Chapter 18, Problem 74E

Calculate ΔG° and K at 25°C for the reactions in Exercises 38 and 42.

Interpretation Introduction

Interpretation: The standard Gibbs free energy $\left(\Delta G^{\text{ο}}\right)$ and the value of equilibrium constant $\left(K\right)$ for the given reactions are to be calculated at ${\text{25 }}^{\text{ο}}\text{C}$.

Concept introduction: Gibbs free energy is a thermodynamic quantity that is used to calculate the maximum work of reversible reaction performed by a system. It is equal to the difference between the enthalpy and the product of entropy at absolute temperature. Equilibrium constant is defined as the ratio of the concentration of products and the concentration of the reactants.

To determine: The standard Gibbs free energy $\left(\Delta G^{\text{ο}}\right)$ and the value of equilibrium constant $\left(K\right)$ for the given reaction at ${\text{25 }}^{\text{ο}}\text{C}$.

Answer to Problem 74E

Answer

The standard Gibbs free energy $\Delta G_{\text{cell}}^{\text{ο}}$ is $-\underset{\_}{414950J}$ and the equilibrium constant $\left(K\right)$ is $\underset{\_}{5.62\times 10^{72}}$.

Explanation of Solution

Explanation

Given

The half cell reactions for the cell are,

${\text{Anode: Fe}}^{2+}\left(aq\right)\stackrel{}{\to }{\text{Fe}}^{3+}\left(aq\right)+{\text{e}}^{-}{E}_{\text{anode}}^{\text{ο}}=0.\text{77V}$ (1)

${\text{Cathode: 4IO}}_3{}^{-}\left(aq\right)+1{\text{2H}}^{+}\left(aq\right)+10{\text{e}}^{-}\stackrel{}{\to }{\text{I}}_2\left(aq\right)+6{\text{H}}_{\text{2}}\text{O}\left(l\right)E_{\text{cathode}}^{\text{ο}}=1.20\text{ V}$ (2)

Where,

• $E_{\text{anode}}^{\text{ο}}$ is the standard electrode potential of equation (1).
• $E_{\text{cathode}}^{\text{ο}}$ is the standard electrode potential of equation (2).

The overall cell reaction is,

${\text{10Fe}}^{+2}\left(aq\right)+{\text{ 4IO}}_3{}^{-}\left(aq\right)+1{\text{2H}}^{+}\left(aq\right)\stackrel{}{\to }{\text{10Fe}}^{3+}\left(aq\right)+{\text{I}}_2\left(aq\right)+6{\text{H}}_{\text{2}}\text{O}\left(l\right)$ (3)

The standard electrode potential of equation (3) is calculated by the formula,

$E_{\text{cell}}{}^{\text{ο}}=E_{\text{cathode}}^{\text{ο}}-E_{\text{anode}}^{\text{ο}}$

Where,

• $E_{\text{cell}}{}^{\text{ο}}$ is the standard electrode potential of equation (3).

Substitute the values of $E_{\text{cathode}}^{\text{ο}}$ and $E_{\text{anode}}^{\text{ο}}$ in the above formula.

$\begin{array}{c}{E}_{\text{cell}}{}^{\text{ο}}=1.2\text{0 V}-0.7\text{7 V}\\ =0.4\text{3 V}\end{array}$

The standard cell potential of the given cell is $0.4\text{3 V}$.

The reaction between $\Delta G_{\text{rxn}}^{\text{ο}}$ and $E_{\text{cell}}^{\text{ο}}$ is given as,

$\Delta G_{\text{cell}}^{\text{ο}}=-nF{E}_{\text{cell}}^{\text{ο}}$

Where,

• $\text{n}$ is the number of electrons involved in the cell reaction.
• F is the faradays constant $\left(9650\text{0 C}\right)$.
• $E_{\text{cell}}^{\text{ο}}$ is the standard cell potential.

Substitute the value of $n$, $F$ and $E_{\text{cell}}^{\text{ο}}$ in the above expression.

$\begin{array}{c}\Delta G_{\text{cell}}^{\text{ο}}=-nF{E}_{\text{cell}}^{\text{ο}}\\ =-\left(10\right)\left(9650\text{0 C}\right)\left(0.43\text{ V}\right)\\ =-\underset{\_}{414950J}\end{array}$

Therefore, $\Delta G_{\text{cell}}^{\text{ο}}$ is $-\underset{\_}{414950J}$.

Equilibrium constant $\left(K\right)$ is calculated by using the Nernst equation.

$E_{\text{cell}}=E_{\text{cell}}^{\text{ο}}-\frac{0.0591}{n}\log \left(K\right)$

When equilibrium is existed in the cell then,

$E_{\text{cell}}=0$

Hence, the Nernst equation becomes,

$\begin{array}{c}{E}_{\text{cell}}=E_{\text{cell}}^{\text{ο}}-\frac{0.0591}{n}\log \left(K\right)\\ 0=E_{\text{cell}}^{\text{ο}}-\frac{0.0591}{n}\log K\\ E_{\text{cell}}^{\text{ο}}=\frac{0.0591}{n}\log K\end{array}$

Substitute the values of $n$ and $E_{\text{cell}}^{\text{ο}}$ in the above expression to calculate the equilibrium constant $\left(K\right)$.

$\begin{array}{c}{E}_{\text{cell}}^{\text{ο}}=\frac{0.0591}{n}\log K\\ 0.4\text{3 V}=\frac{0.0591}{10}\log K\\ \log K=72.75\\ K={10}^{72.75}\end{array}$

Simplify the above expression,

$K=\underset{\_}{5.62\times 10^{72}}$.

Hence, the equilibrium constant $\left(K\right)$ is $\underset{\_}{5.62\times 10^{72}}$.

Interpretation Introduction

Interpretation: The standard Gibbs free energy $\left(\Delta G^{\text{ο}}\right)$ and the value of equilibrium constant $\left(K\right)$ for the given reactions are to be calculated at ${\text{25 }}^{\text{ο}}\text{C}$.

Concept introduction: Gibbs free energy is a thermodynamic quantity that is used to calculate the maximum work of reversible reaction performed by a system. It is equal to the difference between the enthalpy and the product of entropy at absolute temperature. Equilibrium constant is defined as the ratio of the concentration of products and the concentration of the reactants.

To determine: The standard Gibbs free energy $\left(\Delta G^{\text{ο}}\right)$ and the value of equilibrium constant $\left(K\right)$ for the given reaction at ${\text{25 }}^{\text{ο}}\text{C}$.

Answer to Problem 74E

Answer

The standard Gibbs free energy $\Delta G_{\text{cell}}^{\text{ο}}$ is $-\underset{\_}{301080J}$ and the equilibrium constant $\left(K\right)$ is $\underset{\_}{6.165\times 10^{52}}$.

Explanation of Solution

Explanation

Given

The half cell reactions for the cell are,

$\text{Anode: Zn}\left(s\right)\stackrel{}{\to }{\text{Zn}}^{2+}\left(aq\right)+2{\text{e}}^{-}{E}_{\text{anode}}^{\text{ο}}=-0.\text{76V}$ (1)

${\text{Cathode: 2Ag}}^{+}\left(aq\right)+2{\text{e}}^{-}\stackrel{}{\to }2\text{Ag}\left(s\right)E_{\text{cathode}}^{\text{ο}}=0.80\text{ V}$ (2)

Where,

• $E_{\text{anode}}^{\text{ο}}$ is the standard electrode potential of equation (1).
• $E_{\text{cathode}}^{\text{ο}}$ is the standard electrode potential of equation (2).

The overall cell reaction is,

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

The standard electrode potential of equation (3) is calculated by the formula,

$E_{\text{cell}}{}^{\text{ο}}=E_{\text{cathode}}^{\text{ο}}-E_{\text{anode}}^{\text{ο}}$

Where,

• $E_{\text{cell}}{}^{\text{ο}}$ is the standard electrode potential of equation (3).

Substitute the values of $E_{\text{cathode}}^{\text{ο}}$ and $E_{\text{anode}}^{\text{ο}}$ in the above formula.

$\begin{array}{c}{E}_{\text{cell}}{}^{\text{ο}}=0.80\text{ V}-\left(-0.7\text{7 V}\right)\\ =1.56\text{ V}\end{array}$

The standard cell potential of the given cell is $\text{1}\text{.56 V}$.

The reaction between $\Delta G_{\text{rxn}}^{\text{ο}}$ and $E_{\text{cell}}^{\text{ο}}$ is given as,

$\Delta G_{\text{cell}}^{\text{ο}}=-nF{E}_{\text{cell}}^{\text{ο}}$

Where,

• $\text{n}$ is the number of electrons involved in the cell reaction.
• F is the faradays constant $\left(9650\text{0 C}\right)$.
• $E_{\text{cell}}^{\text{ο}}$ is the standard cell potential.

Substitute the value of $n$, $F$ and $E_{\text{cell}}^{\text{ο}}$ in the above expression.

$\begin{array}{c}\Delta G_{\text{cell}}^{\text{ο}}=-nF{E}_{\text{cell}}^{\text{ο}}\\ =-\left(2\right)\left(9650\text{0 C}\right)\left(\text{1}\text{.56 V}\right)\\ =-\underset{\_}{301080J}\end{array}$

Therefore, $\Delta G_{\text{cell}}^{\text{ο}}$ is $-\underset{\_}{301080J}$.

Equilibrium constant $\left(K\right)$ is calculated by using the Nernst equation.

$E_{\text{cell}}=E_{\text{cell}}^{\text{ο}}-\frac{0.0591}{n}\log \left(K\right)$

When equilibrium is existed in the cell then,

$E_{\text{cell}}=0$

Hence, the Nernst equation becomes,

$\begin{array}{c}{E}_{\text{cell}}=E_{\text{cell}}^{\text{ο}}-\frac{0.0591}{n}\log \left(K\right)\\ 0=E_{\text{cell}}^{\text{ο}}-\frac{0.0591}{n}\log K\\ E_{\text{cell}}^{\text{ο}}=\frac{0.0591}{n}\log K\end{array}$

Substitute the values of $n$ and $E_{\text{cell}}^{\text{ο}}$ in the above expression to calculate the equilibrium constant $\left(K\right)$.

$\begin{array}{c}{E}_{\text{cell}}^{\text{ο}}=\frac{0.0591}{n}\log K\\ \text{1}\text{.56 V}=\frac{0.0591}{2}\log K\\ \log K=52.79\\ K={10}^{52.79}\end{array}$

Simplify the above expression,

$K=\underset{\_}{6.165\times 10^{52}}$

Hence, the equilibrium constant $\left(K\right)$ is $\underset{\_}{6.165\times 10^{52}}$.