Chapter 8, Problem 8.77E

Approximate the expected voltage for the following electrochemical reaction using (a) the given molal concentrations and (b) the calculated activities using simple $\text{Debye-Hückel}$ theory. The value $\text{å}$ for both ${\text{Zn}}^{2+}$ and ${\text{Cu}}^{2+}$ is $6\times {10}^{-10}\text{ m}$.

$\text{Zn}\left(\text{s}\right)+{\text{Cu}}^{2+}\left(\text{aq,}\text{0}\text{.05}\text{molal}\right)\to {\text{Zn}}^{2+}\left(\text{aq,}\text{0}\text{.1}\text{molal}\right)+\text{Cu}\left(\text{s}\right)$

Explain why you get the answers you do.

Interpretation Introduction

(a)

Interpretation:

The expected voltage for the given electrochemical reaction is to be calculated by using the molal concentration.

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 8.77E

The expected voltage for the given electrochemical reaction is $1.0948\text{ V}$.

Explanation of Solution

The given electrochemical reaction is represented as,

$\text{Zn}\left(\text{s}\right)+{\text{Cu}}^{2+}\left(\text{aq,}\text{0}\text{.05}\text{molal}\right)\to {\text{Zn}}^{2+}\left(\text{aq,}\text{0}\text{.1}\text{molal}\right)+\text{Cu}\left(\text{s}\right)$

From Table $8.2$, the reduction half reaction of ${\text{Zn}}^{2+}$ and the standard reduction potential of ${\text{Zn}}^{2+}$ is represented as,

${\text{Zn}}^{2+}+2{\text{e}}^{-}\to \text{Zn }E°=-0.7618\text{V}$ …(1)

From Table $8.2$, the reduction half reaction of ${\text{Cu}}^{+}$ and the standard reduction potential of ${\text{Cu}}^{+}$ is represented as,

${\text{Cu}}^{2+}+2{\text{e}}^{-}\to \text{Cu }E°=0.3419\text{V}$ …(2)

The number of electrons transfer in the reaction is $2\text{mol}$.

The standard electrical potential of the reaction is calculated as,

$E°=E{°}_{\mathrm{Re}\text{d}}-E{°}_{\text{oxd}}$

Where,

• $E{°}_{\text{oxd}}$ represents the standard electrical potential of the oxidation half-reaction.

• $E{°}_{\text{red}}$ represents the standard electrical potential of the reduction half-reaction.

Substitute the value of the standard electrical of the oxidation half-reaction of zinc-ion and reduction half-reaction of the copper ion in the above equation.

$\begin{array}{rll}E°& =& 0.3419\text{V}-\left(-0.7618\text{V}\right)\\ & =& 1.1037\text{V}\end{array}$

The Nernst equation for the given reaction is represented as,

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

Where,

• $R$ is the gas constant with value $8.314\text{J/mol}\cdot \text{K}$.

• $F$ is the Faraday constant with value $8.314\text{J/mol}\cdot \text{K}$.

• $T$ is the temperature

• $E°$ is the cell potential.

• $\left[{\text{Zn}}^{2+}\right]$ represents the concentration of zinc ion.

• $\left[{\text{Cu}}^{2+}\right]$ represents the concentration of copper ion.

• $N$ represents the number of electrons transfer.

Substitute the values of $n$, $R$, $F$, $T$, $E°$, $\left[{\text{Zn}}^{2+}\right]$ and $\left[{\text{Cu}}^{2+}\right]$ in the above equation.

$\begin{array}{rll}E& =& 1.1037\text{V}-\frac{\left(8.314\text{J/mol}\cdot \text{K}\right)\left(298\text{K}\right)}{\left(2\right)\left(96,485\text{C/mol}\right)}\ln \left(\frac{\text{0}\text{.1}\text{molal}}{\text{0}\text{.05}\text{molal}}\right)\\ & =& 1.1037\text{V}-\left(0.0128\right)\left(0.693\right)\\ & =& 1.1037\text{V}-0.0089\\ & =& 1.0948\text{ V}\end{array}$

Therefore, the expected voltage for the given electrochemical reaction is $1.0948\text{ V}$.

Conclusion

The expected voltage for the given electrochemical reaction is $1.0948\text{ V}$.

Interpretation Introduction

(b)

Interpretation:

The activity of the given ions is to be calculated. The expected voltage for the given electrochemical reaction is to be calculated by using the activity of ions. The explanation for the value of voltage for the given electrochemical reaction is to be stated.

Concept introduction:

Activity coefficient depends upon the charge and size of the ion, and on the properties of the solvent. Debye-Huckel limiting law, describes the relation between the ionic strength and activity coefficient of a dilute solution.

Answer to Problem 8.77E

The expected voltage for the given electrochemical reaction is $1.0913\text{ V}$.

The voltage calculated by the activity of the ions is lower than the voltage calculated by the molal concentration of ions. The value of voltage is more accurate to calculate with activity of individual ions.

Explanation of Solution

The value $\text{å}$ for both ${\text{Zn}}^{2+}$ and ${\text{Cu}}^{2+}$ is $6\times {10}^{-10}\text{ m}$.

At $\text{25°C}$, $A$ of aqueous solution is equal to $1.171{\text{molal}}^{-1/2}$.

At $\text{25°C}$, $B$ of aqueous solution is equal to $2.32\times {10}^9{\text{m}}^{-1}\cdot {\text{molal}}^{-1/2}$.

The given electrochemical reaction is represented as,

$\text{Zn}\left(\text{s}\right)+{\text{Cu}}^{2+}\left(\text{aq,}\text{0}\text{.05}\text{molal}\right)\to {\text{Zn}}^{2+}\left(\text{aq,}\text{0}\text{.1}\text{molal}\right)+\text{Cu}\left(\text{s}\right)$

The standard electrical potential of the reaction is $1.1037\text{V}$.

The ionic strength of the solution is represented as,

$I=\frac{1}{2}{\displaystyle \sum _{\text{i=1}}^{\begin{array}{l}\text{number of}\\ \text{ ions}\end{array}}{m}_{\text{i}}{z}_{\text{i}}^2}$ …(3)

Where,

• $m_{\text{i}}$ represents the molality of the ${\text{i}}^{\text{th}}$ ion.

• $z_{\text{i}}$ represents the charge on the ${\text{i}}^{\text{th}}$ ion.

It is assumed that a monovalent anion in present in the solution. Then the charge on the ion will be $-1$. The concentration of anion with cation will be twice as the concentration of ions. Then the concentration of anion with copper ion will be $0.01\text{ molal}$ and the concentration of anion with zinc ion will be $0.2\text{ molal}$.

Substitute the value of charges and molality of copper ion and anion in the equation (2).

$\begin{array}{l}=\frac{1}{2}\left(\left(\text{0}\text{.05}\text{molal}\right){\left(2\right)}^2+\left(\text{0}\text{.1}\text{molal}\right){\left(-1\right)}^2\right)\\ =0.15\text{molal}\end{array}$

Substitute the value of charges and molality of zinc ion and anion in the equation (3).

$\begin{array}{l}=\frac{1}{2}\left(\left(\text{0}\text{.1}\text{molal}\right){\left(2\right)}^2+\left(\text{0}\text{.2}\text{molal}\right){\left(-1\right)}^2\right)\\ =0.3\text{molal}\end{array}$

The extended Debye-Huckel equation is shown below,

$\ln \gamma =-\frac{A{z}^2{I}^{1/2}}{I+B\text{å}{I}^{1/2}}$ …(4)

Where,

• $\gamma$ represents the activity coefficient of the ion.

• $A$ and $B$ represent constants.

• $z$ represents the charge on the ion.

• $I$ represents the ionic strength.

• $\text{å}$ represents the ionic diameter.

Substitute the value of $z$, $I$, $\text{å}$, $A$ and $B$ in the equation (4) for the activity coefficient of copper ion.

$\begin{array}{rll}\ln \gamma & =& -\frac{\left(1.171{\text{molal}}^{-1/2}\right){\left(2\right)}^2{\left(0.15\text{molal}\right)}^{1/2}}{1+\left(2.32\times {10}^9{\text{m}}^{-1}\cdot {\text{molal}}^{-1/2}\right)\left(6\times {10}^{-10}\text{ m}\right){\left(0.15\text{molal}\right)}^{1/2}}\\ & =& 1.179\end{array}$

Antilog is taken on both side of the above equation.

$\begin{array}{rll}\gamma & =& \exp \left(1.179\right)\\ & =& 3.25\end{array}$

Substitute the value of $z$, $I$, $\text{å}$, $A$ and $B$ in the equation (4) for the activity coefficient of zinc ion.

$\begin{array}{rll}\ln \gamma & =& -\frac{\left(1.171{\text{molal}}^{-1/2}\right){\left(2\right)}^2{\left(0.3\text{molal}\right)}^{1/2}}{1+\left(2.32\times {10}^9{\text{m}}^{-1}\cdot {\text{molal}}^{-1/2}\right)\left(6\times {10}^{-10}\text{ m}\right){\left(0.3\text{molal}\right)}^{1/2}}\\ & =& 1.456\end{array}$

Antilog is taken on both side of the above equation.

$\begin{array}{rll}\gamma & =& \exp \left(1.456\right)\\ & =& 0.375\end{array}$

The activity of the ion is represented as,

$a_{\text{i}}={\gamma }_{\text{i}}{m}_{\text{i}}$ …(5)

Substitute the value of activity coefficient and molality of copper ion in the equation (5).

$\begin{array}{rll}{a}_{{\text{Cu}}^{2+}}& =& \left(3.25\right)\left(\text{0}\text{.05}\text{molal}\right)\\ & =& 0.0082\text{ molal}\end{array}$

Substitute the value of activity coefficient and molality of zinc ion in the equation (5).

$\begin{array}{rll}{a}_{{\text{Zn}}^{2+}}& =& \left(4.287\right)\left(\text{0}\text{.1}\text{molal}\right)\\ & =& 0.4287\text{ molal}\end{array}$

The Nernst equation for the given reaction is represented as,

$E=E°-\frac{RT}{nF}\ln \left(\frac{a_{{\text{Zn}}^{2+}}}{a_{{\text{Cu}}^{2+}}}\right)$

Where,

• $R$ is the gas constant with value $8.314\text{J/mol}\cdot \text{K}$.

• $F$ is the Faraday constant with value $8.314\text{J/mol}\cdot \text{K}$.

• $T$ is the temperature

• $E°$ is the cell potential.

• $a_{{\text{Zn}}^{2+}}$ represents the activity of zinc ion.

• $a_{{\text{Cu}}^{2+}}$ represents the activity of copper ion.

• $N$ represents the number of electrons transfer.

Substitute the value of $n$, $R$, $F$, $T$, $E°$, $a_{{\text{Zn}}^{2+}}$ and $a_{{\text{Cu}}^{2+}}$ in the above equation.

$\begin{array}{rll}E& =& 1.1037\text{V}-\frac{\left(8.314\text{J/mol}\cdot \text{K}\right)\left(298\text{K}\right)}{\left(2\right)\left(96,485\text{C/mol}\right)}\ln \left(\frac{0.4287\text{ molal}}{0.1625\text{ molal}}\right)\\ & =& 1.1037\text{V}-\left(0.0128\right)\left(0.97\right)\\ & =& 1.1037\text{V}-0.012416\\ & =& 1.0913\text{ V}\end{array}$

The expected voltage for the given electrochemical reaction is $1.0913\text{ V}$.

The voltage calculated by the activity of the ions is lower than the voltage calculated by the molal concentration of ions. The value of voltage is more accurate to calculate with activity of individual ions.

Conclusion

The expected voltage for the given electrochemical reaction is $1.0913\text{ V}$.