Chapter 6, Problem 6.3IA

(a)

Interpretation Introduction

Interpretation:

The Nernst equation for the cell, $\text{Zn}\left(\text{s}\right)|{\text{ZnCl}}_2\left(\text{0}\text{.0050}\text{mol}{\text{kg}}^{-1}\right)|{\text{Hg}}_2{\text{Cl}}_2|\text{Hg}\left(\text{l}\right)$ has to be stated.

Concept introduction:

Nernst equation is the relation between standard electrode potential and the electrode potential at given conditions of pressures, temperatures and concentrations.  The expression for Nernst equation is,

    $E_{\text{cell}}=E_{\text{cell}}^{\text{ο}}-\frac{RT}{vF}\ln Q$

Where,

• $E_{\text{cell}}$ is the electrode potential of the cell.
• $E_{\text{cell}}^{\text{ο}}$ is the standard electrode potential of the cell.
• $R$ is the gas constant.
• $T$ is the temperature.
• $v$ is the number of electrons.
• $F$ is the Faraday’s constant.
• $Q$ is the reaction quotient.

Answer to Problem 6.3IA

The Nernst equation for the cell is shown below.

    $E_{\text{cell}}=E_{\text{cell}}^{\text{ο}}-38.55\times {10}^{-3}\text{V}\ln \left(4{\gamma }_{\pm }b\right)$

Explanation of Solution

The given cell is shown below.

  $\text{Zn}\left(\text{s}\right)|{\text{ZnCl}}_2\left(\text{0}\text{.0050}\text{mol}{\text{kg}}^{-1}\right)|{\text{Hg}}_2{\text{Cl}}_2|\text{Hg}\left(\text{l}\right)$

The cell reaction for this cell is shown below.

  ${\text{Hg}}_2{\text{Cl}}_2\left(\text{s}\right)+\text{Zn}\left(\text{s}\right)\to \text{Hg}\left(\text{l}\right)+2{\text{Cl}}^{-}\left(\text{aq}\right)+{\text{Zn}}^{2+}\left(\text{aq}\right)$

The expression for the Nernst equation for a cell is shown below.

    $E_{\text{cell}}=E_{\text{cell}}^{\text{ο}}-\frac{RT}{vF}\ln Q$

Where,

• $E_{\text{cell}}$ is the electrode potential of the cell.
• $E_{\text{cell}}^{\text{ο}}$ is the standard electrode potential of the cell.
• $R$ is the gas constant.
• $T$ is the temperature.
• $v$ is the number of electrons.
• $F$ is the Faraday’s constant.
• $Q$ is the reaction quotient.

The activity for the solid and the liquid phase is constant.

Therefore, the Nernst equation for the reaction is shown below.

    $E_{\text{cell}}=E_{\text{cell}}^{\text{ο}}-\frac{RT}{vF}\ln {\left({\text{a}}_{-}{\text{Cl}}^{-}\right)}^2\left({\text{a}}_{+}{\text{Zn}}^{2+}\right)$                                                            (1)

Where,

• ${\text{a}}_{-}$ represents the activity of ${\text{Cl}}^{-}$.
• ${\text{a}}_{+}$ represents the activity of ${\text{Zn}}^{2+}$.

The value of $\frac{RT}{F}$ is $25.7\times {10}^{-3}\text{V}$.

The value of $v$ is $2$.

The general formula for the activity is shown below.

    $a_{\pm }={\gamma }_{\pm }b$                                                                                                          (2)

Where,

• $a_{\pm }$ is activity of the ion.
• ${\gamma }_{\pm }$ is the mean activity coefficient of the salt.
• $b$ is the molality of the ion.

The molality of ${\text{Zn}}^{2+}$ is $b$ and the molality of ${\text{Cl}}^{-}$ is $2b$.

Substitute the values of molality of ${\text{Zn}}^{2+}$ to calculate the activity of ${\text{Zn}}^{2+}$ as shown below.

    $a_{+}\left({\text{Zn}}^{2+}\right)={\gamma }_{\pm }b$

Substitute the values of molality of ${\text{Cl}}^{-}$ to calculate the activity of ${\text{Cl}}^{-}$ as shown below.

    $a_{-}\left({\text{Cl}}^{-}\right)={\gamma }_{\pm }2b$

Substitute the values of activity of ${\text{Zn}}^{2+}$ and ${\text{Cl}}^{-}$, $\frac{RT}{F}$ and $v$ in the equation (1) as shown below.

    $\begin{array}{l}{E}_{\text{cell}}=E_{\text{cell}}^{\text{ο}}-\frac{25.7\times {10}^{-3}\text{V}}{2}\ln {\left({\gamma }_{\pm }2b\right)}^2\left({\gamma }_{\pm }b\right)\\ E_{\text{cell}}=E_{\text{cell}}^{\text{ο}}-\frac{25.7\times {10}^{-3}\text{V}}{2}\ln \left(4{\gamma }_{\pm }^3{b}^3\right)\\ E_{\text{cell}}=E_{\text{cell}}^{\text{ο}}-\frac{3\times 25.7\times {10}^{-3}\text{V}}{2}\ln \left(4{\gamma }_{\pm }b\right)\\ E_{\text{cell}}=E_{\text{cell}}^{\text{ο}}-38.55\times {10}^{-3}\text{V}\ln \left(4{\gamma }_{\pm }b\right)\end{array}$                                                    (3)

The equation (3) represents the Nernst equation for the cell.

(b)

Interpretation Introduction

Interpretation:

The standard cell potential for the cell has to be stated.

Concept introduction:

Nernst equation is the relation between standard electrode potential and the electrode potential at given conditions of pressures, temperatures and concentrations. Standard electrode potential is the electrode potential at standard temperature, pressure and concentration.

Answer to Problem 6.3IA

The standard electrode potential is $\underset{\_}{1.0304V}$.

Explanation of Solution

The given cell is shown below.

  $\text{Zn}\left(\text{s}\right)|{\text{ZnCl}}_2\left(\text{0}\text{.0050}\text{mol}{\text{kg}}^{-1}\right)|{\text{Hg}}_2{\text{Cl}}_2|\text{Hg}\left(\text{l}\right)$

The cell reaction for this cell is shown below.

  ${\text{Hg}}_2{\text{Cl}}_2\left(\text{s}\right)+\text{Zn}\left(\text{s}\right)\to \text{Hg}\left(\text{l}\right)+2{\text{Cl}}^{-}\left(\text{aq}\right)+{\text{Zn}}^{2+}\left(\text{aq}\right)$

The right hand electrode represents the cathode and the left hand electrode represents the anode.

The formula to calculate the standard potential of the cell is,

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

The electrode potential of the cathode is $E_{\text{cathode}}^{\text{ο}}=+0.2676\text{V}$.

The electrode potential of the anode is $E_{\text{anode}}^{\text{ο}}=-0.7628\text{V}$.

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

    $\begin{array}{c}{E}_{\text{cell}}^{\text{ο}}=+0.2676\text{V}-\left(-0.7628\text{V}\right)\\ =\underset{\_}{1.0304V}\end{array}$

Therefore, the standard electrode potential is $\underset{\_}{1.0304V}$.

(c)

Interpretation Introduction

Interpretation:

The values of ${\Delta }_{\text{f}}G$, ${\Delta }_{\text{f}}G°$ and $K$ for the cell reaction has to be stated.

Concept introduction:

The change in Gibbs free energy for a reaction when reactants and products are present in their standard states of pressure, forms and concentration is represented by ${\Delta }_{\text{r}}G°$ whereas ${\Delta }_{\text{r}}G$ depends on the reaction condition and extent of reaction.

Answer to Problem 6.3IA

The values of ${\Delta }_{\text{f}}G$, ${\Delta }_{\text{f}}G°$ and $K$ for the cell reaction are $\underset{\_}{236812.784Jmo{l}^{-1}}$, $\underset{\_}{198836.29Jmo{l}^{-1}}$ and $\underset{\_}{6.835\times 10^{34}}$ respectively.

Explanation of Solution

The given cell is shown below.

  $\text{Zn}\left(\text{s}\right)|{\text{ZnCl}}_2\left(\text{0}\text{.0050}\text{mol}{\text{kg}}^{-1}\right)|{\text{Hg}}_2{\text{Cl}}_2|\text{Hg}\left(\text{l}\right)$

The cell reaction for this cell is shown below.

  ${\text{Hg}}_2{\text{Cl}}_2\left(\text{s}\right)+\text{Zn}\left(\text{s}\right)\to \text{Hg}\left(\text{l}\right)+2{\text{Cl}}^{-}\left(\text{aq}\right)+{\text{Zn}}^{2+}\left(\text{aq}\right)$

The standard electrode potential for the cell is $1.0304\text{V}$.

The formula to calculate standard Gibbs free energy is shown below.

  ${\Delta }_{\text{f}}{G}_{\text{cell}}^{\text{ο}}=vF{E}_{\text{cell}}^{\text{ο}}$

Where,

• $E_{\text{cell}}^{\text{ο}}$ is the standard electrode potential of the cell.
• $v$ is the number of electrons.
• $F$ is the Faraday’s constant.

Substitute the value of number of electrons, Faraday’s constant and standard electrode potential in equation (5).

  $\begin{array}{c}{\Delta }_{\text{f}}{G}_{\text{cell}}^{\text{ο}}=\left(2\right)\left(96485\text{C}{\text{mol}}^{-1}\right)\left(1.0304\text{V}\right)\\ =198836.29\text{C}\text{V}{\text{mol}}^{-1}\end{array}$

The unit joule can be represented in the form of Coulombs and Volts as shown below.

    $\text{1}\text{C}\text{V}=1\text{}\text{J}$

The standard Gibbs free energy is $\underset{\_}{198836.29Jmo{l}^{-1}}$.

The electrode potential for the given cell is $+1.2272\text{V}$.

The formula to calculate Gibbs free energy is shown below.

  ${\Delta }_{\text{f}}G=vF{E}_{\text{cell}}$

Where,

• $E_{\text{cell}}$ is the electrode potential of the cell.
• $v$ is the number of electrons.
• $F$ is the Faraday’s constant.

Substitute the value of number of electrons, Faraday’s constant and standard electrode potential in equation (5).

  $\begin{array}{c}{\Delta }_{\text{f}}G=\left(2\right)\left(96485\text{C}{\text{mol}}^{-1}\right)\left(+1.2272\text{V}\right)\\ =236812.784\text{C}\text{V}{\text{mol}}^{-1}\end{array}$

The unit joule can be represented in the form of Coulombs and Volts as shown below.

    $\text{1}\text{C}\text{V}=1\text{}\text{J}$

The standard Gibbs free energy is $\underset{\_}{236812.784Jmo{l}^{-1}}$.

The equilibrium constant can be calculated by using the formula shown below.

  $\ln K=-\frac{{\Delta }_{\text{rxn}}{G}_{\text{cell}}^{\text{ο}}}{RT}$

Where,

• $G_{\text{cell}}^{\text{ο}}$ is the standard Gibbs free energy of the cell.
• $R$ is the gas constant.
• $T$ is the temperature.

Substitute the values of standard Gibbs free energy, gas constant and temperature in the above formula.

  $\begin{array}{l}\ln K=-\frac{\left(-198836.29\text{J}{\text{mol}}^{-1}\right)}{\left(8.314\text{JK}{\text{mol}}^{-1}\right)\left(298.15\text{K}\right)}\\ \ln K=80.21\end{array}$

The equilibrium constant is calculated by taking inverse on both sides.

  $\begin{array}{c}K=e^{80.21}\\ =\underset{\_}{6.835\times 10^{34}}\end{array}$

Therefore, the equilibrium constant for the cell reaction is $\underset{\_}{6.835\times 10^{34}}$.

(d)

Interpretation Introduction

Interpretation:

The mean ionic activity and activity coefficient of ${\text{ZnCl}}_2$ from the measured cell potential has to be calculated.

Concept introduction:

Nernst equation is the relation between standard electrode potential and the electrode potential at given conditions of pressures, temperatures and concentrations.  The expression for Nernst equation is,

    $E_{\text{cell}}=E_{\text{cell}}^{\text{ο}}-\frac{RT}{vF}\ln Q$

Where,

• $E_{\text{cell}}$ is the electrode potential of the cell.
• $E_{\text{cell}}^{\text{ο}}$ is the standard electrode potential of the cell.
• $R$ is the gas constant.
• $T$ is the temperature.
• $v$ is the number of electrons.
• $F$ is the Faraday’s constant.
• $Q$ is the reaction quotient.

Answer to Problem 6.3IA

The mean activity coefficient of ${\text{ZnCl}}_2$ and mean ionic activity is $\underset{\_}{1.3097}$ and $\underset{\_}{0.006548molk{g}^{-1}}$ respectively.

Explanation of Solution

The Nernst equation for the cell is shown below.

    $E_{\text{cell}}=E_{\text{cell}}^{\text{ο}}-38.55\times {10}^{-3}\text{V}\ln \left(4^{1/3}{\gamma }_{\pm }b\right)$                                                           (4)

The standard electrode potential for the cell is $1.0304\text{V}$.

The electrode potential for the given cell is $+1.2272\text{V}$.

Substitute the value of standard electrode potential and electrode potential in equation (4).

    $\begin{array}{c}+1.2272\text{V}=1.0304\text{V}-\left(38.55\times {10}^{-3}\text{V}\right)\ln \left(4^{1/3}{\gamma }_{\pm }0.0050\right)\\ +1.2272\text{V}=1.0304\text{V}-\left(38.55\times {10}^{-3}\text{V}\right)\ln \left(4^{1/3}\times 0.0050\right)-\left(38.55\times {10}^{-3}\text{V}\right)\ln {\gamma }_{\pm }\\ +1.2272\text{V}=1.0304\text{V}-\left(-0.1864\text{V}\right)-\left(38.55\times {10}^{-3}\text{V}\right)\times \ln {\gamma }_{\pm }\\ \ln {\gamma }_{\pm }=\frac{+1.2272\text{V}-1.0304\text{V}+\left(-0.1864\text{V}\right)}{\left(38.55\times {10}^{-3}\text{V}\right)}\end{array}$

The above expression is solved as shown below.

    $\begin{array}{c}\ln {\gamma }_{\pm }=0.2698\\ {\gamma }_{\pm }=e^{0.2698}\\ =\underset{\_}{1.3097}\end{array}$

Therefore, the mean activity coefficient of ${\text{ZnCl}}_2$ is $\underset{\_}{1.3097}$.

The general formula for the activity is shown below.

    $a_{\pm }={\gamma }_{\pm }b$                                                                                                          (2)

Where,

• $a_{\pm }$ is mean activity of the ion.
• ${\gamma }_{\pm }$ is the mean activity coefficient of the salt.
• $b$ is the molality of the ion.

Substitute the value of molality and mean activity coefficient of ${\text{ZnCl}}_2$ in the formula (2).

    $\begin{array}{c}{a}_{\pm }=1.3097\left(\text{0}\text{.0050}\text{mol}{\text{kg}}^{-1}\right)\\ =\underset{\_}{0.006548molk{g}^{-1}}\end{array}$

Therefore, the mean activity is $\underset{\_}{0.006548molk{g}^{-1}}$.

(e)

Interpretation Introduction

Interpretation:

The mean ionic activity coefficient of ${\text{ZnCl}}_2$ has to be calculated.

Concept introduction:

To determine the activity coefficient of an ion in a dilute solution of known ionic strength, the Debye-Hückel limiting law is used.  The expression for the Debye-Hückel limiting law is given below.

    $\log {\gamma }_{\pm }=-A\left|z_{+}{z}_{-}\right|I^{1/2}$

The value of $A$ is $0.509$.

Where,

• $I$ is the ionic strength.
• $z_{+}$ and $z_{-}$ are the charges for the cation and anion, respectively.
• $A$ is the constant.
• ${\gamma }_{\pm }$ is the mean ionic coefficient.

Answer to Problem 6.3IA

The mean ionic activity coefficient of ${\text{ZnCl}}_2$ is $\underset{\_}{0.88}$.

Explanation of Solution

The given cell is shown below.

  $\text{Zn}\left(\text{s}\right)|{\text{ZnCl}}_2\left(\text{0}\text{.0050}\text{mol}{\text{kg}}^{-1}\right)|{\text{Hg}}_2{\text{Cl}}_2|\text{Hg}\left(\text{l}\right)$

The expression for the Debye-Hückel limiting law is shown below.

    $\log {\gamma }_{\pm }=-A\left|z_{+}{z}_{-}\right|I^{1/2}$                                                                                    (5)

The value of $A$ is $0.509$.

Where,

• $I$ is the ionic strength.
• $z_{+}$ and $z_{-}$ are the charges for the cation and anion, respectively.
• $A$ is the constant.
• ${\gamma }_{\pm }$ is the mean ionic coefficient.

The value of $I$ is calculated as shown below.

    $I=\frac{1}{2}{\displaystyle \sum _i{z}_i^2\left(\frac{b}{b°}\right)}$                                                                                             (6)

Where,

• $b$ is the molarity of ${\text{Zn}}^{2+}$ ions.
• $b°$ is the molarity of ${\text{Cl}}^{-}$ ions.

The value of $b$ is given as $\text{0}\text{.0050}\text{mol}{\text{kg}}^{-1}$.  There are two ions of chlorine involved in the reaction.  Therefore, the value of $b°$ for ${\text{Cl}}^{-}$ ions is double of the value of $b$.  The value of $b°$ for ${\text{Cl}}^{-}$ ions is calculated as shown below.

  $\begin{array}{c}b°\left({\text{Cl}}^{-}\right)=2b\\ =2\left(\text{0}\text{.0050}\text{mol}{\text{kg}}^{-1}\right)\\ =0.010\text{mol}{\text{kg}}^{-1}\end{array}$

Substitute value of $b\left({\text{Zn}}^{2+}\right)$ and $b°\left({\text{Cl}}^{-}\right)$ in equation (6).

  $\begin{array}{c}I=\frac{1}{2}\left[4\times \left(\text{0}\text{.0050}\right)+\left(\text{0}\text{.010}\right)\right]\\ =0.015\end{array}$

Substitute value of $I$ in equation (5)

    $\begin{array}{c}\log {\gamma }_{\pm }=\left(-2\right)\left(0.509\right){\left(0.015\right)}^{1/2}\\ =-0.125\\ {\gamma }_{\pm }=e^{-0.125}\\ =\underset{\_}{0.88}\end{array}$

Therefore, the mean ionic activity coefficient of ${\text{ZnCl}}_2$ is $\underset{\_}{0.88}$.

(f)

Interpretation Introduction

Interpretation:

The values of ${\Delta }_{r}S$ and ${\Delta }_{\text{r}}H$ has to be calculated.

Concept introduction:

Enthalpy is the internal heat content of the system.  Under constant pressure, a system can do expansion work under the effect of energy supplied as heat known as enthalpy change of the system.

The entropy of a system can also be defined as the measure of disorder in a closed system. Entropy is an extensive property.

Answer to Problem 6.3IA

The values of ${\Delta }_{r}S$ and ${\Delta }_{\text{r}}H$ is $\underset{\_}{-87.2mo{l}^{-1}{K}^{-1}}$ and $\underset{\_}{-262.8kJmo{l}^{-1}}$ respectively.

Explanation of Solution

The given cell is shown below.

  $\text{Zn}\left(\text{s}\right)|{\text{ZnCl}}_2\left(\text{0}\text{.0050}\text{mol}{\text{kg}}^{-1}\right)|{\text{Hg}}_2{\text{Cl}}_2|\text{Hg}\left(\text{l}\right)$

The change in enthalpy is given by the formula shown below.

  ${\Delta }_{\text{r}}S=-{\left(\frac{\partial {\Delta }_{\text{r}}G}{\partial T}\right)}_P$

The above formula can be written as given below.

  ${\Delta }_{r}S=vF\left(\frac{\partial \Delta F}{\partial T}\right)$                                                                                            (7)

Where,

• $F$ is the faraday constant.
• $T$ is the temperature.
• $v$ is the number of electrons.

Substitute the value of $v$ in the equation (7).

    $\begin{array}{c}{\Delta }_{r}S=\left(2\right)\left(9.6485\times {10}^4\text{C}{\text{mol}}^{-1}\right)\left(-4.52\times {10}^{-4}\text{V}{\text{K}}^{-1}\right)\\ =-87.2\text{C}{\text{mol}}^{-1}\text{V}{\text{K}}^{-1}\\ =\underset{\_}{-87.2mo{l}^{-1}{K}^{-1}}\end{array}$

The Gibb’s free energy is given by the formula shown below.

    ${\Delta }_{\text{r}}G={\Delta }_{\text{r}}H-T{\Delta }_{\text{r}}S$                                                                                         (8)

Where,

• ${\Delta }_{\text{r}}H$ is the change in the enthalpy.
• ${\Delta }_{\text{r}}S$ is the change in entropy.

The value of ${\Delta }_{\text{r}}G$, ${\Delta }_{\text{r}}S$ and $T$ are $-236.81\text{kJ}{\text{mol}}^{-1}$, $\underset{\_}{-87.2mo{l}^{-1}{K}^{-1}}$ and $\text{298}\text{.15}\text{K}$ respectively.

Substitute the value of ${\Delta }_{\text{r}}G$, ${\Delta }_{\text{r}}S$ and $T$ in equation (8).

    $\begin{array}{c}-236.81\text{kJ}{\text{mol}}^{-1}={\Delta }_{\text{r}}H-\text{298}\text{.15}\text{K}\left(-87.2{\text{JK}}^{-1}{\text{mol}}^{-1}\right)\\ {\Delta }_{\text{r}}H=-236.81\text{kJ}{\text{mol}}^{-1}+\text{298}\text{.15}\text{K}\left(-87.2{\text{JK}}^{-1}{\text{mol}}^{-1}\times \frac{1\text{kJ}}{{10}^3\text{J}}\right)\\ =\underset{\_}{-262.8kJmo{l}^{-1}}\end{array}$

The value of ${\Delta }_{\text{r}}H$ is $\underset{\_}{-262.8kJmo{l}^{-1}}$.