{\text{anode}}$ is the standard reduction potential occurring at the cathode and anode, respectively.

The relation between the cell potential $\left(E^0{}_{cell}\right)$ and the equilibrium constant $\left(K\right)$ is given as:

$E^0{}_{cell}\text{ = }\frac{0.0592}{n}\times \log \text{K}$

The relation between standard Gibbs free energy $\left(\Delta G^o\right)$ and the standard cell potential $\left(E^0{}_{cell}\right)$ is given as:

$\Delta G^0=-n\text{F}{E}^o{}_{cell}$

Answer to Problem 24QP

Solution:

(a) $\Delta G^o=-\text{432 kJ/mol and K = 5}\times {\text{10}}^{\text{75}}$

(b) $\Delta G^o=-\text{178 kJ/mol and K}=\text{1}\times {\text{10}}^{31}$

(c) $\Delta G^o=-\text{1}\text{.27 }\times {\text{ 10}}^{\text{3}}\text{ kJ/mol and K = 9}\times {\text{10}}^{221}$

Explanation of Solution

a) $\text{Mg}(s){\text{ + Pb}}^{\text{2+}}\text{(}aq\text{) }\rightleftharpoons {\text{Mg}}^{\text{2+}}\text{(}aq\text{) + Pb (}s\text{) }$

This reaction can be written astwo half-reactions, as follows:

$\begin{array}{l}\text{At cathode (reduction):}\\ {\text{Pb}}^{2+}(aq)+{\text{2e}}^{-}\to \text{Pb}(s)\end{array}$

$\begin{array}{l}\text{At anode(oxidation):}\\ \text{Mg (}s\text{) }\to {\text{ Mg}}^{\text{2+}}(aq)+{\text{2e}}^{-}\end{array}$

Fromtable $19.1$, the reduction potential of lead is $-\text{0}\text{.13 V}$ and the reduction potential of magnesium is $-\text{2}\text{.37 V}$.

Now, the standard cell potential of the cell is given as:

$E^0{}_{cell}=E^0{}_{cathode}-E^0{}_{anode}$

Substitute the values of half-cell potential in the above expression.

$\begin{array}{c}{E}^o{}_{cell}\text{=}\left(-\text{0}\text{.13 V }-\text{(}-\text{2}\text{.37 V)}\right)\\ \text{ = 2}\text{.24 V}\end{array}$

The relation between standard Gibbs free energy $\left(\Delta G^o\right)$ and the standard cell potential $\left(E^0{}_{cell}\right)$ is as follows:

$\Delta G^{\text{o}}=-n\text{F}{E}^o{}_{cell}$

Here, the number of exchange of electrons is two;

Thus, the value of $n\text{ = 2}$.

Substitute the values of Faraday constant, the number of electrons transferred in the half-cell and the standard cell potential.

$\begin{array}{c}\Delta G^o=\left(-2\times 96500\frac{\text{J}}{\text{V}\text{.mol}}\times \text{ 2}\text{.24 V}\right)\\ =-\text{432320 J/mol}\\ =-\text{432 kJ/mol}\end{array}$

The relation between the cell potential $\left(E^0{}_{cell}\right)$ and the equilibrium constant $\left(K\right)$ is given as follows:

$\begin{array}{l}{E}^0{}_{cell}\text{ = }\frac{0.0592}{n}\times \log \text{K}\\ \Rightarrow {\text{ K = 10}}^{\frac{n{E}^o{}_{cell}}{0.0592}V}\end{array}$

Substitute the values $n$ and $E^0{}_{cell}$ to the above expression,

$\begin{array}{c}{\text{K = 10}}^{\frac{\text{2}\times \text{ 2}\text{.24 V}}{0.0592}}\\ {\text{K = 10}}^{75.7}\\ \text{K}=\text{ 5 }\times {\text{ 10}}^{75}\end{array}$

Thus, the equilibrium constant of this reaction is $\text{5 }\times {\text{ 10}}^{75}$.

b) ${\text{O}}_{\text{2}}(g)+{\text{4H}}^{\text{+}}\text{(}aq\text{)}+{\text{4Fe}}^{\text{2+}}\text{(}aq\text{)}\rightleftharpoons {\text{2H}}_{\text{2}}\text{O(}l\text{)}+{\text{4Fe}}^{\text{3+}}\text{(}aq\text{) }$

This reaction can be written astwo half-reactions, as follows:

$\begin{array}{l}\text{At cathode (reduction):}\\ {\text{O}}_{\text{2}}(g){\text{ + 4H}}^{\text{+}}\text{(}aq{\text{) + 4e}}^{-}\to 2{\text{H}}_{\text{2}}\text{O }(l)\end{array}$

$\begin{array}{l}\text{At anode(oxidation):}\\ {\text{4Fe}}^{\text{2+}}\text{(}aq\text{)}\to {\text{ 4Fe}}^{\text{3+}}\text{(}aq{\text{) + 4e}}^{-}\end{array}$

Fromtable $19.1$, the reduction potential of oxygen gas is $\text{1}\text{.23 V}$ and the reduction potential of iron is $\text{0}\text{.77 V}$.

The standard cell potential of a cell is given as follows:

$E^0{}_{cell}=E^0{}_{cathode}-E^0{}_{anode}$

Substitute the values of half-cell potential in the above expression:

$\begin{array}{c}{E}^o{}_{cell}=\text{1}\text{.23 V }-\text{0}\text{.77 V}\\ =\text{ 0}\text{.46 V}\end{array}$

The relation between standard Gibbs free energy $\left(\Delta G^o\right)$ and the standard cell potential $\left(E^0{}_{cell}\right)$ is given as:

$\Delta G^0=-n\text{F}{E}^o{}_{cell}$

Here, the number of exchange of electrons is four;

Thus, the value of $n\text{ = 4}$.

The Gibb’s energy is calculated by the expression as follows:

$\begin{array}{c}\Delta G^o=\left(-4\times 96500\frac{\text{J}}{\text{V}\text{.mol}}\times \text{ 0}\text{.46 V}\right)\\ =-\text{177560 J/mol}\\ =-178\text{ kJ/mol}\end{array}$

The relation between the cell potential $\left(E^0{}_{cell}\right)$ and the equilibrium constant $\left(K\right)$ is given as follows:

$\begin{array}{c}{E}^0{}_{cell}\text{ = }\frac{0.0592}{n}\times \log \text{K}\\ {\text{ K = 10}}^{\frac{n{E}^o{}_{cell}}{0.0592}V}\end{array}$

Substitute the values of $n$ and $E_{cell}^0$ in the above expression,

$\begin{array}{c}\text{K }={\text{10}}^{\frac{4\times \text{ 0}\text{.46 V}}{0.0592}}\\ \text{K }={\text{10}}^{31.1}\\ \text{K}=\text{1 }\times {\text{ 10}}^{31}\end{array}$

Thus, the equilibrium constant of this reaction is $\text{1 }\times {\text{ 10}}^{31}$.

c) $\text{2Al}(s)+{\text{3I}}_{\text{2}}\text{(}s\text{) }\rightleftharpoons {\text{2Al}}^{\text{3+}}\text{(}aq\text{)}+{\text{6I}}^{-}\text{(}aq\text{) }$

This reaction can be written astwo half-reactions as follows:

$\begin{array}{l}\text{At cathode (reduction):}\\ 3{\text{I}}_2(s){\text{ + 6e}}^{-}\to 6{\text{I}}^{-}(aq)\end{array}$

$\begin{array}{l}\text{At anode(oxidation):}\\ \text{2Al (}s\text{) }\to {\text{ 2Al}}^{3+}(aq){\text{ + 6e}}^{-}\end{array}$

Fromtable $19.1$, the reduction potential of aluminum is $-\text{1}\text{.66 V}$ and the reduction potential of iodine is $\text{0}\text{.53 V}$.

The standard cell potential of a cell is given as follows:

$E^0{}_{cell}=E^0{}_{cathode}-E^0{}_{anode}$

Substitute the values of half-cell potential in the above expression,

$\begin{array}{c}{E}^o{}_{cell}\text{ = 0}\text{.53 V }-\text{ (}-1.66\text{ V)}\\ \text{ = 2}\text{.19 V}\end{array}$

The relation between standard Gibbs free energy $\left(\Delta G^o\right)$ and the standard cell potential $\left(E^0{}_{cell}\right)$ is given as follows:

$\Delta G^0=-n\text{F}{E}^o{}_{cell}$

Here, number of exchange of electrons is six;

Thus, the value of $n\text{ = 6}$.

Substitute the values of $n$ and $E_{cell}^0$ in the above expression.

$\begin{array}{c}\Delta G^o=\left(-6\times 96500\frac{\text{J}}{\text{V}\text{.mol}}\times \text{ 2}\text{.19 V}\right)\\ =-\text{1268010 J/mol}\\ =-\text{1}\text{.27 }\times {\text{ 10}}^{\text{3}}\text{ kJ/mol}\end{array}$

The relation between the cell potential $\left(E^0{}_{cell}\right)$ and the equilibrium constant $\left(K\right)$ is given as follows:

$\begin{array}{c}{E}^0{}_{cell}=\left(\frac{0.0592}{n}\times \log \text{K}\right)\\ \text{K}={\text{10}}^{\frac{n{E}^o{}_{cell}}{0.0592}V}\end{array}$

Substitute the values of $n$ and $E_{cell}^0$ in the above expression,

$\begin{array}{c}\text{K}={\text{10}}^{\frac{6\times \text{ 2}\text{.19 V}}{0.0592}}\\ \text{K}={\text{10}}^{221.96}\\ \text{K}=\text{9 }\times {\text{ 10}}^{221}\end{array}$

Thus, the equilibrium constant of this reaction is $\text{9 }\times {\text{ 10}}^{221}$.