Chapter 17, Problem 152CP

(a)

Interpretation Introduction

Interpretation: The standard cell potential $E^{\text{ο}}$ and standard Gibbs free energy $\Delta G^{\text{ο}}$ of the given cell are to be calculated. The Nernst equation for the cell is to be stated. The molar concentration of $\left[{\text{CrO}}_4^{2-}\right]$ for this solution and solubility product $\left(K_{\text{sp}}\right)$ for ${\text{Ag}}_{\text{2}}{\text{CrO}}_{\text{4}}$ is to be calculated.

Concept introduction: Cell potential is defined as the measure of energy per unit charge available from the redox reaction to carry out the reaction. 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.

To determine: The standard cell potential $E^{\text{ο}}$ and standard Gibbs free energy $\Delta G^{\text{ο}}$ at $2{\text{5 }}^{\text{ο}}\text{C}$.

Answer to Problem 152CP

Answer

The standard cell potential of reaction is $\underset{\_}{0.204V}$ and the free energy $\Delta G^{\text{ο}}$ of the reaction is $-\underset{\_}{3.93\times 10^{4}J}$.

Explanation of Solution

Explanation

The half cell reactions are,

$\text{Anode: Hg}\stackrel{}{\to }{\text{Hg}}^{2+}+{\text{2e}}^{-}{E}_{\text{anode}}^{\text{ο}}=0.242\text{ V}$

${\text{Cathode: 2Ag}}^{+}+2{\text{e}}^{-}\underset{}{\to }2\text{Ag }{E}_{\text{cathode}}^{\text{ο}}=0.446\text{ V}$

Where,

• $E_{\text{anode}}^{\text{ο}}$ is the standard reduction potential of anode.
• $E_{\text{cathode}}^{\text{ο}}$ is the standard reduction potential of cathode.

Cell potential of the reaction is the difference in voltage between the cathode and anode. The cell potential is calculated by using the expression,

$E^{\text{ο}}=E^{\text{ο}}\left(\text{cathode}\right)-E^{\text{ο}}\left(\text{anode}\right)$.

Substitute the value of standard reduction potential of anode and cathode in the above expression to calculate the value of cell potential of the reaction.

$\begin{array}{c}{E}^{\text{ο}}=E^{\text{ο}}\left(\text{cathode}\right)-E^{\text{ο}}\left(\text{anode}\right)\\ =0.44\text{6 V}-0.24\text{2 V}\\ =\underset{\_}{0.204V}\end{array}$

Hence, the standard cell potential of reaction is $\underset{\_}{0.204V}$.

The relationship between the standard free energy of a cell $\Delta G^{\text{ο}}$ and the cell potential $E^{\text{ο}}$ is given by the expression,

$\Delta G^{\text{ο}}=-nF{E}^{\text{ο}}$.

Where,

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

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

$\begin{array}{c}\Delta G^{\text{ο}}=-nF{E}^{\text{ο}}\\ =-\left(2\right)\left(9650\text{0 C}\right)\left(\text{0}\text{.204 V}\right)\\ =-\underset{\_}{3.93\times 10^{4}J}\end{array}$

Therefore, the free energy $\Delta G^{\text{ο}}$ of the reaction is $-\underset{\_}{3.93\times 10^{4}J}$.

(b)

Interpretation Introduction

Interpretation: The standard cell potential $E^{\text{ο}}$ and standard Gibbs free energy $\Delta G^{\text{ο}}$ of the given cell are to be calculated. The Nernst equation for the cell is to be stated. The molar concentration of $\left[{\text{CrO}}_4^{2-}\right]$ for this solution and solubility product $\left(K_{\text{sp}}\right)$ for ${\text{Ag}}_{\text{2}}{\text{CrO}}_{\text{4}}$ is to be calculated.

Concept introduction: Cell potential is defined as the measure of energy per unit charge available from the redox reaction to carry out the reaction. 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.

To determine: The Nernst equation for the cell.

Answer to Problem 152CP

Answer

The Nernst equation for the cell is rightfully stated.

Explanation of Solution

Explanation

Given

The cell reaction is,

${\text{Ag}}_{\text{2}}{\text{CrO}}_{\text{4}}+{\text{2e}}^{-}\underset{}{\to }2\text{Ag}+{\text{CrO}}_4^{2-}$.

The general form of the Nernst equation for the cell is,

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

Where,

• $n$ is the number of electrons in the cell reaction.
• $Q$ is the equilibrium expression with the initial concentrations or $\frac{\left[{\text{CrO}}_4^{-}\right]}{\left[{\text{Ag}}^{+}\right]}$.

As the SCE concentrations is constant, $Q$ from the denominator is eliminated.

Therefore, the Nernst equation for the cell is,

$E_{\text{cell}}=0.24\text{2 V}-\frac{0.0591}{2}\log \left(Q\right)$

Where, $E_{\text{anode}}^{\text{ο}}=0.242\text{ V}$

(c)

Interpretation Introduction

Interpretation: The standard cell potential $E^{\text{ο}}$ and standard Gibbs free energy $\Delta G^{\text{ο}}$ of the given cell are to be calculated. The Nernst equation for the cell is to be stated. The molar concentration of $\left[{\text{CrO}}_4^{2-}\right]$ for this solution and solubility product $\left(K_{\text{sp}}\right)$ for ${\text{Ag}}_{\text{2}}{\text{CrO}}_{\text{4}}$ is to be calculated.

Concept introduction: Cell potential is defined as the measure of energy per unit charge available from the redox reaction to carry out the reaction. 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.

To determine: The expected cell potential.

Answer to Problem 152CP

Answer

The cell potential of the given cell is. $\underset{\_}{0.352V}$.

Explanation of Solution

Explanation

Given

The value of $E^{\text{ο}}$ is $0.242\text{ V}$.

The molar concentration of $\left[{\text{CrO}}_4^{2-}\right]=1.00\times {10}^{-5}\text{ mol/L}$.

The molar concentration of $\left[{\text{Ag}}^{+}\right]=1.0\text{0 mol/L}$.

The cell potential of the given cell is calculated by using Nernst equation.

The Nernst equation for the cell is,

$E_{\text{cell}}=E_{}^{\text{ο}}-\frac{0.0591}{n}\log \left(\frac{\left[{\text{CrO}}_4^{2-}\right]}{\left[{\text{Ag}}^{+1}\right]}\right)$

Where,

• $n$ is the number of electrons in the cell reaction.

Substitute the Substitute the values of $n$, molar concentrations of anode and cathode and $E_{\text{cell}}^{\text{ο}}$ in the above expression,

$\begin{array}{c}{E}_{\text{cell}}=E_{}^{\text{ο}}-\frac{0.0591}{n}\log \left(\frac{\left[{\text{CrO}}_4^{2-}\right]}{\left[{\text{Ag}}^{+1}\right]}\right)\\ =0.242V-\frac{0.0591}{2}\log \left(\frac{\left[1.00\times {10}^{-5}\text{ mol/L}\right]}{\left[1.0\text{0 mol/L}\right]}\right)\\ =\underset{\_}{0.352V}\end{array}$

Hence, the cell potential of the given cell is. $\underset{\_}{0.352V}$.

(d)

Interpretation Introduction

Interpretation: The standard cell potential $E^{\text{ο}}$ and standard Gibbs free energy $\Delta G^{\text{ο}}$ of the given cell are to be calculated. The Nernst equation for the cell is to be stated. The molar concentration of $\left[{\text{CrO}}_4^{2-}\right]$ for this solution and solubility product $\left(K_{\text{sp}}\right)$ for ${\text{Ag}}_{\text{2}}{\text{CrO}}_{\text{4}}$ is to be calculated.

Concept introduction: Cell potential is defined as the measure of energy per unit charge available from the redox reaction to carry out the reaction. 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.

To determine: The molar concentration of $\left[{\text{CrO}}_4^{2-}\right]$ for the given Answer.

Answer to Problem 152CP

Answer

The molar concentration of $\left[{\text{CrO}}_4^{2-}\right]$ is $\underset{\_}{1.36\times 10^{-9}mol/L}$.

Explanation of Solution

Explanation

Given

$E^{\text{ο}}=0.242\text{ V}$

$E=0.504\text{ V}$

$n=2$.

The molar concentration of $\left[{\text{Ag}}^{+}\right]=1.0\text{0 mol/L}$.

The concentration of $\left[{\text{CrO}}_4^{2-}\right]$ is calculated by using Nernst equation.

$E_{\text{cell}}=E_{}^{\text{ο}}-\frac{0.0591}{n}\log \left(\frac{\left[{\text{CrO}}_4^{2-}\right]}{\left[{\text{Ag}}^{+1}\right]}\right)$

Substitute the Substitute the values of $n$, molar concentrations of anode, $E_{\text{cell}}$ and $E_{\text{cell}}^{\text{ο}}$ in the above expression,

$\begin{array}{c}{E}_{\text{cell}}=E_{}^{\text{ο}}-\frac{0.0591}{n}\log \left(\frac{\left[{\text{CrO}}_4^{2-}\right]}{\left[{\text{Ag}}^{+1}\right]}\right)\\ 0.50\text{4 V}=0.242V-\frac{0.0591}{2}\log \left(\frac{\left[{\text{CrO}}_4^{2-}\right]}{\left[1.0\text{0 mol/L}\right]}\right)\\ \left(0.504-0.242\right)=-\frac{0.0591}{2}\log \left(\frac{\left[{\text{CrO}}_4^{2-}\right]}{\left[1.0\text{0 mol/L}\right]}\right)\\ 0.262=-\frac{0.0591}{2}\log \left(\frac{\left[{\text{CrO}}_4^{2-}\right]}{\left[1.0\text{0 mol/L}\right]}\right)\end{array}$

Simply the above expression,

$\begin{array}{c}{10}^{-8.866}=\left[{\text{CrO}}_4^{2-}\right]\\ \left[{\text{CrO}}_4^{2-}\right]=\underset{\_}{1.36\times 10^{-9}mol/L}\end{array}$

Hence, the molar concentration of $\left[{\text{CrO}}_4^{2-}\right]$ is $\underset{\_}{1.36\times 10^{-9}mol/L}$.

(e)

Interpretation Introduction

Interpretation: The standard cell potential $E^{\text{ο}}$ and standard Gibbs free energy $\Delta G^{\text{ο}}$ of the given cell are to be calculated. The Nernst equation for the cell is to be stated. The molar concentration of $\left[{\text{CrO}}_4^{2-}\right]$ for this solution and solubility product $\left(K_{\text{sp}}\right)$ for ${\text{Ag}}_{\text{2}}{\text{CrO}}_{\text{4}}$ is to be calculated.

Concept introduction: Cell potential is defined as the measure of energy per unit charge available from the redox reaction to carry out the reaction. 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.

To determine: The solubility product $\left(K_{\text{sp}}\right)$ for ${\text{Ag}}_{\text{2}}{\text{CrO}}_{\text{4}}$.

Answer to Problem 152CP

Answer

The solubility product is $K_{\text{sp}}$ for ${\text{Ag}}_{\text{2}}{\text{CrO}}_{\text{4}}$ is $\underset{\_}{1.55\times 10^8}$.

Explanation of Solution

Explanation

Given

The cell reaction is,

${\text{Ag}}_{\text{2}}{\text{CrO}}_{\text{4}}\underset{}{\to }2{\text{Ag}}^{+}+{\text{CrO}}_4^{2-}$.

$E^{\text{ο}}=0.242\text{ V}$

The expression for the solubility product is $K_{\text{sp}}={\left[{\text{Ag}}^{\text{+}}\right]}^2\left[{\text{CrO}}_4^{-}\right]$.

According to Nernst equation,

$E_{\text{cell}}=E_{\text{cell}}^{\text{ο}}-\frac{0.0591}{n}\log \left(\frac{\left[{\text{Ag}}^{+}\right]\left[{\text{CrO}}_4^{2-}\right]}{\left[{\text{Ag}}_{\text{2}}{\text{CrO}}_{\text{4}}\right]}\right)$

When equilibrium is existed in the cell then,

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

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

$\begin{array}{c}{E}_{\text{cell}}=E_{\text{cell}}^{\text{ο}}-\frac{0.0591}{n}\log \left(\frac{\left[{\text{Ag}}^{+}\right]\left[{\text{CrO}}_4^{2-}\right]}{\left[{\text{Ag}}_{\text{2}}{\text{CrO}}_{\text{4}}\right]}\right)\\ 0.0\text{0 V}=0.2\text{42 V}-\frac{0.0591}{2}\log \left(\frac{\left[{\text{Ag}}^{+}\right]\left[{\text{CrO}}_4^{2-}\right]}{\left[{\text{Ag}}_{\text{2}}{\text{CrO}}_{\text{4}}\right]}\right)\\ 0.2\text{42 V}=\frac{0.0591}{2}\log \left(\frac{\left[{\text{Ag}}^{+}\right]\left[{\text{CrO}}_4^{2-}\right]}{\left[{\text{Ag}}_{\text{2}}{\text{CrO}}_{\text{4}}\right]}\right)\\ \log \left(\frac{\left[{\text{Ag}}^{+}\right]\left[{\text{CrO}}_4^{2-}\right]}{\left[{\text{Ag}}_{\text{2}}{\text{CrO}}_{\text{4}}\right]}\right)=8.1895\end{array}$

Simplify the above expression.

$\begin{array}{c}\log K_{\text{sp}}={10}^{8.1895}\\ =\underset{\_}{1.55\times 10^8}\end{array}$.

Hence, the solubility product is $K_{\text{sp}}$ for ${\text{Ag}}_{\text{2}}{\text{CrO}}_{\text{4}}$ is $\underset{\_}{1.55\times 10^8}$.