Suppose that a hydrogen fuel cell, as described in the text, is to be operated at $75°\text{C}$and atmospheric pressure. We wish to estimate the maximum electrical work done by the cell, using only the room temperature data at the back of this book. It is convenient to first establish a zero-point for each of the three substances, ${\text{H}}_2{\text{,O}}_2{\text{,andH}}_2\text{O}$. Let us take $\text{G}$for both ${\text{H}}_2{\text{andO}}_2$to be zero at $25°\text{C}$, so that G for a mole of ${\text{H}}_2\text{O}$is $-237\text{KJ}$at $25°\text{C}$.

(a) Using these conventions, estimate the Gibbs free energy of a mole of ${\text{H}}_2$at $75°\text{C}$. Repeat for ${\text{O}}_2{\text{andH}}_2\text{O}$.

(b) Using the results of part (a), calculate the maximum electrical work done by the cell at $75°\text{C}$, for one mole of hydrogen fuel. Compare to the ideal performance of the cell at$25°\text{C}$.

The chemical reaction in hydrogen fuel cell is ${\text{H}}_2{\text{O→H}}_2\text{+}\frac{1}{2}{\text{O}}_2$.

The Gibbs free energy for ${\text{H}}_2$and ${\text{O}}_2$at $25°\text{C}$is zero each and for ${\text{H}}_2\text{O}$at $25°\text{C}$is -237 KJ.

Here, Gibbs free energy,

$$\begin{gathered} \text{G=U-TS+PV} \\ \text{where} \\ \text{G=Gibbsfreeenergy} \\ \text{T=absolutetemperature} \\ \text{S=entropy} \\ \text{P=Pressure} \\ \text{V=Volume} \end{gathered}$$

The infinitesimal change in U is

$\text{dU=TdS-PdV+μdN·····}\left(1\right)$

The infinitesimal change in $G$is

localid="1647330926956" $\text{dG=dU-TdS-SdT+PdV+VdP·····}\left(2\right)$

From (1) and (2)

$$\begin{gathered} \text{dG=TdS-PdV+μdN-TdS-SdT+PdV+VdP} \\ =\text{μdN-SdT+VdP·····}\left(3\right) \end{gathered}$$

AS, P and N are constant.

So, equation (3) reduces to

$\text{dG=-SdT}$.

And relation between Gibbs energy at two different temperatures $25°\text{C}$and $75°\text{C}$is:

${\text{dG}}_{75}{\text{=dG}}_{25}\text{-SdT······}\left(4\right)$

Using

$$\begin{gathered} {\text{dG}}_{25}\left({\text{H}}_2\right)\text{=0} \\ {\text{S=131kJ·K}}^{-1} \\ \text{dT=50K} \end{gathered}$$

in equation (4)

$$\begin{gathered} {\text{dG}}_{75}\left({\text{H}}_2\right)\text{=0-}\left(131{\text{kJ·K}}^{-1}\right)\left(50\text{K}\right) \\ \text{=-6550kJ} \end{gathered}$$

Similarly

$$\begin{gathered} {\text{dG}}_{25}\left({\text{O}}_2\right)\text{=0} \\ {\text{S=205kJ·K}}^{-1} \\ \text{dT=50K} \end{gathered}$$

So, equation (4) becomes

$$\begin{gathered} {\text{dG}}_{75}\left({\text{O}}_2\right)\text{=0-}\left(205{\text{kJ·K}}^{-1}\right)\left(50\text{K}\right) \\ \text{=-10250kJ} \end{gathered}$$

Now,

$$\begin{gathered} {\text{dG}}_{25}\left({\text{O}}_2\right){\text{=-237×10}}^3\text{J} \\ {\text{S=70kJ·K}}^{-1} \\ \text{dT=50K} \end{gathered}$$

Hence equation (4) becomes

$$\begin{gathered} {\text{dG}}_{75}\left({\text{H}}_2\text{O}\right)\text{=}\left(-237\times {10}^3\text{J}\right)\text{-}\left(70{\text{kJ·K}}^{-1}\right)\left(50\text{K}\right) \\ \text{=-240500kJ} \end{gathered}$$

The Gibbs free energy for ${\text{H}}_2$is -6550 kJ, for ${\text{O}}_2$is -10250 kJ and for ${\text{H}}_2\text{O}$is -240500 kJ.

The chemical reaction in hydrogen fuel cell is ${\text{H}}_2{\text{O→H}}_2\text{+}\frac{1}{2}{\text{O}}_2$.

The Gibbs free energy for ${\text{H}}_2$and ${\text{O}}_2$at $25°\text{C}$is zero each and for ${\text{H}}_2\text{O}$at $25°\text{C}$is -237 KJ.

Gibbs energy at$75°\text{C}$is

As,

So,

$$\begin{gathered} {\text{dG}}_{75}\text{=-240500kJ+6550kJ+}\frac{1}{2}\left(10250\text{kJ}\right) \\ \text{=-228825kJ} \end{gathered}$$

Hence, the electrical work done for each mole of hydrogen fuel is 228825 kJ.