= Solution
The reversible voltage follows from the two-electron hydrogen reaction:
$$
E_{\rm rev}=-\frac{\Delta_rG^\circ}{2F}=\frac{237000}{193000}
=1.22798\ \mathrm V.
$$
At the specified geometric <current density>, the magnitude of each activation loss follows its <Tafel equation>, $\eta=b\log_{10}(j/j_0)$:
$$
\eta_{\rm H}=0.0295\log_{10}(10/10^{-4})=0.1475\ \mathrm V,
\qquad
\eta_{\rm O}=0.09\log_{10}(10/10^{-7})=0.7200\ \mathrm V.
$$
Both losses reduce the delivered <fuel cell> voltage. Therefore \b[$V=0.36048\ \mathrm V$ and $P/A=jV=3.60\ \mathrm{W\,m^{-2}}$]. The very small oxygen exchange current is responsible for most of the activation loss. The calculation uses <current densities> on a common area basis, unit water activity and the stated gas conditions, and neglects the losses the problem excludes.
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