Half cell method of voltage calculation in an electrochemical cell A cell consists of a copper electrode $\ce{Cu}$, in contact with $1.800\cdot10^{-2}~\mathrm{M}$ cupric ion and a platinum electrode, $\ce{Pt}$, in contact with $1.400\cdot10^{-1}~\mathrm{M}$ ferrous ion and $3.700\cdot10^{-2}~\mathrm{M}$ ferric ion in $1~\mathrm{M}$ $\ce{HCl}$.
  What is the potential of cell?


I would have an idea of what to do with this question if it had a concentration for copper in the cupric half reaction; I would just use the nernst equation, rearranged to solve for $E$: 

$$E = E^\circ - \frac{\mathrm{R}T}{n\mathrm{F}}\log Q$$
Then plug in the concentration values for the cupric half-reaction into $Q$ to get the $E$ value for this half reaction. I'm assuming that the half reaction taking place here is:
\begin{align}
\ce{Cu^{2+} + 2e^- &amp;-&gt; Cu^0} &amp;E^\circ &amp;= +0.337~\mathrm{V}
\end{align}

and the other half reaction taking place would be:
\begin{align}
\ce{Fe^{3+} + e^{-} &amp;-&gt; Fe^{2+}} &amp;E^\circ &amp;= +0.771~\mathrm{V}
\end{align}

Then, I would plug the concentration values, along with their standard potentials, into the rearranged nernst equation to get the $E$ values for each half, then subtract the larger value from the smaller. I already did this, assuming that the concentrations of $\ce{Cu^0}$ and $\ce{Cu^{2+}}$ were the same, giving a logarithm of $0$ and making $E = E^\circ$. The other equation remained the same. I got an answer of $0.399~\mathrm{V}$, which is wrong.

Also, I'm not sure how the concentration for $\ce{HCl}$ factors into this question, because the half reaction for $\ce{Fe}$ doesn't have any hydrogens in it.

Can anyone give me a hand with this one?
Think about what state $\ce{Cu^0}$ is in. $\ce{Cu^2+}$ is dissolved in the solution, but is this true of the unionized copper? Try to write out the half-reactions with the physical states included. What activity (effective concentration) do solids have in an equilibrium expression?

Also, though there are some reactions that could happen involving the acid, and I don't know whether they actually contribute to the cell potential in this case, the way the question is written, it looks like the two half-cells you have are the significant ones.