A given type of fuel cell produces electrical energy $W$ by the interaction of $\mathrm{O}_{2}$ fed into one electrode and $\mathrm{H}_{2}$ to the other. These gases are fed in at 1 atmosphere pressure and $298 \mathrm{~K}$, and react isothermally and isobarically to form water. Assuming that the reaction occurs reversibly and that the internal resistance of the cell is negligible, calculate the e.m.f. of the cell. Given: one Faraday $=96,500$ coulombs $/ \mathrm{g}$ mole.

Enthalpies in joules/g mole at 1 atmospheric and $298 \mathrm{~K}$ for oxygen, hydrogen, and water are respectively $17,200,8,100$ and $-269,300$.

Entropies in joules/mole. $\mathrm{K}$ at 1 atmosphere and $298 \mathrm{~K}$ for oxygen, hydrogen, and water are respectively 201,128 and $66.7$.
The chemical equation is

$$
\mathrm{H}_{2}+\frac{1}{2} \mathrm{O}_{2}=\mathrm{H}_{2} \mathrm{O} \text {. }
$$

In the reversible process at constant temperature and pressure, the decrease of Gibbs function of the system is equal to the difference between the total external work and the work the system does because of the change of volume. Thus

$$
-\Delta g=\varepsilon \Delta q,
$$

or

$$
-\sum_{\mathrm{i}}\left(\Delta h_{\mathrm{i}}-T \Delta S_{\mathrm{i}}\right)=\varepsilon \Delta q \text {. }
$$

If 1 mole of water forms, there must have been electric charges of $2 \mathrm{~F}$ flowing in the circuit, i.e., $\Delta g=2 \mathrm{~F}$. Thus the e.m.f. is

$$
\varepsilon=\frac{1}{2 \mathrm{~F}}\left[T S_{\mathrm{W}}-h_{\mathrm{W}}-T\left(S_{\mathrm{H}}+\frac{1}{2} S_{0}\right)+\left(h_{\mathrm{H}}+\frac{1}{2} h_{0}\right)\right]
$$

As given $, S_{0}=201 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K}, S_{\mathrm{H}}=128 \mathrm{~J} / \mathrm{mol} \mathrm{K}$, $S_{\mathrm{W}}=66.7 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K}, \dot{h}_{0}=-17200 \mathrm{~J} / \mathrm{mol}$ $h_{\mathrm{H}}=8100 \mathrm{~J} / \mathrm{mol}, h_{\mathrm{W}}=-269300 \mathrm{~J} / \mathrm{mol}$, and $T=298 \mathrm{~K}$, We have $\varepsilon=1.23 \mathrm{~V}$