Chapter 19, Problem 30QP

Calculate $E°,E,\text{ and }\Delta G$ for the following cell reactions.

$\begin{array}{l}\text{(a)}\\ {\text{Mg(s)+Sn}}^{\text{2+}}(aq)\underset{}{\rightleftarrows }{\text{Mg}}^{2+}(aq)+\text{Sn}(s)\\ \left[{\text{Mg}}^{2+}\right]=0.045M\text{, }\left[{\text{Sn}}^{2+}\right]=0.035M\end{array}$

$\begin{array}{l}\text{(b)}\\ 3\text{Zn}(s)+2{\text{Cr}}^{3+}(aq)\underset{}{\rightleftarrows }3{\text{Zn}}^{2+}(aq)+2\text{Cr}(s)\\ \left[{\text{Cr}}^{3+}\right]=0.010M,\left[{\text{Zn}}^{2+}\right]=0.0085M\end{array}$

Interpretation Introduction

Interpretation:

The standard cell potential, emf of the galvanic cell, and free energy change for the given cell reactions at ${\text{25}}^{\circ }\text{C}$

are to be determined.

Concept introduction:

Gibbs free energy is the change in the enthalpy $\Delta \text{H}$ of the system minus the product of the temperature (Kelvin) and the change in the entropy $\text{ΔS}$ of the system. It is represented as:

$\Delta \text{G=}\Delta \text{H}-\text{TΔS}$

According to the standard reduction potential values, the electrode that has lower negative reduction potential will act as a cathode and will undergo reduction. However, if the negative reduction potential of the electrode is high, it will act as an anode and will undergo oxidation.

The standard reduction potential of a galvanic cell may be calculated in terms of standard reduction potential of cathode and anode, as the relation mentioned below:

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

According to the Nernst equation, the relation between emf, standard cell potential, and reaction quotient at $298\text{K}$

will be as follows:

$E\text{ = }{E}_{\text{cell}}^{\text{o}}-\frac{\text{0}\text{.0592 V}}{n}\log Q$

Here, $E$ is the cell potential, $E_{cell}^{\circ }$ is the standard cell potential, Q is the reaction quotient, and $n$ is the transfer of electrons in the half-cell reactions.

The relation between cell potential and free energy change is as follows:

$\Delta G=-n\text{F}{E}_{cell}^{\circ }$

Here, $E$ is the cell potential, $\Delta G$

is the Gibb’s free energy change, $\text{F}$

is the Faraday constant $\left(\text{1F = 96500 J/V }{\text{. mol e}}^{-}\right)$, and $n$ is the transfer of electrons in the half-cell reactions.

Answer to Problem 30QP

Solution:

(a)

$\begin{array}{l}{E}_{\text{cell}}^{\text{o}}=+2.23\text{V}\\ E\text{ =}2.23\text{V}\\ \Delta \text{G =}-430\text{kJ/mol}\end{array}$

(b)

$\begin{array}{l}{E}_{\text{cell}}^{\text{o}}=+0.02\text{V}\\ E\text{ =}0.04\text{V}\\ \Delta \text{G =}-23\text{kJ/mol}\end{array}$

Explanation of Solution

a) $\text{Mg}(s)+{\text{ Sn}}^{\text{2+}}(aq)\underset{}{\overset{}{\rightleftarrows }}{\text{ Mg}}^{\text{2+}}(aq)+\text{Sn}(s),\left[{\text{Mg}}^{\text{2+}}\right]=0.045M,\left[{\text{Sn}}^{\text{2+}}\right]=0.035M$

The overall reaction of galvanic cell is as follows:

$\text{Mg}(s)+{\text{ Sn}}^{\text{2+}}(aq)\underset{}{\overset{}{\rightleftarrows }}{\text{ Mg}}^{\text{2+}}(aq)+\text{Sn}(s)$

The half-cell reactions of $\text{Mg}$ electrode and $\text{Sn}$ electrode along with their standard reduction potentials are as follows:

Oxidation half reaction (Anode): $\text{Mg}(s)\to {\text{Mg}}^{\text{2+}}(aq)+{\text{2 e}}^{-}$

$E^{\text{o}}=-2.37\text{V}$

Reduction half reaction (Cathode): ${\text{Sn}}^{\text{2+}}(aq)+{\text{2e}}^{-}\to \text{Sn}(s)$

$E^{\text{o}}=-0.14\text{V}$

As the standard reduction potential of $\text{Sn}$ half-cell reaction is greater (less negative), it will act as a cathode and will undergo reduction. However, the standard reduction potential of $\text{Mg}$

half-cell reaction is less (more negative); therefore, it will act as an anode and will undergo oxidation.

Therefore, $E_{{\text{Mg}}^{2+}\text{/Mg}}^{\text{o}}=-2.37\text{V}$ and $E_{{\text{Sn}}^{2+}\text{/Sn}}^{\text{o}}=-0.14\text{V}$.

The cell representation of a galvanic cell is as follows:

$\text{Mg}(s){\text{|Mg}}^{\text{2+}}(aq){\text{|| Sn}}^{\text{2+}}(aq)\text{|Sn}(s)$

The standard cell potential that is $E_{cell}^o$

(at ${\text{25}}^{\text{o}}\text{C}$

) for the above galvanic cell is calculated as follows:

$\begin{array}{l}{E}_{\text{cell}}^{\text{o}}=E_{{\text{Sn}}^{2+}\text{/Sn}}^{\text{o}}-E_{{\text{Mg}}^{\text{2+}}\text{/Mg}}^{\text{o}}\\ E_{\text{cell}}^{\text{o}}=\left(-0.\text{14}\text{V}-\text{(}-2.37\text{V)}\right)\\ E_{\text{cell}}^{\text{o}}=+2.23\text{V}\end{array}$

The reaction quotient for the reaction is given by the following expression:

$Q=\frac{{\text{[Mg}}^{\text{2+}}\text{]}}{{\text{[Sn}}^{\text{2+}}\text{]}}$

Concentration of ${\text{Mg}}^{\text{2+}}(aq)$

and ${\text{Sn}}^{\text{2+}}(aq)$

is $\text{0}\text{.45 M}$

and $\text{0}\text{.35 M}$, respectively.

Substitute all the values in the above equation,

$\begin{array}{l}Q=\left(\frac{\text{0}\text{.45}\text{M}}{\text{0}\text{.35}\text{M}}\right)\\ Q=\text{1}\text{.285}\end{array}$

In the given galvanic cell, the total loss and total gain of electrons is two.

Now, according to the Nernst equation, the relation between emf, standard cell potential, and reaction quotient at ${\text{25}}^{\circ }\text{C}$

will be as follows:

$E\text{ = }{E}_{\text{cell}}^{\text{o}}-\frac{\text{0}\text{.0592V}}{n}\log Q$

Substitute all the values in the above equation,

$\begin{array}{l}E\text{ =}2.23\text{V}-\frac{\text{0}\text{.0592V}}{\text{2}}\text{log1}\text{.285}\\ E=2.226V\\ E\cong 2.23\text{V}\end{array}$

The relation between cell potential and free energy change is as follows:

$\Delta G=-n\text{F}{E}_{cell}^{\circ }$

Substitute all the values in the above equation,

$\begin{array}{l}\Delta G=\left(-2\times 96500\text{J/V mol}{\text{e}}^{-}\times 2.23\text{V}\right)\\ \Delta G=\left(-430390\text{J/mol}\times \frac{1\text{kJ/mol}}{1000\text{J/mol}}\right)\\ \Delta \text{G }\cong -430.390\text{kJ/mol}\\ \Delta \text{G }=-430\text{kJ/mol}\end{array}$

b) $\text{ 3Zn}(s)+{\text{2Cr}}^{\text{3+}}(aq)\underset{}{\overset{}{\rightleftarrows }}{\text{ 3Zn}}^{\text{2+}}(aq)+\text{2Cr}(s),\left[{\text{Zn}}^{\text{2+}}\right]=0.0085M,\left[{\text{Cr}}^{\text{3+}}\right]=0.010M$

The overall reaction of galvanic cell is as follows:

$\text{3Zn}(s)+{\text{2Cr}}^{\text{3+}}(aq)\underset{}{\overset{}{\rightleftarrows }}{\text{ 3Zn}}^{\text{2+}}(aq)+\text{2Cr}(s)$

The half-cell reactions of $\text{Zn}$ electrode and $\text{Cr}$ electrode along with their standard reduction potentials are as follows:

Oxidation half reaction (Anode): $\text{3Zn}(s)\to {\text{3Zn}}^{\text{2+}}+{\text{6e}}^{-}$

$E^{\text{o}}=-0.76\text{V}$

Reduction half reaction (Cathode): ${\text{2Cr}}^{\text{3+}}(aq)+{\text{6e}}^{-}\to \text{ 2Cr}(s)$

$E^{\text{o}}=-0.74\text{V}$

As the standard reduction potential of $\text{Cr}$

half-cell reaction is greater (less negative), it will act as a cathode and will undergo reduction. However, the standard reduction potential of $\text{Zn}$

half-cell reaction is less (more negative); therefore, it will act as an anode and will undergo oxidation.

Therefore, $E_{{\text{Zn}}^{2+}\text{/Zn}}^{\text{o}}=-0.76\text{V}$ and $E_{{\text{Cr}}^{3+}\text{/Cr}}^{\text{o}}=-0.74\text{V}$.

The cell representation of a galvanic cell is as follows:

$\text{Zn}(s)|{\text{Zn}}^{\text{2+}}(aq){\text{|| Cr}}^{\text{3+}}(aq)\text{|Cr}(s)$

The standard cell potential that is ${\text{E}}_{\text{cell}}^{\text{o}}$

(at ${\text{25}}^{\text{o}}\text{C}$

) for the above galvanic cell is calculated as follows:

$\begin{array}{l}{E}_{\text{cell}}^{\text{o}}=E_{{\text{Cr}}^{3+}\text{/Cr}}^{\text{o}}-E_{{\text{Zn}}^{\text{2+}}\text{/Zn}}^{\text{o}}\\ E_{\text{cell}}^{\text{o}}\text{=}\left(-0.\text{74}\text{V}-\text{(}-0.76\text{V)}\right)\\ E_{\text{cell}}^{\text{o}}=+0.02\text{V}\end{array}$

The reaction quotient for the reaction is as follows:

$Q=\frac{{{\text{[Zn}}^{\text{2+}}\text{]}}^3}{{{\text{[Cr}}^{\text{3+}}\text{]}}^2}$

Concentration of ${\text{Zn}}^{\text{2+}}\text{(aq)}$

and ${\text{Cr}}^{\text{3+}}\text{(aq)}$

is $\text{0}\text{.0085 M}$

and $\text{0}\text{.010 M}$, respectively.

Substitute all the values in the above expression,

$\begin{array}{l}Q=\left(\frac{{\text{(0}\text{.0085}\text{M)}}^3}{{\text{(0}\text{.010}\text{M)}}^2}\right)\\ Q=\text{0}\text{.00614}\end{array}$

In the given galvanic cell, the total loss of electron and total gain of electron is six.

Now, according to the Nernst equation, the relation between emf, standard cell potential, and reaction quotient at ${\text{25}}^{\circ }\text{C}$

will be as follows:

$E\text{ = }{E}_{\text{cell}}^{\text{o}}-\frac{\text{0}\text{.0592V}}{n}\log Q$

On substituting all the values in the above equation,

$\begin{array}{l}E=\left(0.02\text{V}-\frac{\text{0}\text{.0592V}}{6}\text{log0}\text{.0061}\right)\\ E=0.04\text{V}\end{array}$

The relation between cell potential and free energy change is as follows:

$\Delta G=-n\text{F}{E}_{cell}^{\circ }$

Substitute the values of $n$, F, and $E_{cell}^{\circ }$ to the above expression,

$\begin{array}{l}\Delta G=\left(-6\times 96500\text{J/V }\text{mol}{\text{e}}^{-}\times 0.04\text{V}\right)\\ \Delta G\text{ =}\left(-23160\text{J/mol}\times \frac{1\text{kJ/mol}}{1000\text{J/mol}}\right)\\ \Delta G\text{ =}-23.160\text{kJ/mol}\\ \Delta G\simeq -23\text{kJ/mol}\end{array}$