Chapter 18, Problem 75E

Calculate ΔG° and K at 25°C for the reactions in Exercises 37 and 41.

Interpretation Introduction

Interpretation:

The equations for different chemical reactions are given. The value of $\Delta G°$ and $K$ for different reactions is to be calculated.

Concept introduction:

The relationship between reduction potential and standard reduction potential value and activities of species present in an electrochemical cell at a given temperature is given by the Nernst equation.

The value of $E_{\text{cell}}$ is calculated using Nernst formula,

$E=E°-\left(\frac{RT}{nF}\right)\ln \left(Q\right)$

At room temperature the above equation is specifies as,

$E=E°-\left(\frac{0.0591}{n}\right)\log \left(Q\right)$

This relation is further used to determine the relation between $\Delta G°$ and $K$ , $\Delta G°$ and $E{°}_{\text{cell}}$.

To determine: The value of $\Delta G°$ and $K$ for different reactions.

Explanation of Solution

The given cell reaction is,

${\text{Cr}}^{3+}\left(aq\right)+{\text{Cl}}_{\text{2}}\left(g\right)\rightleftharpoons {\text{Cr}}_2{\text{O}}_7{}^{2-}\left(aq\right)+{\text{Cl}}^{-}\left(aq\right)$

The reaction taking place on cathode,

${\text{Cl}}_{\text{2}}\left(g\right)+2{\text{e}}^{-}\to 2{\text{Cl}}^{-}\left(aq\right)E{°}_{\text{red}}=1.36\text{V}$

The reaction taking place at anode,

${\text{2Cr}}^{3+}\left(aq\right)+7{\text{H}}_2\text{O}\left(l\right)\to {\text{Cr}}_2{\text{O}}_7{}^{2-}\left(aq\right)+14{\text{H}}^{+}\left(aq\right)+6{\text{e}}^{-}E{°}_{\text{ox}}=-1.33\text{V}$

Multiply reduction half-reaction with a coefficient of $3$ and then add both the reduction half and oxidation half-reaction.

$\begin{array}{c}{\text{Cl}}_{\text{2}}\left(g\right)+2{\text{e}}^{-}\to 2{\text{Cl}}^{-}\left(aq\right)\\ {\text{2Cr}}^{3+}\left(aq\right)+7{\text{H}}_2\text{O}\left(l\right)\to {\text{Cr}}_2{\text{O}}_7{}^{2-}\left(aq\right)+14{\text{H}}^{+}\left(aq\right)+6{\text{e}}^{-}\end{array}$

The overall cell reaction is,

${\text{2Cr}}^{3+}\left(aq\right)+7{\text{H}}_2\text{O}\left(l\right)+{\text{3Cl}}_{\text{2}}\left(g\right)\to {\text{Cr}}_2{\text{O}}_7{}^{2-}\left(aq\right)+14{\text{H}}^{+}\left(aq\right)+6{\text{Cl}}^{-}\left(aq\right)$

The overall cell potential is calculated as,

$\begin{array}{c}E{°}_{\text{cell}}=E{°}_{\text{ox}}+E{°}_{\text{red}}\\ =-1.33\text{V}+\left(1.36\text{V}\right)\\ =\underset{\_}{0.03V}\end{array}$

The relationship between cell potential and Gibbs free energy change is given by the formula,

$\Delta G°=-nFE{°}_{\text{cell}}$

Where,

• $\Delta G°$ is the Gibbs free energy change at the standard conditions.
• $n$ is the number of electrons involved in the reaction.
• $F$ is the Faraday’s constant.
• $E{°}_{\text{cell}}$ is the cell potential at the standard condition.

The number of moles of electrons involved in the reaction is $6$.

Substitute the values in above equation,

$\begin{array}{c}\Delta G°=-\left(6\text{mol}{\text{e}}^{-}\right)\left(96,485\frac{\text{C}}{\text{mol}\text{e-}}\right)\left(0.03\frac{\text{J}}{\text{C}}\right)\\ =-17,367.3\text{J}\\ \text{=}\underset{\_}{-17.36kJ}\end{array}$

The value of $\Delta G°$ is calculated as $\underset{\_}{-17.36kJ}$

The relationship between Gibbs free energy change and equilibrium constant is given by the formula,

$\Delta G°=-RT\ln K=-nFE{°}_{\text{cell}}$

Where,

• $\Delta G°$ is the Gibbs free energy change at the standard conditions.
• $R$ is the universal gas constant.
• $T$ is the temperature in Kelvin.
• $K$ is the equilibrium constant.

Rearranged equation at $25°\text{C}$ is obtained as,

$\log K=\left(-\frac{\left(n\right)\left(E°\right)}{0.0591}\right)$

Substitute the obtained values in above equation,

$\begin{array}{c}\log K=\left(-\frac{\left(6\right)\left(0.03\right)}{0.0591}\right)\\ =3.04\\ K={10}^{3.04}\\ =\underset{\_}{1096.47}\end{array}$

The value of $K$ is calculated as $\underset{\_}{1096.47}$.

To determine: The standard line notation for the given cell reaction.

Explanation:

The given cell reaction is,

${\text{Cu}}^{2+}\left(aq\right)+\text{Mg}\left(s\right)\rightleftharpoons \text{Cu}\left(s\right)+{\text{Mg}}^{2+}\left(aq\right)$

The reaction taking place on cathode,

${\text{Cu}}^{2+}\left(aq\right)+2{\text{e}}^{-}\to \text{Cu}\left(s\right)E{°}_{\text{red}}=+0.34\text{V}$

The reaction taking place at anode,

$\text{Mg}\left(s\right)\to {\text{Mg}}^{2+}\left(aq\right)+2{\text{e}}^{-}E{°}_{\text{ox}}=+2.37\text{V}$

Add both the reduction half and oxidation half-reaction.

$\begin{array}{c}{\text{Cu}}^{2+}\left(aq\right)+2{\text{e}}^{-}\to \text{Cu}\left(s\right)\\ \text{Mg}\left(s\right)\to {\text{Mg}}^{2+}\left(aq\right)+2{\text{e}}^{-}\end{array}$

The overall cell reaction is,

${\text{Cu}}^{\text{2+}}\left(aq\right)+\text{Mg}\left(s\right)\rightleftharpoons \text{Cu}\left(s\right)+{\text{Mg}}^{2+}\left(aq\right)$

The overall cell potential is calculated as,

$\begin{array}{c}E{°}_{\text{cell}}=E{°}_{\text{ox}}+E{°}_{\text{red}}\\ =2.37\text{V}+\left(0.34\text{V}\right)\\ =\underset{\_}{2.71V}\end{array}$

The relationship between cell potential and Gibbs free energy change is given by the formula,

$\Delta G°=-nFE{°}_{\text{cell}}$

The reaction involves the transfer of $2$ moles of electrons.

Substitute the values in above equation,

$\begin{array}{c}\Delta G°=-\left(2\text{mol}{\text{e}}^{-}\right)\left(96,485\frac{\text{C}}{\text{mol}\text{e-}}\right)\left(2.71\frac{\text{J}}{\text{C}}\right)\\ =-522,948.7\text{J}\\ =\underset{\_}{-522.949kJ}\end{array}$

The value of $\Delta \text{G}°$ is calculated as $\underset{\_}{-522.949kJ}$.

The relationship between Gibbs free energy change and equilibrium constant is given by the formula,

$\Delta G°=-RT\ln K=-nFE{°}_{\text{cell}}$

Rearranged equation at $25°\text{C}$ is obtained as,

$\log K=\left(-\frac{\left(n\right)\left(E°\right)}{0.0591}\right)$

Substitute the obtained values in above equation,

$\begin{array}{c}\log K=\left(-\frac{\left(2\right)\left(2.71\right)}{0.0591}\right)\\ =91.70\\ K={10}^{91.70}\\ =\underset{\_}{5.01\times 10^{91}}\end{array}$

The value of $K$ is calculated as $\underset{\_}{5.01\times 10^{91}}$.

Interpretation Introduction

Interpretation:

The equations for different chemical reactions are given. The value of $\Delta G°$ and $K$ for different reactions is to be calculated.

Concept introduction:

The relationship between reduction potential and standard reduction potential value and activities of species present in an electrochemical cell at a given temperature is given by the Nernst equation.

The value of $E_{\text{cell}}$ is calculated using Nernst formula,

$E=E°-\left(\frac{RT}{nF}\right) \ln \left(Q\right)$

At room temperature the above equation is specifies as,

$E=E°-\left(\frac{0.0591}{n}\right) \log \left(Q\right)$

This relation is further used to determine the relation between $\Delta G°$ and $K$ , $\Delta G°$ and $E{°}_{\text{cell}}$.

To determine: The standard line notation for the given cell reaction.

Explanation of Solution

Explanation

(a)

The given cell reaction is,

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

The reaction taking place on cathode,

${\text{Cl}}_{\text{2}}\left(g\right)+2{\text{e}}^{-}\to 2{\text{Cl}}^{-}\left(aq\right)E{°}_{\text{red}}=1.36\text{V}$

The reaction taking place at anode,

$2{\text{Br}}^{-}\left(aq\right)\to {\text{Br}}_2\left(g\right)+2{\text{e}}^{-}E{°}_{\text{ox}}=-1.09\text{V}$

Add both the reduction half and oxidation half-reaction.

$\begin{array}{c}{\text{Cl}}_{\text{2}}\left(g\right)+2{\text{e}}^{-}\to 2{\text{Cl}}^{-}\left(aq\right)\\ 2{\text{Br}}^{-}\left(aq\right)\to {\text{Br}}_2\left(g\right)+2{\text{e}}^{-}\end{array}$

The overall cell reaction is,

${\text{Cl}}_2\left(g\right)+2{\text{Br}}^{-}\left(aq\right)\to 2{\text{Cl}}^{-}\left(aq\right)+{\text{Br}}_2\left(g\right)$

The overall cell potential is calculated as,

$\begin{array}{c}E{°}_{\text{cell}}=E{°}_{\text{ox}}+E{°}_{\text{red}}\\ =-1.09\text{V}+\left(1.36\text{V}\right)\\ =\underset{\_}{0.27V}\end{array}$

The relationship between cell potential and Gibbs free energy change is given by the formula,

$\Delta G°=-nFE{°}_{\text{cell}}$

The number of moles of electrons involved in the reaction is $2$.

Substitute the values in above equation,

$\begin{array}{c}\Delta G°=-\left(2\text{mol}{\text{e}}^{-}\right)\left(96,485\frac{\text{C}}{\text{mol}\text{e-}}\right)\left(0.27\frac{\text{J}}{\text{C}}\right)\\ =-52101.9\text{J}\\ =\underset{\_}{-52.10kJ}\end{array}$

The value of $\Delta G°$ is calculated as $\underset{\_}{-52.10kJ}$.

The relationship between Gibbs free energy change and equilibrium constant is given by the formula,

$\Delta G°=-RT\ln K=-nFE{°}_{\text{cell}}$

Rearranged equation at $25°\text{C}$ is obtained as,

$\log K=\left(-\frac{\left(n\right)\left(E°\right)}{0.0591}\right)$

Substitute the obtained values in above equation,

$\begin{array}{c}\log K=\left(-\frac{\left(2\right)\left(0.27\right)}{0.0591}\right)\\ =9.137\\ K={10}^{9.137}\\ =\underset{\_}{1.37\times 10^9}\end{array}$

The value of $K$ is calculated as $\underset{\_}{1.37\times 10^9}$.

To determine: The standard line notation for the given cell reaction.

Explanation:

(b)

The given cell reaction is,

${\text{2Mn}}^{2+}\left(aq\right)+{\text{5IO}}_4{}^{-}\left(g\right)+3{\text{H}}_2\text{O}\left(l\right)+2{\text{e}}^{-}\to 2{\text{MnO}}_4{}^{-}\left(aq\right)+5{\text{IO}}_3{}^{-}\left(aq\right)+6{\text{H}}^{+}\left(aq\right)$

The reaction taking place on cathode,

${\text{IO}}_4{}^{-}\left(g\right)+2{\text{H}}^{+}\left(aq\right)+2{\text{e}}^{-}\to {\text{IO}}_3{}^{-}\left(aq\right)+{\text{H}}_2\text{O}\left(l\right)E{°}_{\text{red}}=+1.60\text{V}$

The reaction taking place at anode,

${\text{Mn}}^{2+}\left(aq\right)+4{\text{H}}_2\text{O}\left(l\right)\to {\text{MnO}}_4{}^{-}\left(aq\right)+5{\text{e}}^{-}+8{\text{H}}^{+}\left(aq\right)E{°}_{\text{ox}}=-1.51\text{V}$

Multiply reduction half-reaction with a coefficient of 5 and oxidation half-reaction with a coefficient of $2$ and then add both the reduction half and oxidation half-reaction.

$\begin{array}{c}{\text{5IO}}_4{}^{-}\left(g\right)+10{\text{H}}^{+}\left(aq\right)+10{\text{e}}^{-}\to 5{\text{IO}}_3{}^{-}\left(aq\right)+5{\text{H}}_2\text{O}\left(l\right)\\ {\text{2Mn}}^{2+}\left(aq\right)+8{\text{H}}_2\text{O}\left(l\right)\to 2{\text{MnO}}_4{}^{-}\left(aq\right)+10{\text{e}}^{-}+16{\text{H}}^{+}\left(aq\right)\end{array}$

The overall cell reaction is,

${\text{2Mn}}^{2+}\left(aq\right)+{\text{5IO}}_4{}^{-}\left(g\right)+3{\text{H}}_2\text{O}\left(l\right)+2{\text{e}}^{-}\to 2{\text{MnO}}_4{}^{-}\left(aq\right)+5{\text{IO}}_3{}^{-}\left(aq\right)+6{\text{H}}^{+}\left(aq\right)$

The overall cell potential is calculated as,

$\begin{array}{c}E{°}_{\text{cell}}=E{°}_{\text{ox}}+E{°}_{\text{red}}\\ =-1.51\text{V}+\left(1.60\text{V}\right)\\ =\underset{\_}{0.09V}\end{array}$

The relationship between cell potential and Gibbs free energy change is given by the formula,

$\Delta G°=-nFE{°}_{\text{cell}}$

The reaction involves the transfer of $10$ moles of electrons.

Substitute the values in above equation,

$\begin{array}{c}\Delta G°=-\left(10\text{mol}{\text{e}}^{-}\right)\left(96,485\frac{\text{C}}{\text{mol}\text{e-}}\right)\left(0.09\frac{\text{J}}{\text{C}}\right)\\ =-86,836.5\text{J}\\ =\underset{\_}{-86.83kJ}\end{array}$

The value of $\Delta G°$ is calculated as $\underset{\_}{-86.83kJ}$.

The relationship between Gibbs free energy change and equilibrium constant is given by the formula,

$\Delta G°=-RT\ln K=-nFE{°}_{\text{cell}}$

Rearranged equation at $25°\text{C}$ is obtained as,

$\log K=\left(-\frac{\left(n\right)\left(E°\right)}{0.0591}\right)$

Substitute the obtained values in above equation,

$\begin{array}{c}\log K=\left(-\frac{\left(10\right)\left(0.09\right)}{0.0591}\right)\\ =15.23\\ K={10}^{15.23}\\ =\underset{\_}{1.69\times 10^{15}}\end{array}$

The value of $K$ is calculated as $\underset{\_}{1.69\times 10^{15}}$.

Conclusion

The Nernst equation is used to find various relationships between different terms like $\Delta G°$ and $K$ , $\Delta G°$ and $E{°}_{\text{cell}}$.