Chapter 19, Problem 126AP

The zinc-air battery shows much promise for electric cars because it is lightweight and rechargeable:

The net transformation is $\text{Zn(}s\text{)+}\frac{1}{2}{\text{O}}_{\text{2}}(g)\stackrel{}{\to }\text{ZnO}(s)$ (a) Write the half-reactions at the zinc-air electrodes, and calculate the standard emf of the battery at 25*C. (b) Calculate the emf under actual operating conditions when the partial pressure of oxygen is 0.21 atm. (c) What is the energy density (measured as the energy in kilojoules that can be obtained from 1 kg of the metal) of the zinc electrode? (d) If a current of $\text{2}\text{.1 }\times \text{ 1}{\text{0}}^{\text{5}}$ A is to be drawn from a zinc-air battery system, what volume of air (in liters) would need to be supplied to the battery every second? Assume that the temperature is $\text{25°C}$ and the partial pressure of oxygen is 0.21 atm.

Interpretation Introduction

Interpretation:

The half-reaction, the standard emf of battery, the emf under actual operating conditions, the energy density, and the volume of air essential to supply the battery every second is to be calculated.

Concept introduction:

The Nernst equation is the reduction potential of an electrochemical reaction to the standard electrode potential, temperature, and activities of the chemical species undergoing oxidation and reduction. 

The Nernst equation is an important equation of electrochemistry. The equation is $E=E^{\text{o}}-\frac{0.0592\text{ V}}{n}\log Q$

Here, $E$ is the emf, $E^{\text{o}}$ is the standard emf of the cell, $n$ is the number of moles, and $Q$ is the reaction quotient.

The standard free energy change is the difference of the sum of standard free energy change of products and the sum of standard free energy change of reactants.

$\Delta {\text{G}}^{\text{o}}\text{=}{\displaystyle \sum \Delta {\text{G}}_{product}^{\circ }}-{\displaystyle \sum \Delta {\text{G}}_{reac\tan t}^{\circ }}$

Answer to Problem 126AP

Solution:

(a)

The half-reaction is as follows:

$\begin{array}{l}\underset{\_}{\begin{array}{l}\text{ At anode: Zn}\stackrel{}{\to }{\text{ Zn}}^{2+}+2e^{-}\\ \text{At cathode: }\frac{1}{2}{\text{O}}_2+2{\text{e}}^{-}\stackrel{}{\to }{\text{ O}}^{2-}\end{array}}\\ \text{overall: Zn +} \frac{1}{2}{\text{O}}_2\stackrel{}{\to }\text{ZnO}\end{array}$

The standard emf of battery is $1.65\text{ V}$.

(b)

The emf is $1.63\text{ V}$.

(c)

The energy density of Zn is $4.86\times {10}^3\text{ kJ/kg}$.

(d)

The volume of air essential to supply the battery every second is $64\text{ L}$.

Explanation of Solution

Given information: The given reaction is as follows:

$\text{Zn}\left(s\right)+\frac{1}{2}{\text{O}}_2\left(g\right)\stackrel{}{\to }\text{ ZnO}\left(s\right)$

The current is $2.1\times {10}^5\text{ A}$.

a) The half-reaction at zinc-air electrode and the standard emf of the battery.

The half-reaction is as follows:

$\begin{array}{l}\underset{\_}{\begin{array}{l}\text{ At anode: Zn}\stackrel{}{\to }{\text{ Zn}}^{2+}+2e^{-}\\ \text{At cathode: }\frac{1}{2}{\text{O}}_2+2{\text{e}}^{-}\stackrel{}{\to }{\text{ O}}^{2-}\end{array}}\\ \text{overall: Zn +} \frac{1}{2}{\text{O}}_2\stackrel{}{\to }\text{ZnO}\end{array}$

The standard emf is calculated with the help of $\Delta G^{\text{o}}$ for the reaction as follows:

$\Delta {\text{G}}^{\text{o}}\text{=}{\displaystyle \sum \Delta {\text{G}}_{product}^{\circ }}-{\displaystyle \sum \Delta {\text{G}}_{reac\tan t}^{\circ }}$

$\Delta G^{\text{o}}=\Delta G_{\text{f}}^{\text{°}}\left(\text{ZnO}\right)-[\Delta G_{\text{f}}^{\text{°}}\left(\text{Zn}\right)+\frac{1}{2}\Delta G_{\text{f}}^{\text{°}}\left({\text{O}}_2\right)]$

Here, $\Delta G_{\text{f}}^{°}$ is the standard free energy of formation.

The standard free energy changes for the formation of $\text{ZnO}$, $\text{Zn}$, and ${\text{O}}_2$ are as follows:

$\begin{array}{c}\Delta {\text{G}}_f^{\circ }\left(\text{ZnO}\right)=-318.2kJ/mol\\ \Delta {\text{G}}_f^{\circ }\left(\text{Zn}\right)=0\\ \Delta {\text{G}}_f^{\circ }\left({\text{O}}_2\right)=\text{ 0}\end{array}$

Substitute the values of standard Gibbs energy of formation of reactants and products in the equation above,

$\begin{array}{l}\Delta G^{\text{o}}=-318.2\text{ kJ/mol }-[0+0]\\ \Delta G^{\text{o}}=-318.2\text{ kJ/mol}\end{array}$

Thus, the standard free energy change $\left(\Delta G^{\text{o}}\right)$ is $-318.2\text{ kJ/mol}$.

The standard emf of battery is calculated as follows:

$\begin{array}{l}\Delta G^{°}=-n\text{F}{E}_{\text{cell}}^{\text{o}}\\ E_{\text{cell}}^{°}=\frac{-\Delta G^{°}}{n\text{F}}\end{array}$

Here, $\Delta G^{°}$ is the standard free energy change, $\text{F}$ is the Faraday constant $\left(96500{\text{ C/mol e}}^{-}\right)$, $E_{\text{cell}}^{°}$ is the standard emf of the cell, and $n$ is the numbers of electrons transferred.

Substitute the values of $\Delta G^{°}$, $F$, and $n$

$\left(2\text{ moles}\right)$ in the equation above:

$\begin{array}{l}{E}_{\text{cell}}^{°}=\frac{-318.2\times {10}^3\text{}\cancel{\text{J/mol}}}{-\left(2\right)\left(96500\cancel{\text{J}}\text{/V}\text{.}\cancel{\text{mol}}\right)}\\ E_{\text{cell}}^{°}=1.65\text{ V}\end{array}$

Therefore, the standard emf of battery is $1.65\text{ V}$.

b) The emf under actual operating conditions when the partial pressure of oxygen is $0.21\text{ atm}$.

The Nernst equation is as follows:

$E=E^{\text{o}}-\frac{0.0592\text{ V}}{n}\log Q$

Substitute the values of $E^{\text{o}}$, $n$, and $Q$ in the equation above:

$\begin{array}{l}E=1.65\text{ V}-\frac{0.0592\text{ V}}{2}\log \frac{1}{p_{{\text{O}}_2}}\\ E=1.65\text{ V}-\frac{0.0592\text{ V}}{2}\log \frac{1}{0.21}\\ E=1.65\text{ V}-0.020\text{ V}\\ E=1.63\text{ V}\end{array}$

Therefore, the emf is $1.63\text{ V}$.

c) The energy density of zinc electrode.

The maximum energy obtained from the reaction is the free energy. The energy density of the zinc electrode is calculated with the help of free energy.

The energy density of the zinc electrode is calculated as follows:

$\text{Energy density }=\text{ Free energy}\times \text{Number of moles of Zn}$

Substitute the values of free energy $\left(318.2\text{ kJ}\right)$ and number of moles in the equation above:

$\begin{array}{l}\text{Energy density}=\left(\frac{318.2\text{ kJ}}{\cancel{1\text{ mol zn}}}\times \frac{\cancel{1\text{ mol Zn}}}{65.41\cancel{\text{g Zn}}}\times \frac{1000\cancel{\text{g Zn}}}{1\text{ kg Zn}}\right)\\ \text{Energy density}=4.86\times {10}^3\text{ kJ/kg Zn}\end{array}$

Therefore, the energy density is $4.86\times {10}^3\text{ kJ/kg Zn}$.

d) The volume of air would need to be supplied to the battery every second.

The number of moles can be calculated with the help of charge as follows:

$\text{charge }=n\text{F}$

Here, $n$ is the number of moles and F is the faraday constant.

Substitute the values in the equation above:

$\begin{array}{c}2.1\times {10}^5\text{ C}=n\left(96500{\text{ C/mol e}}^{-}\right)\\ \frac{2.1\times {10}^5\cancel{\text{C}}}{96500\cancel{\text{C}}{\text{/mol e}}^{-}}=n\\ 2.2{\text{ mol e}}^{-}=n\end{array}$

Therefore, the number of moles is $2.2{\text{ mol e}}^{-}$.

In the balanced reaction, $4$ moles of electron reduce $1$ mole of oxygen. Thus, the number of moles of oxygen reduced by $2.2$ moles is calculated as follows:

$\begin{array}{l}{\text{mol O}}_2=2.2\cancel{{\text{mol e}}^{-}}\times \frac{1{\text{ mol O}}_2}{4\cancel{{\text{mol e}}^{-}}}\\ {\text{mol O}}_2=0.55{\text{ mol O}}_2\end{array}$

Therefore, the number of moles of oxygen reduced by $2.2$ moles is $0.55$.

The volume of oxygen at $1$ atm pressure is calculated with the help of ideal gas equation as follows:

$\begin{array}{l}PV=n\text{R}T\\ V=\frac{n\text{R}T}{P}\end{array}$

Here, $P$ is the pressure, $n$ is the number of moles, $\text{R}$ is the gas constant, $T$ is the temperature, and $V$ is the volume of the gas.

Substitute the values of pressure, moles, temperature, and gas constant in the equation above,

$\begin{array}{l}{V}_{{\text{O}}_2}=\frac{\left(0.55\cancel{\text{mol}}\right)\left(0.0821\text{ L}\text{.}\cancel{\text{atm}}\text{/}\cancel{\text{mol}}\text{.}\cancel{\text{K}}\right)\left(298\cancel{\text{K}}\right)}{1\cancel{\text{atm}}}\\ V_{{\text{O}}_2}=13.5\text{ L}\end{array}$

Therefore, the volume of oxygen is $13.5\text{ L}$.

As the air is $21\%$ of oxygen by its volume.

The volume of air essential to supply the battery every second is calculated as follows:

$\begin{array}{l}{V}_{\text{air}}=\left(13.5\text{ L}\times \frac{\text{100\% air}}{21\%{\text{ O}}_2}\right)\\ V_{\text{air}}=64\text{ L}\end{array}$

Therefore, the volume of air essential to supply the battery every second is $64\text{ L}$.