Rather than pointing whether the assumptions are correct or incorrect, I will show you a general method to obtain the Nernst's equation for any system. It is a more systematic and formal approach, generally found in electrochemistry books, but has the advantage that all the magnitudes pop up so you can clearly identify what each letter means.
The condition of electrochemical equilibrium for a single electrochemical reaction of $N$ species is
$$ \sum_{j = 1}^N \nu_j \tilde{\mu}_j = 0 \tag1 $$
where $\nu_j$ is the stoichiometric coefficient, and $\tilde{\mu}_j$ the electrochemical potential defined as
$$ \tilde{\mu}_j \equiv \mu_j^\circ + RT\ln a_j + z_j F \phi \tag2 $$
where $\mu_j^\circ$ is the standard chemical potential, $a_j$ is the activity of species $j$, and $\phi$ the electric potential.
The important rules are:
- For electrons, we drop out its standard chemical potential and the activity, so that $\tilde{\mu}_\ce{e^-} = -F \phi^\mathrm{M}$. The letter $\mathrm{M}$ is because the electron as a species belongs to the electrode.
- For hydrogen ions, we assign its standard chemical potential as zero for an aqueous solution at all temperatures, i.e. $\mu_\ce{H+}^0 \equiv 0$. However, in a fuel cell the liquid phase is not water, so we cannot eliminate it along the derivations.
- For elements in their most stable state of aggregation, their standard chemical potential is zero. For our case, $\mu_\ce{H2}^0 = \mu_\ce{O2}^0 \equiv 0$.
The electrochemical reaction in the negative electrode, written as a reduction, is
$$ \ce{2H^+(aq) + 2e^-(M,n) <=> H2(g)} \tag{3} $$
and application of Eq. (1) leads to
\begin{align}
2\mu_\ce{H+}^\circ + 2RT\ln(a_\ce{H+}) + 2F\phi^\mathrm{S,n}
- 2F\phi^\mathrm{M,n} &= RT \ln (a_\ce{H2}) \\
2F\phi^\mathrm{S,n} - 2F\phi^\mathrm{M,n} &=
-2\mu_\ce{H+}^\circ - RT\ln(a_\ce{H+}^2) + RT\ln(a_\ce{H2}) \\
-2F(\phi^\mathrm{M,n} - \phi^\mathrm{S,n}) &= -2\mu_\ce{H+}^\circ
+ RT\ln\left(\frac{a_\ce{H2}}{a_\ce{H+}^2}\right) \\
\phi^\mathrm{M,n} - \phi^\mathrm{S,n} &= \frac{\mu_\ce{H+}^\circ}{F}
- \frac{RT}{2F}\ln\left(\frac{a_\ce{H2}}{a_\ce{H+}^2}\right) \\
E_\text{n} &= {E_\text{n}}^\circ -
\frac{RT}{2F}\ln\left(\frac{a_\ce{H2}}{a_\ce{H+}^2}\right) \tag{4} \\
\end{align}
where we defined the electrode potential $E_\text{n} \equiv \phi^\mathrm{M,n} - \phi^\mathrm{S,n}$ and the standard reduction potential ${E_\text{n}}^\circ \equiv \mu_\ce{H+}^\circ /F$ for the reaction in the negative electrode.
The electrochemical reaction in the positive electrode is
$$ \ce{1/2 O2(g) + 2H^+(aq) + 2e^-(M,p) <=> H2O(g)} \tag5 $$
and application of Eq. (1) leads to
\begin{align}
&\frac{1}{2}RT \ln (a_\ce{O2}) + 2\mu_\ce{H+}^\circ + 2RT\ln(a_\ce{H+}) +
2F\phi^\mathrm{S,p} - 2F\phi^\mathrm{M,p}
= \mu_\ce{H2O}^\circ + RT\ln(a_\ce{H2O}) \\
& 2F\phi^\mathrm{S,p} - 2F\phi^\mathrm{M,p}
= \mu_\ce{H2O}^\circ - 2\mu_\ce{H+}^\circ +
RT\ln(a_\ce{H2O}) - RT \ln (a_\ce{O2}^{1/2})- RT\ln(a_\ce{H+}^2) \\
&-2F(\phi^\mathrm{M,p} - \phi^\mathrm{S,p})
= \mu_\ce{H2O}^\circ - 2\mu_\ce{H+}^\circ +
RT\ln\left(\frac{a_\ce{H2O}}{a_\ce{O2}^{1/2}a_\ce{H+}^2}\right) \\
&\phi^\mathrm{M,p} - \phi^\mathrm{S,p}
= -\frac{\mu_\ce{H2O}^\circ - 2\mu_\ce{H+}^\circ}{2F} - \frac{RT}{2F}
\ln\left(\frac{a_\ce{H2O}}{a_\ce{O2}^{1/2}a_\ce{H+}^2}\right) \\
&E_\mathrm{p} = {E_\mathrm{p}}^\circ - \frac{RT}{2F}
RT\ln\left(\frac{a_\ce{H2O}}{a_\ce{O2}^{1/2}a_\ce{H+}^2}\right) \tag6 \\
\end{align}
where we defined the electrode potential $E_\text{p} \equiv \phi^\mathrm{M,p} - \phi^\mathrm{S,p}$ and the standard reaction potential ${E_\text{p}}^\circ \equiv -(\mu_\ce{H2O}^\circ - 2\mu_\ce{H+}^\circ)/2F$ for the reaction in the positive electrode.
Subtracting Eq. (6) from Eq. (5)
\begin{align}
\require{cancel}
E_\mathrm{p} - E_\mathrm{n} &=
{E_\mathrm{p}}^\circ - {E_\mathrm{n}}^\circ - \frac{RT}{2F}
\ln\left(\frac{a_\ce{H2O}}{a_\ce{O2}^{1/2}a_\ce{H+}^2}\right) +
\frac{RT}{2F}\ln\left(\frac{a_\ce{H2}}{a_\ce{H+}^2}\right) \\
E_\mathrm{p} - E_\mathrm{n} &=
{E_\mathrm{p}}^\circ - {E_\mathrm{n}}^\circ -
\frac{RT}{2F}\ln\left(\frac{a_\ce{H2O}\cancel{a_\ce{H+}^2}}
{a_\ce{O2}^{1/2}\cancel{a_\ce{H+}^2}a_\ce{H2}}\right) \\
\Delta E &= \Delta E^\circ -
\frac{RT}{2F}\ln\left(\frac{a_\ce{H2O}}
{a_\ce{O2}^{1/2}a_\ce{H2}}\right) \tag7\\
\end{align}
where we have defined the cell potential $\Delta E \equiv E_\mathrm{p} - E_\mathrm{n}$, and the standard cell potential as $\Delta E^\circ \equiv {E_\mathrm{p}}^\circ - {E_\mathrm{n}}^\circ$.
The activities of a species has a general form which is the product of the activity coefficient $\gamma_j$ and some measure of the concentration. If the gas phase behaves as an ideal gas and the ionic solution behaves as an ideal ionic solution, we can set $\gamma_j = 1$ and stick with the concentration of the species. The three species are in the gaseous phase, so we consider their measure of concentration as their partial pressures, but you can choose anyone you like. The final equation is
\begin{equation}
\boxed{\Delta E = \Delta E^\circ -
\frac{RT}{2F}\ln\left(\frac{p_\ce{H2O}}
{p_\ce{O2}^{1/2}p_\ce{H2}}\right)} \tag8
\end{equation}
Final remarks:
- We crossed out the hydrogen ion concentration. This is only true if its concentration is homogeneous across the cell. In a fuel cell, $\ce{H+}$ is liberated through the oxidation in the anode, travels through the membrane, and reaches the positive electrode. Outside of equilibrium this will not hold.
- As you can see, $\Delta E$ is a collection of electric potential differences, which in this case is $(\phi^\mathrm{M,p} - \phi^\mathrm{S,p}) -(\phi^\mathrm{M,n} - \phi^\mathrm{S,n})$. In conditions of equilibrium, that is called the open circuit voltage in batteries, and measuring the voltage difference at both electrodes should yield $\Delta E$. This is actually not totally true, but this needs another answer. For practical purposes, $\Delta E$ and what the voltmeter says is the same.
