Question

Consider the fuel cell that accomplishes the overall reaction

${\mathbf{H}}_{\mathbf{2}}\left(\mathbf{g}\right)\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{O}}_{\mathbf{2}}\left(\mathbf{g}\right)\to {\mathbf{H}}_{\mathbf{2}}\mathbf{O}\left(\mathbf{l}\right)$

If the fuel cell operates with 60% efficiency, calculate the amount of electrical work generated per gram of water produced. The gas pressures are constant at 1 atm, and the temperature is 25°C.

Short answer

The maximum amount of electrical work generated per gram of water produced in the fuel cell is 7900 J/g

Step-by-step solution

The fuel cell reaction

${\mathrm{H}}_2\left(\mathrm{g}\right)+\frac{1}{2}{\mathrm{O}}_2\left(\mathrm{g}\right)\to {\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$

The standard free energy change of the water $\left({\mathrm{H}}_2\mathrm{O}\right)$ is,

${\mathrm{\Delta G}}_{\mathrm{f}}^{\mathrm{o}}=-237.8{\mathrm{kJmol}}^{-1}$

Calculate the electrical work generated per mole of water produced in the fuel cell

The efficiency of the fuel cell is 60 %

= 0.60

The standard free energy change of the water$\left({\mathrm{H}}_2\mathrm{O}\right)$is,

${\mathrm{\Delta G}}_{\mathrm{f}}^{\mathrm{o}}=-237.8{\mathrm{kJmol}}^{-1}$

Calculate the maximum amount of electrical work generated per mole of water $\left({\mathrm{H}}_2\mathrm{O}\right)$produced,

$\begin{array}{l}-{\mathrm{W}}_{\max }{=-\mathrm{\epsilon \Delta G}}_{\mathrm{f}}^{\mathrm{o}}\\ =\left(-0.60\right)\times \left(-237.8{\mathrm{kJmol}}^{-1}\right)\end{array}$

$-{\mathrm{W}}_{\max }=142.32{\mathrm{kJmol}}^{-1}$

Therefore, the maximum amount of electrical work generated per mole of water$\left({\mathrm{H}}_2\mathrm{O}\right)$ produced in the fuel cell is 142.32kJ/ mol.

Calculate the electrical work generated per gram of water produced in the fuel cell.

$\left({\mathrm{H}}_2\mathrm{O}\right)$The amount of electrical work generated per mole of water = 142.32 kJ/ mol.

Molar mass of water = 18.02 g/ mol.

Now calculate the amount of electrical work generated per gram of water$\left({\mathrm{H}}_2\mathrm{O}\right)$,

$$\begin{gathered} -{\mathrm{W}}_{\max }=142.32\frac{\mathrm{kJ}}{\mathrm{mol}}\times \frac{1000\mathrm{J}}{1\mathrm{kJ}}\times \frac{1\mathrm{mol}}{18.02\mathrm{g}} \\ \begin{array}{c}=7897\mathrm{J}/\mathrm{g}\\ -{\mathrm{W}}_{\max }=7900\mathrm{J}/\mathrm{g}\end{array} \end{gathered}$$

Hence, the maximum amount of electrical work generated per gram of water is 7900J/ g.