Chapter 17, Problem 2SP

(a)

Interpretation Introduction

Interpretation:

The value of ${\Delta }_{\text{r}}{G}^{\circ }$ for nickel-zinc battery and iron-nickel battery has to be determined.

Concept Introduction:

The Gibbs free energy of a system is defined as the enthalpy of the system minus the product of the temperature times the entropy of the system.  The Gibbs free energy of the system is a state function as it is defined in terms of thermodynamic properties that are state functions.

Answer to Problem 2SP

The value of ${\Delta }_{\text{r}}{G}^{\circ }$ for nickel-zinc battery and iron-nickel battery is $\underset{\_}{-318.40kJ/mol}$ and $\underset{\_}{-264.36kJ/mol}$ respectively.

Explanation of Solution

The half-cell reactions which take place at cathode and anode in nickel-zinc battery are shown below.

  $\begin{array}{cc}\text{Cathode}:& \text{NiO}\left(\text{OH}\right)\left(\text{s}\right)+{\text{H}}_2\text{O}\left(l\right)+{\text{e}}^{-}\to \text{Ni}{\left(\text{OH}\right)}_2\left(\text{s}\right)+{\text{OH}}^{-}\\ \text{Anode}:& \text{Zn}\left(\text{s}\right)+4{\text{OH}}^{-}\left(\text{aq}\right)\to \text{Zn}{\left(\text{OH}\right)}_4{}^{2-}+2{\text{e}}^{-}\end{array}$

The number of electron change in the half cell reactions of nickel-zinc battery is $2$.

As per the given data, the standard cell potential, $E_{\text{cell}}^{°}$ for the nickel-zinc battery is $1.65\text{V}$.

The relation between standard cell potential and standard free energy change is shown below.

    ${\Delta }_{\text{r}}{G}^{\circ }\text{=}-nF{E}_{\text{cell}}^{\circ }$        (1)

Where,

• ${\Delta }_{\text{r}}{G}^{\circ }$ is the standard free energy change.
• $E_{\text{cell}}^{\circ }$ is the standard cell potential.
• $F$ is the Faradays constant.
• $n$ is the number of electron change in the half cell reaction.

The value of $n$ is $2$.

The value of $E_{\text{cell}}^{\circ }$ is $1.65\text{V}$.

The value of $F$ is $96485\text{C/mol}{\text{e}}^{-}$.

Substitute the value of $n$, $E_{\text{cell}}^{\circ }$ and $F$ for nickel-zinc battery in equation (1).

  $\begin{array}{c}{\Delta }_{\text{r}}{G}^{\circ }\text{=}-2{\text{e}}^{-}\times \frac{96485\text{C}}{1\text{mol}{\text{e}}^{-}}\times 1.65\text{V}\\ =-\frac{\text{318400}\text{.5}\text{J}}{1\text{mol}}\times \frac{1\text{kJ}}{1000\text{J}}\\ =-318.40\text{kJ/mol}\end{array}$

Thus, the ${\Delta }_{\text{r}}{G}^{\circ }$ for nickel-zinc battery is $\underset{\_}{-318.40kJ/mol}$.

The half-cell reactions which take place at cathode and anode in iron-nickel battery are shown below.

  $\begin{array}{cc}\text{Cathode}:& \text{NiO}\left(\text{OH}\right)\left(\text{s}\right)+{\text{H}}_2\text{O}\left(l\right)+{\text{e}}^{-}\to \text{Ni}{\left(\text{OH}\right)}_2\left(\text{s}\right)+{\text{OH}}^{-}\\ \text{Anode}:& \text{Fe}\left(\text{s}\right)+2{\text{OH}}^{-}\left(\text{aq}\right)\to \text{Fe}{\left(\text{OH}\right)}_2\left(\text{s}\right)+2{\text{e}}^{-}\end{array}$

The number of electron change in the half cell reactions of iron-nickel battery is $2$. 

As per the given data the standard cell potential, $E_{\text{cell}}^{°}$ for the iron-nickel battery is $1.37\text{V}$.

The value of $n$ is $2$.

The value of $E_{\text{cell}}^{\circ }$ is $1.37\text{V}$.

The value of $F$ is $96485\text{C/mol}{\text{e}}^{-}$.

Substitute the value of $n$, $E_{\text{cell}}^{\circ }$ and $F$ for iron-nickel battery in equation (1).

  $\begin{array}{c}{\Delta }_{\text{r}}{G}^{\circ }\text{=}-2{\text{e}}^{-}\times \frac{96485\text{C}}{1\text{mol}{\text{e}}^{-}}\times 1.37\text{V}\\ =-\frac{\text{264368}\text{.9}\text{J}}{1\text{mol}}\times \frac{1\text{kJ}}{1000\text{J}}\\ =-264.36\text{kJ/mol}\end{array}$

Thus, the ${\Delta }_{\text{r}}{G}^{\circ }$ for iron-nickel battery is $\underset{\_}{-264.36kJ/mol}$.

(b)

Interpretation Introduction

Interpretation:

The battery which is more product favored reaction at standard conditions has to be identified.

Concept Introduction:

If the value of equilibrium constant, $K^{\circ }$ for a battery is larger than $1$, then the reactions occurring in the battery are the product favored reaction.  The relation between standard cell potential and equilibrium constant is shown below.

    $\ln K^{\circ }\text{=}\frac{nF{E}_{\text{cell}}^{\circ }}{RT}$

Answer to Problem 2SP

Nickel-zinc battery has the more product favored reaction at standard conditions.

Explanation of Solution

The half-cell reactions which take place at cathode and anode in nickel-zinc battery are shown below.

  $\begin{array}{cc}\text{Cathode}:& \text{NiO}\left(\text{OH}\right)\left(\text{s}\right)+{\text{H}}_2\text{O}\left(l\right)+{\text{e}}^{-}\to \text{Ni}{\left(\text{OH}\right)}_2\left(\text{s}\right)+{\text{OH}}^{-}\\ \text{Anode:}& \text{Zn}\left(\text{s}\right)+4{\text{OH}}^{-}\left(\text{aq}\right)\to \text{Zn}{\left(\text{OH}\right)}_4{}^{2-}+2{\text{e}}^{-}\end{array}$

The number of electron change in the half cell reactions of nickel-zinc battery is $2$. 

As per the given data the standard cell potential, $E_{\text{cell}}^{°}$ for the nickel-zinc battery is $1.65\text{V}$.

The relation between standard cell potential and equilibrium constant is shown below.

    $\ln K^{\circ }\text{=}\frac{nF{E}_{\text{cell}}^{\circ }}{RT}$        (2)

Where,

• $K^{\circ }$ is the equilibrium constant.
• $E_{\text{cell}}^{\circ }$ is the standard cell potential.
• $F$ is the Faradays constant.
• $n$ is the number of electron change in the half cell reaction.
• $R$ is the gas constant.
• $T$ is the standard temperature.

The value of $n$ is $2$.

The value of $E_{\text{cell}}^{\circ }$ is $1.65\text{V}$.

The value of $F$ is $96485\text{C/mol}{\text{e}}^{-}$.

The value of $R$ is $8.314\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}$.

The value of $T$ is $298\text{K}$.

Substitute the value of $n$, $E_{\text{cell}}^{\circ }$, $F$, $R$ and $T$ for nickel-zinc battery in equation (2).

    $\begin{array}{c}\ln K^{\circ }\text{=}\frac{\left(2\right)\left(\frac{96485\text{C}}{1\text{mol}{\text{e}}^{-}}\right)\left(1.65\text{V}\right)}{\left(8.314\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\right)\left(298\text{K}\right)}\\ K^{\circ }={\text{e}}^{128.5}\\ =6.47\times {10}^{55}\end{array}$

Therefore, the value of equilibrium constant, $K^{\circ }$ for nickel-zinc battery is $6.47\times {10}^{55}$.

The half-cell reactions which take place at cathode and anode in iron-nickel battery are shown below.

  $\begin{array}{cc}\text{Cathode}:& \text{NiO}\left(\text{OH}\right)\left(\text{s}\right)+{\text{H}}_2\text{O}\left(l\right)+{\text{e}}^{-}\to \text{Ni}{\left(\text{OH}\right)}_2\left(\text{s}\right)+{\text{OH}}^{-}\\ \text{Anode}:& \text{Fe}\left(\text{s}\right)+2{\text{OH}}^{-}\left(\text{aq}\right)\to \text{Fe}{\left(\text{OH}\right)}_2\left(\text{s}\right)+2{\text{e}}^{-}\end{array}$

The number of electron change in the half cell reactions of iron-nickel battery is $2$. 

As per the given data the standard cell potential, $E_{\text{cell}}^{°}$ for the iron-nickel battery is $1.37\text{V}$.

The value of $n$ is $2$.

The value of $E_{\text{cell}}^{\circ }$ is $1.37\text{V}$.

The value of $F$ is $96485\text{C/mol}{\text{e}}^{-}$.

The value of $R$ is $8.314\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}$.

The value of $T$ is $298\text{K}$.

Substitute the value of $n$, $E_{\text{cell}}^{\circ }$, $F$, $R$ and $T$ for iron-nickel battery in equation (2).

    $\begin{array}{c}\ln K^{\circ }\text{=}\frac{\left(2\right)\left(\frac{96485\text{C}}{1\text{mol}{\text{e}}^{-}}\right)\left(1.37\text{V}\right)}{\left(8.314\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\right)\left(298\text{K}\right)}\\ K^{\circ }={\text{e}}^{106.70}\\ =2.18\times {10}^{46}\end{array}$

Therefore, the value of equilibrium constant, $K^{\circ }$ for iron-nickel battery is $2.18\times {10}^{46}$.

Since the values of equilibrium constant, $K^{\circ }$ for nickel-zinc battery and iron-nickel battery are larger than $1$.  Therefore both reactions are the product favored reaction.  But the equilibrium constant, $K^{\circ }$ for nickel-zinc battery is greater than that for iron-nickel battery.

Thus, nickel-zinc battery has the more product favored reaction at standard conditions.

(c)

Interpretation Introduction

Interpretation:

The value for equilibrium constant for nickel-zinc battery has to be determined.

Concept Introduction:

If the value of equilibrium constant, $K^{\circ }$ for a battery is larger than $1$, then the reactions occurring in the battery are the product favored reaction.  The relation between standard cell potential and equilibrium constant is shown below.

    $K^{\circ }={\text{e}}^{\frac{nF{E}_{\text{cell}}^{\circ }}{RT}}$

Answer to Problem 2SP

The equilibrium constant, $K^{\circ }$ for nickel-zinc battery is very large.

Explanation of Solution

The half-cell reactions which take place at cathode and anode in nickel-zinc battery are shown below.

  $\begin{array}{cc}\text{Cathode}:& \text{NiO}\left(\text{OH}\right)\left(\text{s}\right)+{\text{H}}_2\text{O}\left(l\right)+{\text{e}}^{-}\to \text{Ni}{\left(\text{OH}\right)}_2\left(\text{s}\right)+{\text{OH}}^{-}\\ \text{Anode}:& \text{Zn}\left(\text{s}\right)+4{\text{OH}}^{-}\left(\text{aq}\right)\to \text{Zn}{\left(\text{OH}\right)}_4{}^{2-}+2{\text{e}}^{-}\end{array}$

The number of electron change in the half cell reactions of nickel-zinc battery is $2$.

As per the given data the standard cell potential, $E_{\text{cell}}^{°}$ for the nickel-zinc battery is $1.65\text{V}$.

The relation between standard cell potential and equilibrium constant is shown below.

    $K^{\circ }={\text{e}}^{\frac{nF{E}_{\text{cell}}^{\circ }}{RT}}$        (3)

Where,

• $K^{\circ }$ is the equilibrium constant.
• $E_{\text{cell}}^{\circ }$ is the standard cell potential.
• $F$ is the Faradays constant.
• $n$ is the number of electron change in the half cell reaction.
• $R$ is the gas constant.
• $T$ is the standard temperature.

The value of $n$ is $2$.

The value of $E_{\text{cell}}^{\circ }$ is $1.65\text{V}$.

The value of $F$ is $96485\text{C/mol}{\text{e}}^{-}$.

The value of $R$ is $8.314\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}$.

The value of $T$ is $298\text{K}$.

Substitute the value of $n$, $E_{\text{cell}}^{\circ }$, $F$, $R$ and $T$ for nickel-zinc battery in equation (3).

    $\begin{array}{c}{K}^{\circ }=\text{e}{}^{\frac{\left(2\right)\left(\frac{96485\text{C}}{1\text{mol}{\text{e}}^{-}}\right)\left(1.65\text{V}\right)}{\left(8.314\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\right)\left(298\text{K}\right)}}\\ ={\text{e}}^{128.5}\\ =6.41\times {10}^{55}\end{array}$

Therefore, the value of equilibrium constant, $K^{\circ }$ for nickel-zinc battery is $6.41\times {10}^{55}$, which is larger than one.