Question

Consider a fuel cell that uses methane ("natural gas") as fuel. The reaction is

${\mathrm{CH}}_4+2{\mathrm{O}}_2⟶2{\mathrm{H}}_2\mathrm{O}+{\mathrm{CO}}_2$

(a) Use the data at the back of this book to determine the values of $\Delta H$and $\Delta G$for this reaction, for one mole of methane. Assume that the reaction takes place at room temperature and atmospheric pressure.

(b) Assuming ideal performance, how much electrical work can you get out of the cell, for each mole of methane fuel?

(c) How much waste heat is produced, for each mole of methane fuel?

(d) The steps of this reaction are

$$\begin{gathered} \mathrm{at}-\mathrm{electrode}:{\mathrm{CH}}_4+2{\mathrm{H}}_2\mathrm{O}\to {\mathrm{CO}}_2+8{\mathrm{H}}^{+}+8{\mathrm{e}}^{-} \\ \mathrm{at}-\mathrm{electrode}:2{\mathrm{O}}_2+8{\mathrm{H}}^{+}+8{\mathrm{e}}^{-}\to 4{\mathrm{H}}_2\mathrm{O} \end{gathered}$$

What is the voltage of the cell?

Short answer

(a) The value in the change of enthalpy is $-890.36\mathrm{kJ}$and the value in the change of Gibbs free energy is $-817.9\mathrm{kJ}$.

(b) The electrical work done for each mole of methane fuel is -817.9 kJ.

(c) The amount of waste heat produced for each mole of methane fuel is 72.46 kJ.

(d) The voltage of the cell is 1.061 V.

Step-by-step solution

Explanation

Given:

The transition is

${\mathrm{CH}}_{4}12{\mathrm{O}}_2,2{\mathrm{H}}_2\mathrm{O}\mid {\mathrm{CO}}_2$

The temperature is 208k and the pressure is 1 bar.

Formula used:

Write the expression for Gibbs energy-

$G=H-TS$

Here, G is Gibbs energy, I is the enthalpy, T is the absolute Write the expression for the infinitesimal change in G.

$\Delta G=\Delta H-T\Delta S_{m\dots }\left(1\right)$

Write the expression for the change in enthalpy for the reaction

$\Delta G=2\Delta G_{{\mathrm{H}}_2\mathrm{O}}+\Delta G_{{\mathrm{CO}}_2}-\Delta G_{{\mathrm{CH}}_4}-2\Delta G_{{\mathrm{O}}_2}\dots \dots ..\text{(3)}$

Calculation

Refer table at the back of the book.

Substitute $-393.51\mathrm{kJ}$for$\Delta H_{{\mathrm{CO}}_2},-285.83\mathrm{kJ}$for $\Delta H_{{\mathrm{H}}_2\mathrm{O}},0$for $\Delta H_{{\mathrm{O}}_2}$and $-74.81\mathrm{kJ}$for $\Delta H_{{\mathrm{CH}}_4}$from the table in expression (2).

Thus, the value in the change of enthalpy is $-890.36\mathrm{kJ}$ and the value in the change of Gibbs free energy is $-817.9\mathrm{kJ}$.

Step 3. (b) Given information

The reaction is ${\mathrm{CH}}_4+2{\mathrm{O}}_2\to 2{\mathrm{H}}_2\mathrm{O}+{\mathrm{CO}}_2$.

Temperature is 298 K.

Pressure is 1 bar.

Formula used:

Work done, $\mathrm{W}=∆\mathrm{G}$

where

G=Gibbs free energy

W=work done

Step 4. Calculation

Here, $∆\mathrm{G}=-817.9\mathrm{kJ}$.

So,$\mathrm{W}=-817.9\mathrm{kJ}$.

Step 5. Conclusion

Hence, the electrical work done for each mole of methane fuel is -817.9 kJ.

Step 6. (c) Given information

The reaction is ${\mathrm{CH}}_4+2{\mathrm{O}}_2\to 2{\mathrm{H}}_2\mathrm{O}+{\mathrm{CO}}_2$.

Temperature is 298 K.

Pressure is 1 bar.

As the reaction is occuring at constant pressure.

So,

$$\begin{gathered} ∆\mathrm{Q}=∆{\mathrm{H}}_{\mathrm{r}}-∆{\mathrm{H}}_{\mathrm{p}} \\ \mathrm{where} \\ ∆\mathrm{Q}=\mathrm{Energy}\mathrm{difference} \\ ∆{\mathrm{H}}_{\mathrm{r}}=\mathrm{Enthalpy}\mathrm{change}\mathrm{for}\mathrm{reactants} \\ ∆{\mathrm{H}}_{\mathrm{p}}=\mathrm{Enthalpy}\mathrm{change}\mathrm{for}\mathrm{the}\mathrm{products} \end{gathered}$$

Step 7. Calculation

As, $∆{\mathrm{H}}_{\mathrm{r}}=890.36\mathrm{kJ}$and $∆{\mathrm{H}}_{\mathrm{p}}=817.9\mathrm{kJ}$.

So,$$\begin{gathered} ∆\mathrm{Q}=890.36\mathrm{kJ}-817.9\mathrm{kJ} \\ =72.46\mathrm{kJ} \end{gathered}$$

Step 8. Conclusion

Hence the amount of waste heat produced for each mole of methane fuel is 72.46 kJ.

Step 9. Given information

The reaction is ${\mathrm{CH}}_4+2{\mathrm{O}}_2\to 2{\mathrm{H}}_2\mathrm{O}+{\mathrm{CO}}_2$.

Temperature is 298 K.

Pressure is 1 bar.

Formula for the work done per each electron is

$$\begin{gathered} {\mathrm{W}}_{\mathrm{e}}=\frac{\mathrm{W}}{8{\mathrm{N}}_{\mathrm{A}}} \\ \mathrm{where} \\ \mathrm{W}=\mathrm{work}\mathrm{done}\mathrm{per}\mathrm{mole} \\ {\mathrm{N}}_{\mathrm{A}}=\mathrm{Avogadro}\text{'}\mathrm{s}\mathrm{number} \end{gathered}$$

Step 10. Calculation

Here

$$\begin{gathered} \mathrm{W}=817.9\mathrm{kJ} \\ {\mathrm{N}}_{\mathrm{A}}=6.623\times {10}^{23} \end{gathered}$$

So,

$$\begin{gathered} {\mathrm{W}}_{\mathrm{e}}=\frac{817.9\mathrm{kJ}}{8\left(6.623\times {10}^{23}\right)} \\ =1.061\mathrm{eV} \end{gathered}$$

Step 11. Conclusion

Hence, the voltage of the cell is 1.061 V.