Chapter 18, Problem 18.118QP

Consider a Daniell cell operating under nonstandard-state conditions. Suppose that the cell’s reaction is multiplied by 2. What effect does this have on each of the following quantities in the Nernst equation: (a) E, (b) E°, (c) Q, (d) ln Q, (e) n?

Interpretation Introduction

Interpretation:

The changes ${\text{E}}_{\text{cell}}$ ,${\text{E}}^{°}{}_{\text{cell}}$, equilibrium constant $\left(\text{Q}\right)$, $\ln \text{Q}$ and number of electrons transferred $\left(\text{n}\right)$ have to be explained when a Daniel cell equation is multiplied by a factor of two.

Concept Introduction:

Nernst equation is one of the important equation 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.

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

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

${\text{E}}_{\text{cell}}{\text{= E}}^{\text{°}}{}_{\text{cell}}\text{-}\frac{0.0591}{\text{n}}\text{lnQ}$

The standard electrode potential ${\text{(E}}^{°}{}_{\text{cell}})$ of a cell is the difference in electrode potential of the cathode and anode.

${\text{E}}^{\text{°}}{}_{\text{cell}}={\text{E}}^{\text{°}}{}_{\text{cathode}}-{\text{E}}^{\text{°}}{}_{\text{anode}}$

Answer to Problem 18.118QP

On multiplying the overall cell equation for Daniel cell by a factor of two

Cell potential $\left({\text{E}}_{\text{cell}}\right)$ remains the same.

Explanation of Solution

To explain the how cell potential ${\text{(E}}_{\text{cell}})$ of Daniel cell changes when the cell equation is multiplied by a factor of two.

The overall reaction in the Daniel cell is found to be,

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

Nernst equation for the above reaction is given below,

${\text{E}}_{\text{cell}}{\text{=E}}^{\text{°}}{}_{\text{cell}}\text{-}\frac{\text{RT}}{\text{2F}}\text{ln}\frac{\left[{\text{Zn}}^{\text{2+}}\text{(aq)}\right]}{\left[{\text{Cu}}^{\text{2+}}\text{(aq)}\right]}$

${\text{E}}_{\text{cell}}{\text{=E}}^{\text{°}}{}_{\text{cell}}\text{-}\frac{\text{RT}}{\text{2F}}\text{lnQ}$ (1)

Lets multiply the overall reaction by two and write the Nernst equation for the new equilibrium.

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

On multiplying the overall cell equation with two, the total number of electrons transferred becomes twice of the initial number and the equilibrium constant is squared.

${\text{E}}_{\text{cell}}{\text{=E}}^{\text{°}}{}_{\text{cell}}\text{-}\frac{\text{RT}}{\text{4F}}{\text{lnQ}}^2$ (2)

Equation number $2$ can be written as,

${\text{E}}_{\text{cell}}{\text{=E}}^{\text{°}}{}_{\text{cell}}\text{-}\frac{\text{RT}}{\text{4F}}\text{2lnQ}$ (3)

On simplifying equation $3$, we get equation $1$

$\begin{array}{l}{\text{E}}_{\text{cell}}{\text{=E}}^{\text{°}}{}_{\text{cell}}\text{-}\frac{\text{RT}}{\cancel{\text{4}}\text{F}}\cancel{\text{2}}\text{lnQ }\\ {\text{= E}}^{\text{°}}{}_{\text{cell}}-\frac{\text{RT}}{\text{2F}}\text{lnQ}\end{array}$

Hence, it is clear that on multiplying the overall cell equation by a factor of two, cell potential remains the same.

Interpretation Introduction

Interpretation:

The changes ${\text{E}}_{\text{cell}}$ ,${\text{E}}^{°}{}_{\text{cell}}$, equilibrium constant $\left(\text{Q}\right)$, $\ln \text{Q}$ and number of electrons transferred $\left(\text{n}\right)$ have to be explained when a Daniel cell equation is multiplied by a factor of two.

Concept Introduction:

Nernst equation is one of the important equation 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.

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

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

${\text{E}}_{\text{cell}}{\text{= E}}^{\text{°}}{}_{\text{cell}}\text{-}\frac{0.0591}{\text{n}}\text{lnQ}$

The standard electrode potential ${\text{(E}}^{°}{}_{\text{cell}})$ of a cell is the difference in electrode potential of the cathode and anode.

${\text{E}}^{\text{°}}{}_{\text{cell}}={\text{E}}^{\text{°}}{}_{\text{cathode}}-{\text{E}}^{\text{°}}{}_{\text{anode}}$

Answer to Problem 18.118QP

On multiplying the overall cell equation for Daniel cell by a factor of two

$\left({\text{E}}^{\text{°}}{}_{\text{cell}}\right)$ remains unchanged.

Explanation of Solution

To explain the how ${\text{E}}^{\text{°}}{}_{\text{cell}}$ of Daniel cell changes when the cell equation is multiplied by a factor of two.

The overall reaction in the Daniel cell is found to be,

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

Nernst equation for the above reaction is given below,

${\text{E}}_{\text{cell}}{\text{=E}}^{\text{°}}{}_{\text{cell}}\text{-}\frac{\text{RT}}{\text{2}\text{.303nF}}\text{log}\frac{\left[{\text{Zn}}^{\text{2+}}\text{(aq)}\right]}{\left[{\text{Cu}}^{\text{2+}}\text{(aq)}\right]}$

The standard electrode potential depends only upon the standard reduction potentials of cathode and anode.  It is independent upon the stoichiometric coefficients of the reactants and products.  Hence, on multiplying the overall equation by two ${\text{E}}^{\text{°}}{}_{\text{cell}}$ remains unchanged.

Interpretation Introduction

Interpretation:

The changes ${\text{E}}_{\text{cell}}$ ,${\text{E}}^{°}{}_{\text{cell}}$, equilibrium constant $\left(\text{Q}\right)$, $\ln \text{Q}$ and number of electrons transferred $\left(\text{n}\right)$ have to be explained when a Daniel cell equation is multiplied by a factor of two.

Concept Introduction:

Nernst equation is one of the important equation 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.

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

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

${\text{E}}_{\text{cell}}{\text{= E}}^{\text{°}}{}_{\text{cell}}\text{-}\frac{0.0591}{\text{n}}\text{lnQ}$

The standard electrode potential ${\text{(E}}^{°}{}_{\text{cell}})$ of a cell is the difference in electrode potential of the cathode and anode.

${\text{E}}^{\text{°}}{}_{\text{cell}}={\text{E}}^{\text{°}}{}_{\text{cathode}}-{\text{E}}^{\text{°}}{}_{\text{anode}}$

Answer to Problem 18.118QP

On multiplying the overall cell equation for Daniel cell by a factor of two

The equilibrium constant $\left(\text{Q}\right)$ remains unchanged.

Explanation of Solution

To explain the how the equilibrium constant $\text{(Q)}$ of Daniel cell changes when the cell equation is multiplied by a factor of two.

Explanation:

The overall reaction in the Daniel cell is found to be,

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

Nernst equation for the above reaction is given below,

${\text{E}}_{\text{cell}}{\text{=E}}^{\text{°}}{}_{\text{cell}}\text{-}\frac{\text{RT}}{\text{2}\text{.303nF}}\text{log}\frac{\left[{\text{Zn}}^{\text{2+}}\text{(aq)}\right]}{\left[{\text{Cu}}^{\text{2+}}\text{(aq)}\right]}$

The equilibrium constant,                  $\text{Q=}\frac{\left[{\text{Zn}}^{\text{2+}}\text{(aq)}\right]}{\left[{\text{Cu}}^{\text{2+}}\text{(aq)}\right]}$

${\text{E}}_{\text{cell}}{\text{=E}}^{\text{°}}{}_{\text{cell}}\text{-}\frac{\text{RT}}{\text{2F}}\text{lnQ}$

On multiplying the equation by two the equilibrium constant becomes,

${\text{E}}_{\text{cell}}{\text{=E}}^{\text{°}}{}_{\text{cell}}\text{-}\frac{\text{RT}}{\text{4F}}{\text{lnQ}}^2$

On multiplying the overall equation by two, the equilibrium constant is squared.

Interpretation Introduction

Interpretation:

The changes ${\text{E}}_{\text{cell}}$ ,${\text{E}}^{°}{}_{\text{cell}}$, equilibrium constant $\left(\text{Q}\right)$, $\ln \text{Q}$ and number of electrons transferred $\left(\text{n}\right)$ have to be explained when a Daniel cell equation is multiplied by a factor of two.

Concept Introduction:

Nernst equation is one of the important equation 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.

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

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

${\text{E}}_{\text{cell}}{\text{= E}}^{\text{°}}{}_{\text{cell}}\text{-}\frac{0.0591}{\text{n}}\text{lnQ}$

The standard electrode potential ${\text{(E}}^{°}{}_{\text{cell}})$ of a cell is the difference in electrode potential of the cathode and anode.

${\text{E}}^{\text{°}}{}_{\text{cell}}={\text{E}}^{\text{°}}{}_{\text{cathode}}-{\text{E}}^{\text{°}}{}_{\text{anode}}$

Answer to Problem 18.118QP

On multiplying the overall cell equation for Daniel cell by a factor of two

The value of $\text{lnQ}$ is doubled

Explanation of Solution

To explain the how $\text{lnQ}$ of Daniel cell changes when the cell equation is multiplied by a factor of two.

The overall reaction in the Daniel cell is found to be,

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

Nernst equation for the above reaction is given below,

${\text{E}}_{\text{cell}}{\text{=E}}^{\text{°}}{}_{\text{cell}}\text{-}\frac{\text{RT}}{\text{nF}}\text{ln}\frac{\left[{\text{Zn}}^{\text{2+}}\text{(aq)}\right]}{\left[{\text{Cu}}^{\text{2+}}\text{(aq)}\right]}$

or

${\text{E}}_{\text{cell}}{\text{=E}}^{\text{°}}{}_{\text{cell}}\text{-}\frac{\text{RT}}{\text{nF}}\text{lnQ}$

On multiplying the equation by two,

${\text{E}}_{\text{cell}}{\text{=E}}^{\text{°}}{}_{\text{cell}}\text{-}\frac{\text{RT}}{\text{nF}}{\text{lnQ}}^2$

This equation can also represented as,

${\text{E}}_{\text{cell}}{\text{=E}}^{\text{°}}{}_{\text{cell}}\text{-}\frac{\text{RT}}{\text{nF}}\text{2lnQ}$

On multiplying the equation by two, the value of $\text{lnQ}$ is doubled

Interpretation Introduction

Interpretation:

The changes ${\text{E}}_{\text{cell}}$ ,${\text{E}}^{°}{}_{\text{cell}}$, equilibrium constant $\left(\text{Q}\right)$, $\ln \text{Q}$ and number of electrons transferred $\left(\text{n}\right)$ have to be explained when a Daniel cell equation is multiplied by a factor of two.

Concept Introduction:

Nernst equation is one of the important equation 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.

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

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

${\text{E}}_{\text{cell}}{\text{= E}}^{\text{°}}{}_{\text{cell}}\text{-}\frac{0.0591}{\text{n}}\text{lnQ}$

The standard electrode potential ${\text{(E}}^{°}{}_{\text{cell}})$ of a cell is the difference in electrode potential of the cathode and anode.

${\text{E}}^{\text{°}}{}_{\text{cell}}={\text{E}}^{\text{°}}{}_{\text{cathode}}-{\text{E}}^{\text{°}}{}_{\text{anode}}$

Answer to Problem 18.118QP

On multiplying the overall cell equation for Daniel cell by a factor of two

The number of electrons $\left(\text{n}\right)$ involved in the given reaction is doubled.

Explanation of Solution

To explain the how number of electrons $\left(\text{n}\right)$ of a cell changes when the cell equation is multiplied by a factor of two.

The overall reaction in the Daniel cell is found to be,

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

The half cell reactions are,

$\begin{array}{l}{\text{Zn}}_{\text{(s)}}\to {\text{ Zn}}^{\text{2+}}{}_{\text{(aq)}}{\text{ + 2e}}^{\text{-}} \\ {\text{Cu}}^{\text{2+}}{}_{\text{(aq)}} +{\text{ 2e}}^{\text{-}} \to {\text{ Cu}}_{\text{(s)}}\end{array}$

On multiplying each half cell equations by two

$\begin{array}{l}{\text{2Zn}}_{\text{(s)}}\to {\text{ 2Zn}}^{\text{2+}}{}_{\text{(aq)}}{\text{ + 4e}}^{\text{-}} \\ {\text{2Cu}}^{\text{2+}}{}_{\text{(aq)}} +{\text{ 4e}}^{\text{-}} \to {\text{ Cu}}_{\text{(s)}}\end{array}$

Hence, it is clear that the number of electrons involved in the given reaction is doubled when the overall reaction is multiplied by a factor of two