# 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^- &-> 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?

• For the reaction quotient Q, solids are omitted. That would mean that the copper electrode concentration would be taken out of the equation. Dec 28 '14 at 6:19

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?
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: $$\mathcal{E}\approx0.451\cdot$$