Chapter 6, Problem 6.2IA

(a)

Interpretation Introduction

Interpretation: Under the given conditions, the ionic strengths of the solutions have to be calculated.

Concept introduction: The concentrations of ions in solution are measured in terms of ionic strength.  The unit of ionic strength can be molar or molal.  Ionic strength also describes the dissociation constant as well as solubility of the salts.

Answer to Problem 6.2IA

The ionic strengths of the solutions are $\underset{\_}{4.0\times 10^{-3}mol\text{}k{g}^{-1}}$ and $\underset{\_}{12.0\times 10^{-3}mol\text{}k{g}^{-1}}$ respectively.

Explanation of Solution

The ionic strength of a salt is expressed as,

  $I=\frac{1}{2}{\displaystyle \sum _{\text{i}=1}^n{m}_{\text{i}}{z}_{\text{i}}^{\text{2}}}$                                                                                                  (1)

Where,

• $I$ is the  ionic strength of any salt.
• $m_{\text{i}}$ is the molar concentration of ions $\text{i}$.
• $z_{\text{i}}$ is the number of charge on cations and anions.

It is given that,

The molality of ${\text{CuSO}}_{\text{4}}$ is $\text{1}\text{.0}\times {\text{10}}^{-\text{3}}\text{mol}\text{}{\text{kg}}^{-1}$.

The molality of ${\text{ZnSO}}_{\text{4}}$ is $\text{3}\text{.0}\times {\text{10}}^{-\text{3}}\text{mol}\text{}{\text{kg}}^{-1}$.

When ${\text{CuSO}}_{\text{4}}$ dissolves in the solution, it dissociates into ${\text{Cu}}^{\text{2+}}$ and ${\text{SO}}_{\text{4}}{}^{2-}$ ions.  Therefore the value of $z$ for copper and sulphate ions is $2$ for each.

The ionic strength of ${\text{CuSO}}_{\text{4}}$ is calculated as,

  $\begin{array}{c}I=\frac{1}{2}\left[\text{1}\text{.0}\times {\text{10}}^{-\text{3}}\times 2^2+\text{1}\text{.0}\times {\text{10}}^{-\text{3}}\times 2^2\right]\text{ mol}\text{}{\text{kg}}^{-1}\\ =\frac{1}{2}\left[\text{8}\text{.0}\times {\text{10}}^{-\text{3}}\right]\text{}\text{mol}\text{}{\text{kg}}^{-1}\\ =\underset{\_}{4.0\times 10^{-3}mol\text{}k{g}^{-1}}\end{array}$

When ${\text{ZnSO}}_{\text{4}}$ dissolve in the solution, it dissociates into ${\text{Zn}}^{\text{2+}}$ and ${\text{SO}}_{\text{4}}{}^{2-}$ ions.  Therefore the value of $z$ for zinc and sulphate ions is $2$ for each.

The ionic strength of ${\text{ZnSO}}_{\text{4}}$ is calculated as,

  $\begin{array}{c}I=\frac{1}{2}\left[\text{3}\text{.0}\times {\text{10}}^{-\text{3}}\times 2^2+3.\text{0}\times {\text{10}}^{-\text{3}}\times 2^2\right]\text{mol}\text{}{\text{kg}}^{-1}\\ =\frac{1}{2}\left[\text{24}\text{.0}\times {\text{10}}^{-\text{3}}\right]\text{}\text{mol}\text{}{\text{kg}}^{-1}\\ =\underset{\_}{12.0\times 10^{-3}mol\text{}k{g}^{-1}}\end{array}$

Therefore, the ionic strengths of the solutions are $\underset{\_}{4.0\times 10^{-3}mol\text{}k{g}^{-1}}$ and $\underset{\_}{12.0\times 10^{-3}mol\text{}k{g}^{-1}}$ respectively.

(b)

Interpretation Introduction

Interpretation: Under the given conditions, the mean ionic activity coefficients have to be calculated.

Concept introduction: Activity coefficient of substances is defined as the ratio of activity of substances to the molar concentrations of substances.  The average rational activity coefficient of the ions of an electrolyte that dissociates in solution into cations of charge $z^{+}$ and anions of charge $z^{-}$ is measured in terms of mean ionic coefficient.

Answer to Problem 6.2IA

The mean ionic activity coefficient of ${\text{CuSO}}_{\text{4}}$ and ${\text{ZnSO}}_{\text{4}}$ are $\underset{\_}{0.744}$ and $\underset{\_}{0.60}$ respectively.

Explanation of Solution

The mean ionic coefficient of electrolytes is calculated by the formula,

  $\log \left({\gamma }_{\pm }\right)=-A\left|z^{+}\times z^{-}\right|I^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$                                                                               (1)

Where,

• ${\gamma }_{\pm }$ is the mean ionic coefficient.
• $z^{+}$ is the charge on cations.
• $z^{-}$ is the charge on anions.
• $I$ is the  ionic strength.
• $A$ is a constant $\left(0.509\right)$.

When ${\text{CuSO}}_{\text{4}}$ dissolves in the solution, it dissociates into ${\text{Cu}}^{\text{2+}}$ and ${\text{SO}}_{\text{4}}{}^{2-}$ ions.  Therefore the value of $z$ for cation is $+2$ and the value of $z$ for anion is $-2$ for each.

The mean ionic coefficient of ${\text{CuSO}}_{\text{4}}$ is calculated as,

  $\begin{array}{c}\log \left({\gamma }_{\pm }\right)=-0.509\left|\left(+2\right)\times \left(-2\right)\right|{\left(4\times {10}^{-3}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\\ =-0.128\\ {\gamma }_{\pm }={10}^{-0.128}\\ =\underset{\_}{0.744}\end{array}$

When ${\text{ZnSO}}_{\text{4}}$ dissolve in the solution, it dissociates into ${\text{Zn}}^{\text{2+}}$ and ${\text{SO}}_{\text{4}}{}^{2-}$ ions.  Therefore the value of $z$ for cation is $+2$ and the value of $z$ for anion is $-2$ for each.

The mean ionic coefficient of ${\text{ZnSO}}_{\text{4}}$ is calculated as,

  $\begin{array}{c}\log \left({\gamma }_{\pm }\right)=-0.509\left|\left(+2\right)\times \left(-2\right)\right|{\left(12\times {10}^{-3}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\\ =-0.223\\ {\gamma }_{\pm }={10}^{-0.223}\\ =\underset{\_}{0.60}\end{array}$

Therefore, the mean ionic activity coefficient of ${\text{CuSO}}_{\text{4}}$ and ${\text{ZnSO}}_{\text{4}}$ are $\underset{\_}{0.744}$ and $\underset{\_}{0.60}$ respectively.

(c)

Interpretation Introduction

Interpretation: Under the given conditions, the reaction quotient has to be calculated.

Concept introduction: The reaction quotient is an entity that describes the direction in which a reaction proceeds, under the given concentrations of the reactants and the products.  It is also defined as the function of the activities or concentrations of the species.

Answer to Problem 6.2IA

The reaction coefficient for the reaction is $\underset{\_}{5.9}$.

Explanation of Solution

The reaction quotient is the function of the activities or concentrations of the species.

The given reaction is,

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

Therefore, the reaction quotient $\left(Q\right)$ of given chemical reaction in terms of mean ionic activity coefficient is,

  $Q=\frac{{\left(m_{{\text{Zn}}^{\text{2+}}}\times {\gamma }_{{\text{Zn}}^{\text{2+}}}\right)}^2}{{\left(m_{{\text{Cu}}^{\text{2+}}}\times {\gamma }_{{\text{Cu}}^{\text{2+}}}\right)}^2}$                                                                                        (1)

Where,

• $m_{{\text{Zn}}^{\text{2+}}}$ is the molar concentration of ${\text{Zn}}^{\text{2+}}$ ion.
• $m_{{\text{Cu}}^{\text{2+}}}$ is the molar concentration of ${\text{Cu}}^{\text{2+}}$ ion.
• ${\gamma }_{{\text{Zn}}^{\text{2+}}}$ is the mean ionic activity coefficient of ${\text{ZnSO}}_{\text{4}}$.
• ${\gamma }_{{\text{Cu}}^{\text{2+}}}$ is the mean ionic activity coefficient of ${\text{CuSO}}_{\text{4}}$.

Substitute the values of $m_{{\text{Zn}}^{\text{2+}}}$, $m_{{\text{Cu}}^{\text{2+}}}$, ${\gamma }_{{\text{Zn}}^{\text{2+}}}$ and ${\gamma }_{{\text{Cu}}^{\text{2+}}}$ in equation (1).

  $\begin{array}{c}Q=\frac{{\left(\left(3\times {10}^{-3}\right)\times 0.60\right)}^2}{{\left(\left(1\cdot {10}^{-3}\right)\times 0.74\right)}^2}\\ =\frac{{\left(1.8\times {10}^{-3}\right)}^2}{{\left(0.74\times {10}^{-3}\right)}^2}\\ ={\left(2.43\right)}^2\\ =\underset{\_}{5.9}\end{array}$

Therefore, the reaction coefficient for the given reaction is $\underset{\_}{5.9}$.

(d)

Interpretation Introduction

Interpretation: Under the given conditions, the standard cell potential of the reaction has to be calculated.

Concept introduction: The amount of voltage exists between the two half cell, is measured in term of cell potential.  The cell potential under standard conditions (at $1\text{atm}$ pressure, ${25}^{\circ }\text{C}$ temperature with $1\text{M}$ concentration) is called as standard cell potential.

Answer to Problem 6.2IA

The standard cell potential is $\underset{\_}{1.10V}$.

Explanation of Solution

The relation between standard cell potential and standard Gibbs free energy is expressed as,

  $\Delta G_{\text{cell}}^{\text{o}}=-nF{E}_{\text{cell}}^{\text{o}}$                                                                                            (1)

Where,

• $\Delta G_{\text{cell}}^{\text{o}}$ is the standard Gibbs free energy.
• $E_{\text{cell}}^{\text{o}}$ is the standard cell potential.
• $F$ is the Faraday constant $\left(96485\text{C}{\text{mol}}^{-1}\right)$.
• $n$ is the number of transferred electrons.

It is given that,

The standard Gibbs free energy $\left(\Delta G_{\text{cell}}^{\text{o}}\right)$ is $-212.7\text{kJ}{\text{mol}}^{-1}$.

For the reaction,

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

The number of transferred electron $\left(n\right)$ is $2$.

The conversion of $\text{kJ}{\text{mol}}^{-1}$ to $\text{J}{\text{mol}}^{-1}$ is done as,

  $\text{1}\text{kJ}{\text{mol}}^{-1}={10}^3\text{}\text{J}{\text{mol}}^{-1}$

Therefore, the conversion of $-212.7\text{kJ}{\text{mol}}^{-1}$ to $\text{J}{\text{mol}}^{-1}$ is done as,

  $-212.7\text{kJ}{\text{mol}}^{-1}=-212.7\times {10}^3\text{}\text{J}{\text{mol}}^{-1}$

The conversion of $\text{J}{\text{mol}}^{-1}$ to $\text{C}\text{V}{\text{mol}}^{-1}$ is done as,

  $\text{1}\text{J}{\text{mol}}^{-1}=\text{1}\text{C}\text{V}{\text{mol}}^{-1}$

Therefore, the conversion of $-212.7\times {10}^3\text{}\text{J}{\text{mol}}^{-1}$ to $\text{C}\text{V}{\text{mol}}^{-1}$ is done as,

  $-212.7\times {10}^3\text{}\text{J}{\text{mol}}^{-1}=-212.7\times {10}^3\text{}\text{C}\text{V}{\text{mol}}^{-1}$

Rearrange the equation (1) to calculate $E_{\text{cell}}^{\text{o}}$.

  $E_{\text{cell}}^{\text{o}}=-\frac{\Delta G_{\text{cell}}^{\text{o}}}{nF}$                                                                                                (2)

Substitute the values of $\Delta G_{\text{cell}}^{\text{o}}$, $F$ and $n$ in the equation (2).

  $\begin{array}{c}{E}_{\text{cell}}^{\text{o}}=-\frac{\left(-212.7\times {10}^3\text{}\text{C}\text{V}{\text{mol}}^{-1}\right)}{2\times 96485\text{C}{\text{mol}}^{-1}}\\ =\underset{\_}{1.10V}\end{array}$

Therefore, the standard cell potential is $\underset{\_}{1.10V}$.

(e)

Interpretation Introduction

Interpretation: Under the given conditions, the cell potential of whole cell reaction has to be calculated.

Concept introduction: The amount of voltage exists between the two half cell, is measured in term of cell potential.  The cell potential under standard conditions (at $1\text{atm}$ pressure, ${25}^{\circ }\text{C}$ temperature with $1\text{M}$ concentration) is called as standard cell potential.  Nernst’s equation relates the cell potential, standard cell potential and reaction quotient.

Answer to Problem 6.2IA

The cell potential for the cell reaction is $\underset{\_}{1.077V}$.

Explanation of Solution

The Nernst’s equation of the cell potential is,

  $E_{\text{cell}}=E_{\text{cell}}^{\circ }-\frac{RT}{nF}\ln Q$                                                                                     (1)

Where,

• $E_{\text{cell}}$ is the cell potential.
• $E_{\text{cell}}^{\text{o}}$ is the standard cell potential.
• $F$ is the Faraday constant $\left(96485\text{C}{\text{mol}}^{-1}\right)$.
• $n$ is the number of transferred electrons.
• $R$ is the gas constant $\left(8.314\text{J}\text{}{\text{K}}^{-1}{\text{mol}}^{-1}\right)$
• $T$ is the standard temperature $\left({25}^{\circ }\text{C}\right)$.

Celsius is converted to Kelvin using the equation given below as,

    $T\left(\text{K}\right)=T\left(°\text{C}\right)+273.15$

Substitute the temperature values in the above equation as follows.

    $\begin{array}{c}{T}_{}\left(\text{K}\right)=T\left(°\text{C}\right)+273.15\\ =25+273.15\\ =298.15\text{K}\end{array}$

The conversion of $\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}$ to $\text{C}\text{V}{\text{K}}^{-1}{\text{mol}}^{-1}$ is done as,

  $\text{1}\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}=\text{1}\text{C}\text{V}{\text{K}}^{-1}{\text{mol}}^{-1}$

Therefore, the conversion of $8.314\text{J}\text{}{\text{K}}^{-1}{\text{mol}}^{-1}$ to $\text{C}\text{V}{\text{K}}^{-1}{\text{mol}}^{-1}$ is done as,

  $8.314\text{J}\text{}{\text{K}}^{-1}{\text{mol}}^{-1}=8.314\text{}\text{C}\text{V}{\text{K}}^{-1}{\text{mol}}^{-1}$

Substitute the values of $E_{\text{cell}}^{\text{o}},n,R,T,F$ and $Q$ in equation (1).

  $\begin{array}{c}{E}_{\text{cell}}=1.10\text{V}-\frac{\left(8.314\text{}\text{C}\text{V}{\text{K}}^{-1}{\text{mol}}^{-1}\right)298.15\text{K}}{2\times 96485\text{C}{\text{mol}}^{-1}}\ln \left(5.9\right)\\ =1.10\text{V}-0.0228\text{V}\\ =\underset{\_}{1.077V}\end{array}$

Therefore, the cell potential for the cell reaction is $\underset{\_}{1.077V}$.