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 $1~\mathrm{M}$ $\ce{HCl}$.  
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>  What is the potential of cell?
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> 


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^- &-> Cu^0} &E^\circ &= +0.337~\mathrm{V}
\end{align}


and the other half reaction taking place would be:
\begin{align}
\ce{Fe^{3+} + e^{-} &-> Fe^{2+}} &E^\circ &= +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.


 2

Don't know why but you're just making some value errors, happens to everybody once in a while:
$$\ce{2Fe^{3+} + Cu -> 2Fe^{2+} +Cu^{2+}}$$
$$\mathcal{E}\approx(0.771-0.337)-\underbrace{\frac{0.059}{2}}\_{\large\frac{\mathcal {RT}}{\mathcal F}\ln{10}\approx0.059,\nu=2}\cdot\log\_{10}\underbrace{\overbrace{\frac{0.018\cdot0.14^2}{1\cdot0.037^2}}^{\ce{[Cu^2+][Fe^2+]^2}}}\_{\ce{[Cu][Fe^3+]^2}}$$
So:
$$\mathcal{E}\approx0.451\cdot$$


 1