Chapter 17, Problem 86E

 (a)

Interpretation Introduction

Interpretation:

The information regarding the disproportionation reaction is given. Various questions based on the spontaneity are to be answered and the calculation of $\Delta G°$ and $K$ is to be stated.

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 specified as,

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

To determine: The spontaneity of the given reaction (a) and further calculation of $\Delta G°$ and $K$.

Answer to Problem 86E

1. a. The given reaction (a) is spontaneous and The value of $\Delta G°$ for the given reaction is calculated as $\underset{\_}{-34.73kJ}$ and The value of $K$ is calculated as $\underset{\_}{1.2\times 10^6}$.

Explanation of Solution

Given

The disproportionation reaction is,

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

The reaction taking place at cathode is,

${\text{Cu}}^{+}\left(aq\right)+{\text{e}}^{-}\to \text{Cu}\left(s\right)E{°}_{\text{red}}=0.52\text{V}$

The reaction taking place at anode is,

${\text{Cu}}^{+}\left(aq\right)\to {\text{Cu}}^{\text{2+}}\left(aq\right)+{\text{e}}^{-}E{°}_{\text{ox}}=-0.16\text{V}$

Add both the oxidation and reduction half-reaction,

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

The final equation is,

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

The overall cell potential is calculated as,

$\begin{array}{c}E{°}_{\text{cell}}=E{°}_{\text{ox}}+E{°}_{\text{red}}\\ =-0.16\text{V}+\text{0}\text{.52}\text{V}\\ \text{=0}\text{.36}\text{V}\end{array}$

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.

As the value of $E{°}_{\text{cell}}$ comes out to be positive then $\Delta G°$ becomes negative. Therefore it leads to a spontaneous reaction.

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 $1\text{mol}{\text{e}}^{-}$

Substitute the values of $n$, $E{°}_{\text{cell}}$ and $F$ in the above equation as,

$\begin{array}{c}\Delta G°=-1\times 96,485\times 0.36\\ =-34,734.6\text{J}\\ =\underset{\_}{-34.73kJ}\end{array}$

The value of $\Delta G°$ for the given reaction is calculated as $\underset{\_}{-34.73kJ}$.

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,

$\ln K=\left(-\frac{\left(\Delta G°\right)}{R\times T}\right)$

Substitute the obtained values in above equation,

$\begin{array}{c}2.303\log K=\left(\frac{\left(34.73kJ\right)}{.00831\frac{kJ}{Kmol}\times 298K}\right)\\ \log K=6.08\\ K={10}^{6.08}\\ =\underset{\_}{1.2\times 10^6}\end{array}$

The value of $K$ is calculated as $\underset{\_}{1.2\times 10^6}$.

(b)

Interpretation Introduction

Interpretation:

The information regarding the disproportionation reaction is given. Various questions based on the spontaneity are to be answered and the calculation of $\Delta G°$ and $K$ is to be stated.

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 specified as,

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

To determine: The spontaneity of the given reaction (b) and the value of $\Delta G°$ and $K$ if the reaction is spontaneous.

Answer to Problem 86E

1. b. The given reaction (b) is non-spontaneous in nature.

Explanation of Solution

The disproportionation reaction is,

${\text{3Fe}}^{\text{2+}}\left(aq\right)\to 2{\text{Fe}}^{3+}\left(aq\right)+\text{Fe}\left(s\right)$

The reaction taking place at cathode is,

${\text{Fe}}^{\text{2+}}\left(aq\right)+2{\text{e}}^{-}\to \text{Fe}\left(s\right)E{°}_{\text{red}}=-0.44\text{V}$

The reaction taking place at anode is,

${\text{2Fe}}^{\text{2+}}\left(aq\right)\to 2{\text{Fe}}^{3+}\left(aq\right)+2{\text{e}}^{-}E{°}_{\text{ox}}=-0.77\text{V}$

Add both the oxidation and reduction half-reaction,

$\begin{array}{c}{\text{Fe}}^{\text{2+}}\left(aq\right)+2{\text{e}}^{-}\to \text{Fe}\left(s\right)\\ {\text{2Fe}}^{\text{2+}}\left(aq\right)\to 2{\text{Fe}}^{3+}\left(aq\right)+2{\text{e}}^{-}\end{array}$

The final equation is,

${\text{3Fe}}^{\text{2+}}\left(aq\right)\to 2{\text{Fe}}^{3+}\left(aq\right)+\text{Fe}\left(s\right)$

The overall cell potential is calculated as,

$\begin{array}{c}E{°}_{\text{cell}}=E{°}_{\text{ox}}+E{°}_{\text{red}}\\ =-0.44\text{V}+\left(-0.77\text{V}\right)\\ =\underset{\_}{-1.21V}\end{array}$

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.

As the value of $E{°}_{\text{cell}}$ comes out to be negative then $\Delta G°$ becomes positive. Therefore it leads to a non-spontaneous reaction.

(c)

Interpretation Introduction

Interpretation:

The information regarding the disproportionation reaction is given. Various questions based on the spontaneity are to be answered and the calculation of $\Delta G°$ and $K$ is to be stated.

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 specified as,

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

To determine: The spontaneity of the given reaction (c) and further calculation of $\Delta G°$ and $K$ if the reaction is spontaneous.

Answer to Problem 86E

1. c. The given reaction (c) is spontaneous and The value of $\Delta G°$ for the given reaction is calculated as $\underset{\_}{-84.90kJ}$ and The value of $K$ is calculated as $\underset{\_}{7.5\times 10^{14}}$.

Explanation of Solution

Given

The disproportionation reaction is,

${\text{HClO}}_2\left(aq\right)\to {\text{ClO}}_3{}^{-}\left(aq\right)+\text{HClO}\left(aq\right)$

The reaction taking place at cathode is,

${\text{HClO}}_2\left(aq\right)+2{\text{H}}^{+}\left(aq\right)+2{\text{e}}^{-}\to \text{HClO}\left(aq\right)+{\text{H}}_2\text{O}\left(l\right)E{°}_{\text{red}}=1.65\text{V}$

The reaction taking place at anode is,

${\text{HClO}}_2\left(aq\right)+{\text{H}}_2\text{O}\left(l\right)\to {\text{ClO}}_3{}^{-}\left(aq\right)+3{\text{H}}^{\text{+}}\left(aq\right)+2{\text{e}}^{-}E{°}_{\text{ox}}=-1.21\text{V}$

Add both the oxidation and reduction half-reaction,

$\begin{array}{c}{\text{HClO}}_2\left(aq\right)+2{\text{H}}^{+}\left(aq\right)+2{\text{e}}^{-}\to \text{HClO}\left(aq\right)+{\text{H}}_2\text{O}\left(l\right)\\ {\text{HClO}}_2\left(aq\right)+{\text{H}}_2\text{O}\left(l\right)\to {\text{ClO}}_3{}^{-}\left(aq\right)+3{\text{H}}^{\text{+}}\left(aq\right)+2{\text{e}}^{-}\end{array}$

The final equation is,

${\text{2HClO}}_2\left(aq\right)\to \text{HClO}\left(aq\right){\text{+ClO}}_3{}^{-}\left(aq\right)+{\text{H}}^{\text{+}}\left(aq\right)$

The overall cell potential is calculated as,

$\begin{array}{c}E{°}_{\text{cell}}=E{°}_{\text{ox}}+E{°}_{\text{red}}\\ =-1.21\text{V}+1.65\text{V}\\ =\underset{\_}{0.44V}\end{array}$

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.

As the value of $E{°}_{\text{cell}}$ comes out to be positive then $\Delta G°$ becomes negative. Therefore it leads to a spontaneous reaction.

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 of $n$, $E{°}_{\text{cell}}$ and $F$ in the above equation as,

$\begin{array}{c}\Delta G°=-2\times 96,485\times 0.44\\ =-84,906.8\text{J}\\ =\underset{\_}{-84.90kJ}\end{array}$

The value of $\Delta G°$ for the given reaction is calculated as $\underset{\_}{-84.90kJ}$.

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.

• $n$ 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.

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

$\ln K=\left(-\frac{\left(\Delta G°\right)}{R\times T}\right)$

Substitute the obtained values in above equation,

$\begin{array}{c}2.303\log K=\left(\frac{\left(84.90kJ\right)}{0.00831\frac{kJ}{Kmol}\times 298K}\right)\\ \log K=14.88\\ K={10}^{14.88}\\ =\underset{\_}{7.5\times 10^{14}}\end{array}$

The value of $K$ is calculated as $\underset{\_}{7.5\times 10^{14}}$.