Chapter 17, Problem 85E

(a)

Interpretation Introduction

Interpretation:

Various questions based on the displacement of one ion from its solution by other are to be answered and the $E°$, $\Delta G°$ and $K$ for the possible reaction are 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,

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

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

$\Delta G°=-nFE{°}_{\text{cell}}$

To determine: The reaction, if any, when crystals of ${\text{I}}_2$ are added to solution of $\text{NaCl}$.

Answer to Problem 85E

1. a. On the addition of crystals of iodine to $\text{NaCl}$ no reaction occurs.

Explanation of Solution

The reduction potential value for ${\text{I}}_2$ is,

${\text{I}}_2+2{\text{e}}^{-}\to 2{\text{I}}^{-}E°=0.52\text{V}$

The reduction potential value for ${\text{Cl}}_2$ is,

${\text{Cl}}_2+2{\text{e}}^{-}\to 2{\text{Cl}}^{-}E°=1.36\text{V}$

As the value of reduction potential for iodine is lower than that of chlorine. So, it won’t be possible for iodine to replace ions from its solution.

(b)

Interpretation Introduction

Interpretation:

Various questions based on the displacement of one ion from its solution by other are to be answered and the $E°$, $\Delta G°$ and $K$ for the possible reaction are 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,

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

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

$\Delta G°=-nFE{°}_{\text{cell}}$

To determine: The reaction that takes place when ${\text{Cl}}_2$ gas is bubbled into a solution of $\text{NaI}$.

Answer to Problem 85E

1. b. The value of $E°$ is $\underset{\_}{0.82V}$, the value of $\Delta G°$ is $\underset{\_}{-158.23kJ}$ and the value of $K$ is when ${\text{Cl}}_2$ is bubbled into a solution of $\text{NaI}$ $\underset{\_}{1096.47}$.

Explanation of Solution

Explanation

The reduction potential value for ${\text{I}}_2$ is,

${\text{I}}_2+2{\text{e}}^{-}\to 2{\text{I}}^{-}E°=0.52\text{V}$

The reduction potential value for ${\text{Cl}}_2$ is,

${\text{Cl}}_2+2{\text{e}}^{-}\to 2{\text{Cl}}^{-}E°=1.36\text{V}$

As the reduction potential of chlorine is greater than iodine. Therefore, it is possible for chlorine to replace iodide ion from its solution and this leads to the liberation of iodine gas.

The reaction taking place at cathode is,

${\text{Cl}}_2+2{\text{e}}^{-}\to 2{\text{Cl}}^{-}{E}_{\text{red}}°=1.36\text{V}$

The reaction taking place at anode is,

$2{\text{I}}^{-}\to {\text{I}}_2+2{\text{e}}^{-}{E}_{\text{ox}}°=-0.54\text{V}$

Add both the oxidation and reduction half-reaction,

$\begin{array}{c}{\text{Cl}}_2+2{\text{e}}^{-}\to 2{\text{Cl}}^{-}\\ 2{\text{I}}^{-}\to {\text{I}}_2+2{\text{e}}^{-}\end{array}$

The final equation is,

${\text{Cl}}_2+2{\text{I}}^{-}\to 2{\text{Cl}}^{-}{\text{+I}}_2$

The value of $E°$ is calculated as,

$\begin{array}{c}E°=E_{\text{ox}}+E_{\text{red}}\\ =-0.54\text{V}+1.36\text{V}\\ =\underset{\_}{0.82V}\end{array}$

The value of $E°$ is calculated as $\underset{\_}{0.82V}$.

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

Substitute the values in above equation,

$\begin{array}{c}\Delta G°=-\left(2\text{mol}{\text{e}}^{-}\right)\left(96,485\frac{\text{C}}{\text{mol}\text{e-}}\right)\left(0.82\frac{\text{J}}{\text{C}}\right)\\ =-158,235.4\text{J}\\ =\underset{\_}{-158.23kJ}\end{array}$

The value of $\Delta G°$ is calculated as $\underset{\_}{-158.23kJ}$

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,

$\begin{array}{c}\log K=\left(-\frac{\left(2\right)\left(0.82\right)}{0.0591}\right)\\ =27.75\\ K={10}^{27.75}\\ =\underset{\_}{5.62\times 10^{27}}\end{array}$

The value of $K$ is $\underset{\_}{5.62\times 10^{27}}$.

(c)

Answer:

c. There occurs no reaction when a silver wire is placed in a solution of ${\text{CuCl}}_2$.

Interpretation Introduction

Interpretation:

Various questions based on the displacement of one ion from its solution by other are to be answered and the $E°$, $\Delta G°$ and $K$ for the possible reaction are 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,

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

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

$\Delta G°=-nFE{°}_{\text{cell}}$

To determine: The reaction, if any, when a silver wire is placed into a solution of ${\text{CuCl}}_2$

Answer to Problem 85E

c. There occurs no reaction when a silver wire is placed in a solution of ${\text{CuCl}}_2$.

Explanation of Solution

The reduction potential value for ${\text{Ag}}^{+}$ is,

${\text{Ag}}^{+}+{\text{e}}^{-}\to \text{Ag}E°=0.80\text{V}$

Write it in reverse order,

$\text{Ag}\to {\text{Ag}}^{+}+{\text{e}}^{-}E{°}_{\text{ox}}=-0.80\text{V}$

The reduction potential value for ${\text{Cu}}^{\text{2+}}$ is,

${\text{Cu}}^{\text{2+}}+2{\text{e}}^{-}\to \text{Cu}E{°}_{\text{red}}=0.34\text{V}$

The value of $E°$ is calculated as,

$\begin{array}{c}E°=E{°}_{\text{ox}}+E{°}_{\text{red}}\\ =-0.80\text{V}+0.34\text{V}\\ =\underset{\_}{-0.64V}\end{array}$

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 $E{°}_{\text{cell}}$ is negative, therefore the value of $\Delta G°$ is positive. It means it corresponds to a non-spontaneous reaction.

(d)

Interpretation Introduction

Interpretation:

Various questions based on the displacement of one ion from its solution by other are to be answered and the $E°$, $\Delta G°$ and $K$ for the possible reaction are 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,

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

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

$\Delta G°=-nFE{°}_{\text{cell}}$

To determine: The reaction, if any, an acidic solution of ${\text{FeSO}}_4$ is exposed to air.

Answer to Problem 85E

d. The value of $E°$ is $\underset{\_}{0.46V}$, the value of $\Delta G°$ is $\underset{\_}{-177.53kJ}$ and the value of $K$ is $\underset{\_}{1.34\times 10^{31}}$ when an acidic solution of ${\text{FeSO}}_4$ is exposed to air.

Explanation of Solution

The reduction potential value for ${\text{O}}_2$ is,

${\text{O}}_2+4{\text{H}}^{+}+4{\text{e}}^{-}\to 2{\text{H}}_2\text{O}E°=1.23\text{V}$

The reduction potential value for ${\text{Fe}}^{3+}$ is,

${\text{Fe}}^{3+}+{\text{e}}^{-}\to {\text{Fe}}^{2+}E°=0.77\text{V}$

The reaction taking place at cathode is,

${\text{O}}_2+4{\text{H}}^{+}+4{\text{e}}^{-}\to 2{\text{H}}_2\text{O}E{°}_{\text{red}}=1.23\text{V}$

The reaction taking place at anode is,

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

Multiply the oxidation half-reaction with a coefficient of $4$ and then add both the oxidation and reduction half-reaction,

$\begin{array}{c}{\text{O}}_2+4{\text{H}}^{+}+4{\text{e}}^{-}\to 2{\text{H}}_2\text{O}\\ {\text{4Fe}}^{2+}\to 4{\text{Fe}}^{3+}+4{\text{e}}^{-}\end{array}$

The final equation is,

${\text{4Fe}}^{2+}+{\text{O}}_2+4{\text{H}}^{+}\to 4{\text{Fe}}^{3+}+2{\text{H}}_2\text{O}$

The value of $E°$ is calculated as,

$\begin{array}{c}E°=E_{\text{ox}}+E_{\text{red}}\\ =-0.77\text{V}+1.23\text{V}\\ =\underset{\_}{0.46V}\end{array}$

The value of $E°$ is calculated as $\underset{\_}{0.46V}$.

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 reaction involves the transfer of $4$ moles of electrons.

Substitute the values in above equation,

$\begin{array}{c}\Delta G°=-\left(4\text{mol}{\text{e}}^{-}\right)\left(96,485\frac{\text{C}}{\text{mol}\text{e-}}\right)\left(0.46\frac{\text{J}}{\text{C}}\right)\\ =-17,7532.4\text{J}\\ =\underset{\_}{-177.53kJ}\end{array}$

The value of $\Delta G°$ is calculated as $\underset{\_}{-177.53kJ}$

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,

$\begin{array}{c}\log K=\left(-\frac{\left(4\right)\left(0.46\right)}{0.0591}\right)\\ =31.13\\ K={10}^{31.13}\\ =\underset{\_}{1.34\times 10^{31}}\end{array}$

The value of $K$ is calculated as $\underset{\_}{1.34\times 10^{31}}$.