Chapter 18, Problem 18.102QP

(a)

Interpretation Introduction

Interpretation:

The half-cell reactions and the overall reactions in the given cell has to be calculated.

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.

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

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

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

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

The relation between electrode potential and equilibrium constant: cell potential and equilibrium constant are related by the given equation.

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

Where,

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

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

$\text{K}$ is the formation constant

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

$\text{T}$ is the temperature

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

The standard electrode potential of a cell ${\text{(E}}^{°}{}_{\text{cell}})$ 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 18.102QP

The half-cell reactions are,

$\begin{array}{l}{\text{Sn}}^{\text{4+}}{\text{(aq)+2e}}^{\text{-}}\to {\text{Sn}}^{\text{2+}}\text{(aq)}{\text{E}}^{°}{}_{cathode}=0.13\text{V}\\ \text{2Tl(s)}\to {\text{Tl}}^{\text{+}}{\text{(aq)+2Tl}}^{\text{+}}\text{(aq)}{\text{E}}^{\text{°}}{}_{anode}\text{=}-\text{0}\text{.34V}\end{array}$

The overall reaction,

${\text{Sn}}^{4+}\text{(aq)+2Tl(s)}\to {\text{Sn}}^{\text{2+}}{\text{(aq)+2Tl}}^{+}\text{(aq)}$

Explanation of Solution

To find the completely balanced chemical equation

The overall cell reaction can be written as given below,

$\begin{array}{l}{\text{Sn}}^{\text{4+}}{\text{(aq)+2e}}^{\text{-}}\to {\text{Sn}}^{\text{2+}}\text{(aq)}{\text{E}}^{°}{}_{cathode}=0.13\text{V}\\ \text{2Tl(s)}\to {\text{Tl}}^{\text{+}}{\text{(aq)+2Tl}}^{\text{+}}\text{(aq)}{\text{E}}^{\text{°}}{}_{anode}\text{=}-\text{0}\text{.34V}\end{array}$

Overall reaction ${\text{Sn}}^{4+}\text{(aq)+2Tl(s)}\to {\text{Sn}}^{\text{2+}}{\text{(aq)+2Tl}}^{+}\text{(aq)}$

(b)

Interpretation Introduction

Interpretation:

The equilibrium constant for the reaction has to be calculated in accordance with the given conditions.

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.

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

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

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

The relation between electrode potential and equilibrium constant: cell potential and equilibrium constant are related by the given equation.

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

Where,

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

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

$\text{K}$ is the formation constant

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

$\text{T}$ is the temperature

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

The standard electrode potential of a cell ${\text{(E}}^{°}{}_{\text{cell}})$ 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 18.102QP

The equilibrium constant at ${25}^{°}C$ is found to be, $\text{K}=8\times {10}^{15}$

Explanation of Solution

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

The standard cell potential can be calculated as given below

$\begin{array}{l}{\text{E}}^{\text{°}}{}_{\text{cell}}={\text{E}}^{\text{°}}{}_{\text{cathode}}-{\text{E}}^{\text{°}}{}_{\text{anode}}\\ =0.13\text{V-(-0}\text{.34V)}\\ =0.47\text{V}\end{array}$

To calculate the equilibrium constant.

The equilibrium constant and the electrode potential is related by the given equation.

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

On rearranging we get,

$\begin{array}{l}\text{lnK}=\frac{{\text{nFE}}^{\text{°}}{}_{\text{cell}}}{\text{RT}}\\ =\frac{{\text{(2)(96500JVmol}}^{\text{-1}}\text{)(0}\text{.47V)}}{\text{(8}{\text{.314JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{)}}\\ =36.61\\ \end{array}$

Hence,

$\begin{array}{l}{\text{K}}_{}{\text{ = e}}^{36.61}\\ \text{=8}\times {\text{10}}^{15}\end{array}$

(c)

Interpretation Introduction

Interpretation:

The cell potential has to be calculated in accordance with the given conditions.

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.

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

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

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

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

The relation between electrode potential and equilibrium constant: cell potential and equilibrium constant are related by the given equation.

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

Where,

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

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

$\text{K}$ is the formation constant

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

$\text{T}$ is the temperature

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

The standard electrode potential of a cell ${\text{(E}}^{°}{}_{\text{cell}})$ 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 18.102QP

The cell potential, when the concentration of ${\text{Tl}}^{\text{+}}$ of increased by a factor of ten is found to be $0.41\text{V}$

Explanation of Solution

To record the given data

The concentration of ${\text{Sn}}^{\text{2+}}=1.0\text{M}$

The concentration of ${\text{Sn}}^{\text{4+}}=1.0\text{M}$

The concentration of ${\text{Tl}}^{+}=1.0\text{M}$

Temperature $=298K$

To find the cell voltage when the concentration ${\text{Tl}}^{\text{+}}$of increased by a factor of ten.

The ${\text{E}}_{\text{cell}}$ can be calculated using the Nernst equation.

$\begin{array}{l}{\text{E}}_{\text{cell}}\text{=E°-}\frac{\text{0}\text{.05912V}}{\text{n}}\text{log}\frac{\left[{\text{Sn}}^{\text{2+}}\right]{\left[{\text{10Tl}}^{+}\right]}^2}{\left[{\text{Sn}}^{4+}\right]}\\ \text{=0}\text{.47V-}\frac{\text{0}\text{.0591V}}{2}\text{log}\frac{\left[\text{1}\text{.0}\right]\left[10.0\right]}{\left[1.0\right]}\\ =0.41\text{V}\end{array}$