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I was carrying out an experiment into the effect of temperature changes on the standard potential of a cell.

$$\ce{Zn(s) +2Fe^{3+}->Zn^{2+} +2Fe^{2+}}$$ I used equi-molar concentrations of $\ce{Fe(CN)_6^{4-}}$and $\ce{Fe(CN)_6^{3-}}$. Thus the Nernst equation

$$\mathrm{E^{\theta}_{cell}=E_{cell}-{\frac{RT}{nF}}*ln1}$$ reduces to $$\mathrm{E^{\theta}_{cell}=E_{cell}}$$

We plotted our $\mathrm{E^{\theta}_{cell}}$ values versus T and noticed that our $\mathrm{E^{\theta}_{cell}}$ values were actually increasing with temperature.

I would have expected that these values go down as the concentration of the reactant falls.

I asked my demonstrator and sourced information from different web resources, but can't access a definitive answer.

Any hints even?

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  • $\begingroup$ @JohnSow use E^\circ for $E^\circ$ $\endgroup$ – RE60K Oct 2 '14 at 10:45
  • $\begingroup$ My demonstrator told me that he expected the entropy change to be negative as well. I would have that with a larger atom $$Fe^{2+}$$, there would be less solvation and thus a positive change in entropy? $\endgroup$ – Edward Oct 4 '14 at 21:16
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A few thoughts:

For starters, $\ce{Zn^{2+}}$ is also part of the reaction quotient, so the log term in the Nernst equation will not necessarily be zero if the hexacyanoferrate concentrations are equal, depending on the $\ce{Zn^{2+}}$ concentration:

$$E = E°+\frac{RT}{zF}\ln\frac{[\ce{Zn^2+}][\ce{Fe^2+}]^2}{[\ce{Fe^3+}]^2}$$ Depending on whether the reaction quotient is greater than or less than 1, the potential can have either a positive or negative temperature dependance.

Second, the Nernst equation uses activities not concentrations so depending on the ionic strength and other factors, the activities of the ions in solution may not be what you think. This deviation is greater for multiply charged ions and may be significant for concentrations over 1 mM.

Finally, since the hexacyanoferrate species are both soluble, I'm assuming you're using some kind of reference electrode for the second electrode. This is also temperature dependant in the same way as the rest of the cell.

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  • $\begingroup$ use \ln for $\ln$ $\endgroup$ – RE60K Oct 2 '14 at 10:45

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