Chapter 18, Problem 18.115QP

(a)

Interpretation Introduction

Interpretation:

The $\text{emf}$ of the cell in the given conditions, energy density of the zinc electrode and the volume of air supplied to the cell in each second has to be calculated.

Concept Introduction:

Free energy (Gibbs free energy) is the term that is used to explain the total energy content in a thermodynamic system that can be converted into work.  The free energy is represented by the letter $\text{G}$.  All spontaneous process is associated with the decrease of free energy in the system.  The standard free energy change ${\text{(ΔG}}^{\text{°}}{}_{\text{rxn}})$ is the difference in free energy of the reactants and products in their standard state.

${\text{ΔG}}^{\text{°}}{}_{\text{rxn}}\text{=}\sum {\text{nΔG}}_{\text{f}}{}^{\text{°}}\text{(Products)-}\sum {\text{nΔG}}_{\text{f}}{}^{\text{°}}\text{(Reactants)}$

Where, n is the number of moles

The relation between Gibbs free energy and cell potential: The amount of energy in a system that can be converted into useful energy is defined as free energy in thermodynamics.

Free energy and the cell potential is related by the given equation.

$\text{ΔG=-nFE}$

Where,

$\text{ΔG}$ is the change in free energy

$\text{n}$ is the number of electrons transferred

$\text{F}$ is the Faraday constant $\text{(F}={\text{96500Cmol}}^{\text{-1}})$

$\text{E}$ is the cell potential

Nernst equation is one of the important equations in electrochemistry.  In Nernst equation the electrode potential of a cell reaction is related to the standard electrode potential, concentration or activities of the species that is involved in the chemical reaction and temperature.

${\text{E}}_{\text{cell}}{\text{=E}}^{\text{°}}{}_{\text{cell}}\text{-}\frac{\text{RT}}{\text{2}\text{.303nF}}\text{log}\frac{\left[\text{Red}\right]}{\left[\text{Oxd}\right]}$

Where,

${\text{E}}_{\text{cell}}$ is the potential of the cell at a given temperature

${\text{E}}^{\text{°}}{}_{\text{cell}}$ is the standard electrode potential

$\text{R}$ is the universal gas constant $\text{(R=8}{\text{.314JK}}^{\text{-1}}{\text{mol}}^{\text{-1}})$

$\text{T}$ is the temperature

$\text{n}$ is the number of electrons involved in a reaction

$\text{F}$ is the Faraday constant $\text{(F=9}\text{.64853399}\times {\text{10}}^4{\text{Cmol}}^{\text{-1}})$

$\left[\text{Red}\right]$ is the concentration of the reduced species

$\left[\text{Oxd}\right]$ is the concentration of the oxidised species

At room temperature $({\text{25}}^{\text{°}}\text{C})$, after substituting the values of all the constants the equation can be written as

${\text{E}}_{\text{cell}}{\text{= E}}^{\text{°}}{}_{\text{cell}}\text{-}\frac{0.0591}{\text{n}}\text{log}\frac{\left[\text{Red}\right]}{\left[\text{Oxd}\right]}$

Ideal gas equation is an equation that is describing the state of a imaginary ideal gas.

${\text{PV}}_{}\text{=n RT}$

Where,

$\text{P}$ is the pressure of the gas

$\text{V}$ is the volume

$\text{n}$ is the number of moles of gas

$\text{R}$ is the universal gas constant $\text{(R=0}{\text{.0821LatmK}}^{\text{-1}}{\text{mol}}^{\text{-1}})$

$\text{T}$ is the temperature

Answer to Problem 18.115QP

Answer:

The half cell reactions of the given cell,

$\begin{array}{l}\text{Zn}\to {\text{Zn}}^{\text{2+}}{\text{+2e}}^{\text{-}}\text{Anode}\\ \frac{1}{2}{\text{O}}_{\text{2}}+{\text{2e}}^{\text{-}}\to {\text{O}}^{\text{2-}}\text{Cathode}\end{array}$

The standard $\text{emf}$ of the cell is found to be $\text{1}\text{.65V}$

Explanation of Solution

Explanation:

To record the given data

$\begin{array}{l}{\text{G}}_{\text{f}}{}^{\text{°}}\text{(ZnO)}=\text{-318}{\text{.2KJmol}}^{\text{-1}}\\ \\ {\text{ΔG}}_{\text{f}}{}^{\text{°}}\text{(Zn)}=0\\ \\ {\text{ΔG}}_{\text{f}}{}^{\text{°}}{\text{(O}}_{\text{2}}\text{)}=0\end{array}$

To write the half cell reactions and overall reaction

The half cell reactions are,

$\begin{array}{l}\text{Zn}\to {\text{Zn}}^{\text{2+}}{\text{+2e}}^{\text{-}}\text{Anode}\\ \frac{1}{2}{\text{O}}_{\text{2}}+{\text{2e}}^{\text{-}}\to {\text{O}}^{\text{2-}}\text{Cathode}\end{array}$

Overall reaction,

$\text{Zn+}\frac{1}{2}{\text{O}}_{\text{2}}\to \text{ZnO}$

To find the ${\text{ΔG}}^{\text{°}}{}_{\text{rxn}}$ of the given reaction

The of ${\text{ΔG}}^{\text{°}}{}_{\text{rxn}}$ the given reaction is calculated by plugging in the values of standard free energy of the reactants and products in the given equation.

${\text{ΔG}}^{\text{°}}{}_{\text{rxn}}\text{=}\sum {\text{nΔG}}_{\text{f}}{}^{\text{°}}\text{(Products)-}\sum {\text{nΔG}}_{\text{f}}{}^{\text{°}}\text{(Reactants)}$

$\begin{array}{l}{\text{=ΔG}}_{\text{f}}{}^{\text{°}}\text{(ZnO)-}\left[{\text{ΔG}}_{\text{f}}{}^{\text{°}}\text{(Zn)+}\frac{\text{1}}{\text{2}}{\text{ΔG}}_{\text{f}}{}^{\text{°}}{\text{(O}}_{\text{2}}\text{)}\right]\\ \text{=}\text{-318}{\text{.2KJmol}}^{\text{-1}}\text{-}\left[\text{0+0}\right]\\ \text{=}\text{-318}{\text{.2kJmol}}^{\text{-1}}\end{array}$

To find the ${\text{E}}^{\text{°}}{}_{\text{cell}}$

Using the value of free energy and the number of electrons transferred the ${\text{E}}^{\text{°}}{}_{\text{cell}}$ can be calculated.

${\text{ΔG}}^{°}{\text{=-nFE}}^{°}$

On rearranging the equation we get,

$\begin{array}{l}{\text{E}}^{°}=\frac{{\text{ΔG}}^{°}}{\text{-nF}}\\ \text{=}\frac{\text{-318}{\text{.2×10}}^{\text{3}}{\text{Jmol}}^{\text{-1}}}{{\text{(2)(96500)JV}}^{\text{-1}}\text{mol)}}\\ \text{=1}\text{.65V}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The $\text{emf}$ of the cell in the given conditions, energy density of the zinc electrode and the volume of air supplied to the cell in each second has to be calculated.

Concept Introduction:

Free energy (Gibbs free energy) is the term that is used to explain the total energy content in a thermodynamic system that can be converted into work.  The free energy is represented by the letter $\text{G}$.  All spontaneous process is associated with the decrease of free energy in the system.  The standard free energy change ${\text{(ΔG}}^{\text{°}}{}_{\text{rxn}})$ is the difference in free energy of the reactants and products in their standard state.

${\text{ΔG}}^{\text{°}}{}_{\text{rxn}}\text{=}\sum {\text{nΔG}}_{\text{f}}{}^{\text{°}}\text{(Products)-}\sum {\text{nΔG}}_{\text{f}}{}^{\text{°}}\text{(Reactants)}$

Where, n is the number of moles

The relation between Gibbs free energy and cell potential: The amount of energy in a system that can be converted into useful energy is defined as free energy in thermodynamics.

Free energy and the cell potential is related by the given equation.

$\text{ΔG=-nFE}$

Where,

$\text{ΔG}$ is the change in free energy

$\text{n}$ is the number of electrons transferred

$\text{F}$ is the Faraday constant $\text{(F}={\text{96500Cmol}}^{\text{-1}})$

$\text{E}$ is the cell potential

Nernst equation is one of the important equations in electrochemistry.  In Nernst equation the electrode potential of a cell reaction is related to the standard electrode potential, concentration or activities of the species that is involved in the chemical reaction and temperature.

${\text{E}}_{\text{cell}}{\text{=E}}^{\text{°}}{}_{\text{cell}}\text{-}\frac{\text{RT}}{\text{2}\text{.303nF}}\text{log}\frac{\left[\text{Red}\right]}{\left[\text{Oxd}\right]}$

Where,

${\text{E}}_{\text{cell}}$ is the potential of the cell at a given temperature

${\text{E}}^{\text{°}}{}_{\text{cell}}$ is the standard electrode potential

$\text{R}$ is the universal gas constant $\text{(R=8}{\text{.314JK}}^{\text{-1}}{\text{mol}}^{\text{-1}})$

$\text{T}$ is the temperature

$\text{n}$ is the number of electrons involved in a reaction

$\text{F}$ is the Faraday constant $\text{(F=9}\text{.64853399}\times {\text{10}}^4{\text{Cmol}}^{\text{-1}})$

$\left[\text{Red}\right]$ is the concentration of the reduced species

$\left[\text{Oxd}\right]$ is the concentration of the oxidised species

At room temperature $({\text{25}}^{\text{°}}\text{C})$, after substituting the values of all the constants the equation can be written as

${\text{E}}_{\text{cell}}{\text{= E}}^{\text{°}}{}_{\text{cell}}\text{-}\frac{0.0591}{\text{n}}\text{log}\frac{\left[\text{Red}\right]}{\left[\text{Oxd}\right]}$

Ideal gas equation is an equation that is describing the state of a imaginary ideal gas.

${\text{PV}}_{}\text{=n RT}$

Where,

$\text{P}$ is the pressure of the gas

$\text{V}$ is the volume

$\text{n}$ is the number of moles of gas

$\text{R}$ is the universal gas constant $\text{(R=0}{\text{.0821LatmK}}^{\text{-1}}{\text{mol}}^{\text{-1}})$

$\text{T}$ is the temperature

Answer to Problem 18.115QP

Answer:

The $\text{emf}$ of the cell when the partial pressure of oxygen is $0.21\text{atm}$ is found to be $\text{1}\text{.63V}$

Explanation of Solution

Explanation:

To record the given data

Partial pressure of oxygen $=0.21\text{atm}$

${\text{E}}^{\text{°}}{}_{\text{cell}}=1.65\text{V}$

To calculate the $\text{emf}$ of the cell

The $\text{emf}$ of the cell at the given conditions can be calculated by using the modified Nernst equation.

$\begin{array}{l}{\text{E}}_{\text{cell}}\text{= E°-}\frac{\text{0}\text{.05912V}}{\text{n}}\text{log}\frac{{\text{P}}_{\text{ZnO}}}{{\text{P}}_{\text{Zn}}{\text{P}}_{{\text{O}}_{\text{2}}}}\\ \text{= 1}\text{.65V-}\frac{\text{0}\text{.0591V}}{2}\text{log}\frac{\text{1}}{{\text{P}}_{{\text{O}}_{\text{2}}}}\\ =\text{1}\text{.65V-}\frac{\text{0}\text{.0591V}}{2}\text{log}\frac{\text{1}}{\text{0}\text{.21}}\\ =\text{1}\text{.63V}\end{array}$

(c)

Interpretation Introduction

Interpretation:

The $\text{emf}$ of the cell in the given conditions, energy density of the zinc electrode and the volume of air supplied to the cell in each second has to be calculated.

Concept Introduction:

Free energy (Gibbs free energy) is the term that is used to explain the total energy content in a thermodynamic system that can be converted into work.  The free energy is represented by the letter $\text{G}$.  All spontaneous process is associated with the decrease of free energy in the system.  The standard free energy change ${\text{(ΔG}}^{\text{°}}{}_{\text{rxn}})$ is the difference in free energy of the reactants and products in their standard state.

${\text{ΔG}}^{\text{°}}{}_{\text{rxn}}\text{=}\sum {\text{nΔG}}_{\text{f}}{}^{\text{°}}\text{(Products)-}\sum {\text{nΔG}}_{\text{f}}{}^{\text{°}}\text{(Reactants)}$

Where, n is the number of moles

The relation between Gibbs free energy and cell potential: The amount of energy in a system that can be converted into useful energy is defined as free energy in thermodynamics.

Free energy and the cell potential is related by the given equation.

$\text{ΔG=-nFE}$

Where,

$\text{ΔG}$ is the change in free energy

$\text{n}$ is the number of electrons transferred

$\text{F}$ is the Faraday constant $\text{(F}={\text{96500Cmol}}^{\text{-1}})$

$\text{E}$ is the cell potential

Nernst equation is one of the important equations in electrochemistry.  In Nernst equation the electrode potential of a cell reaction is related to the standard electrode potential, concentration or activities of the species that is involved in the chemical reaction and temperature.

${\text{E}}_{\text{cell}}{\text{=E}}^{\text{°}}{}_{\text{cell}}\text{-}\frac{\text{RT}}{\text{2}\text{.303nF}}\text{log}\frac{\left[\text{Red}\right]}{\left[\text{Oxd}\right]}$

Where,

${\text{E}}_{\text{cell}}$ is the potential of the cell at a given temperature

${\text{E}}^{\text{°}}{}_{\text{cell}}$ is the standard electrode potential

$\text{R}$ is the universal gas constant $\text{(R=8}{\text{.314JK}}^{\text{-1}}{\text{mol}}^{\text{-1}})$

$\text{T}$ is the temperature

$\text{n}$ is the number of electrons involved in a reaction

$\text{F}$ is the Faraday constant $\text{(F=9}\text{.64853399}\times {\text{10}}^4{\text{Cmol}}^{\text{-1}})$

$\left[\text{Red}\right]$ is the concentration of the reduced species

$\left[\text{Oxd}\right]$ is the concentration of the oxidised species

At room temperature $({\text{25}}^{\text{°}}\text{C})$, after substituting the values of all the constants the equation can be written as

${\text{E}}_{\text{cell}}{\text{= E}}^{\text{°}}{}_{\text{cell}}\text{-}\frac{0.0591}{\text{n}}\text{log}\frac{\left[\text{Red}\right]}{\left[\text{Oxd}\right]}$

Ideal gas equation is an equation that is describing the state of a imaginary ideal gas.

${\text{PV}}_{}\text{=n RT}$

Where,

$\text{P}$ is the pressure of the gas

$\text{V}$ is the volume

$\text{n}$ is the number of moles of gas

$\text{R}$ is the universal gas constant $\text{(R=0}{\text{.0821LatmK}}^{\text{-1}}{\text{mol}}^{\text{-1}})$

$\text{T}$ is the temperature

Answer to Problem 18.115QP

Answer:

The energy density of the zinc electrode is found to be $\text{4}\text{.86}\times {\text{10}}^3\text{kJ/kg}$

Explanation of Solution

Explanation:

To record the given data

Amount of zinc $=1\text{kg}$

Molecular weight of zinc $=65.41{\text{gmol}}^{-1}$

To calculate the number of moles of zinc

Number of moles of zinc in $\text{1kg}$ zinc can be calculated as given below.

$\begin{array}{l}\text{Number}\text{of}\text{moles}\text{of}\text{zinc=}\frac{1000g}{65.41{\text{gmol}}^{\text{-1}}}\\ =15.28\text{mol}\end{array}$

To calculate the energy density of zinc electrode

Free energy is the maximum amount of energy in the system that can be converted into useful work.  Energy density can be obtained by multiplying the free energy value with the number of moles of zinc.

$\begin{array}{l}\text{Energy density =}\frac{\text{318}\text{.2kJ}}{\text{1mol}\text{Zn}}\text{×}\frac{\text{1mol}\text{Zn}}{\text{65}\text{.41g}\text{Zn}}\text{×}\frac{\text{1000g}\text{Zn}}{\text{1}\text{kg}\text{Zn}}\\ \text{ =}\text{4}\text{.86}\times {\text{10}}^3\text{kJ/kg}\text{Zn}\end{array}$

(d)

Interpretation Introduction

Interpretation:

The $\text{emf}$ of the cell in the given conditions, energy density of the zinc electrode and the volume of air supplied to the cell in each second has to be calculated.

Concept Introduction:

Free energy (Gibbs free energy) is the term that is used to explain the total energy content in a thermodynamic system that can be converted into work.  The free energy is represented by the letter $\text{G}$.  All spontaneous process is associated with the decrease of free energy in the system.  The standard free energy change ${\text{(ΔG}}^{\text{°}}{}_{\text{rxn}})$ is the difference in free energy of the reactants and products in their standard state.

${\text{ΔG}}^{\text{°}}{}_{\text{rxn}}\text{=}\sum {\text{nΔG}}_{\text{f}}{}^{\text{°}}\text{(Products)-}\sum {\text{nΔG}}_{\text{f}}{}^{\text{°}}\text{(Reactants)}$

Where, n is the number of moles

The relation between Gibbs free energy and cell potential: The amount of energy in a system that can be converted into useful energy is defined as free energy in thermodynamics.

Free energy and the cell potential is related by the given equation.

$\text{ΔG=-nFE}$

Where,

$\text{ΔG}$ is the change in free energy

$\text{n}$ is the number of electrons transferred

$\text{F}$ is the Faraday constant $\text{(F}={\text{96500Cmol}}^{\text{-1}})$

$\text{E}$ is the cell potential

Nernst equation is one of the important equations in electrochemistry.  In Nernst equation the electrode potential of a cell reaction is related to the standard electrode potential, concentration or activities of the species that is involved in the chemical reaction and temperature.

${\text{E}}_{\text{cell}}{\text{=E}}^{\text{°}}{}_{\text{cell}}\text{-}\frac{\text{RT}}{\text{2}\text{.303nF}}\text{log}\frac{\left[\text{Red}\right]}{\left[\text{Oxd}\right]}$

Where,

${\text{E}}_{\text{cell}}$ is the potential of the cell at a given temperature

${\text{E}}^{\text{°}}{}_{\text{cell}}$ is the standard electrode potential

$\text{R}$ is the universal gas constant $\text{(R=8}{\text{.314JK}}^{\text{-1}}{\text{mol}}^{\text{-1}})$

$\text{T}$ is the temperature

$\text{n}$ is the number of electrons involved in a reaction

$\text{F}$ is the Faraday constant $\text{(F=9}\text{.64853399}\times {\text{10}}^4{\text{Cmol}}^{\text{-1}})$

$\left[\text{Red}\right]$ is the concentration of the reduced species

$\left[\text{Oxd}\right]$ is the concentration of the oxidised species

At room temperature $({\text{25}}^{\text{°}}\text{C})$, after substituting the values of all the constants the equation can be written as

${\text{E}}_{\text{cell}}{\text{= E}}^{\text{°}}{}_{\text{cell}}\text{-}\frac{0.0591}{\text{n}}\text{log}\frac{\left[\text{Red}\right]}{\left[\text{Oxd}\right]}$

Ideal gas equation is an equation that is describing the state of a imaginary ideal gas.

${\text{PV}}_{}\text{=n RT}$

Where,

$\text{P}$ is the pressure of the gas

$\text{V}$ is the volume

$\text{n}$ is the number of moles of gas

$\text{R}$ is the universal gas constant $\text{(R=0}{\text{.0821LatmK}}^{\text{-1}}{\text{mol}}^{\text{-1}})$

$\text{T}$ is the temperature

Answer to Problem 18.115QP

Answer:

The amount of air supplied to the battery in each second is found to be $64\text{L}$

Explanation of Solution

Explanation:

To record the given data

Amount of current derived from the cell $=2.1\times {10}^5\text{A}$

To calculate the number of moles of electrons required for producing given amount of charge

Charge produced and the numbers of moles of electrons transferred are related by the following equation.

$\text{Charge=nF}$

The number of moles of electrons transferred,

$\begin{array}{l}\text{n}\text{=}\frac{\text{C}}{\text{F}}\\ =\frac{\text{2}{\text{.1×10}}^5\text{C}}{{\text{96,500Cmol}}^{\text{-1}}}\\ =2.2\text{mol}\end{array}$

To calculate the number of moles of oxygen gas reduced by $2.2\text{mol}$ of electrons.

From the equation for the cell reaction we have seen that $4\text{moles}$ of electrons are required to reduce one mole of oxygen gas.  Hence, the number of moles of oxygen reduced by $2.2\text{mol}$ of electrons can be calculated as given below.

$\begin{array}{l}{\text{Number of moles of O}}_{\text{2}}\text{=2}{\text{.2 mol e}}^{-}\times \frac{{\text{1molO}}_{\text{2}}}{\text{4}\text{mol}{\text{e}}^{\text{-}}}=\text{0}\text{.55}\text{mol}{\text{O}}_{\text{2}}\\ \end{array}$

To calculate the volume of oxygen when the partial pressure is $1\text{atm}$.

The volume of oxygen at $1\text{atm}$ partial pressure can be calculated using ideal gas equation.

$\begin{array}{l}{\text{V}}_{{\text{O}}_{\text{2}}}\text{=}\frac{\text{n RT}}{\text{P}}\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{=}\frac{\text{(0}\text{.55mol}{\text{O}}_{\text{2}}\text{)}\left[\text{0}{\text{.0821LatmK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\right]\text{(298K)}}{1\text{atm}}\\ =\text{13}\text{.5L}\\ \end{array}$

To calculate the volume of air required at each second.

The volume of air required at each second is found as given below.

$\begin{array}{l}{\text{V}}_{\text{air}}\text{=}13.5\text{L}\times \frac{\text{100\%}\text{}\text{air}}{\text{21\%}\text{}{\text{O}}_2}=\text{64L}\\ \end{array}$