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In a book about electrochemistry the situation of a cell is described of which one half cell has an "undefined thermodynamic potential" as no redox couple exists:

$$ Pt/Cu^{2+}\ (0.01\ M),\ Cd^{2+}\ (0.01\ M),\ H_2SO_4\ (1\ M) $$

Estimating the potential where reduction of Cu sets in by thermodynamic data when a potential is applied,

$$ Cu^{2+} + 2e \rightleftharpoons Cu,\ +\ 0.340\ V\ vs. NHE\\ Cd^{2+} + 2e \rightleftharpoons Cu,\ -\ 0.4025\ V\ vs. NHE\\ O_{2+} + 4H^+ + 4e \rightleftharpoons H_2O,\ +\ 1.229\ V\ vs.\ NHE, $$

they argue that there is a rest potential range $0.340\ V ... 1.229\ V\ vs. NHE$ where no current is to be expected. Beyond these values current flow is expected. I don't understand this, because I understand $0.340\ V\ vs. NHE$ to be the standard potential of copper reduction under standard conditions, but the concentration of 0.01 M is not "standard" (= 1 M), right?

On the other hand, I'm afraid I have a general lack of understanding a system where one species of a redox pair is missing, because argueing in terms of Nernst's equation for such systems always leads to infinitely large (or small) potentials. In the case of copper above Nernst is not applicable in my opinion (correct ?), but what does the general relation

$$ \lim_{[R] \rightarrow 0} (E_0 + \frac{RT}{nF} \ln\frac{[O]}{[R]})=\infty $$ for the reaction $O+ne\rightleftharpoons R$ mean? And how can one figure out the rest potential range for a half cell on thermodynamic basis (neglecting kinetics)? What does the situation for copper looks like where [R] does not appear in Nernst's equation? When the solution in the book is right, why?

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  • $\begingroup$ Nernst's equation is not applicable for systems where are more ions in the solution. If the concentrations of ions are different, than there will be potential shift. If no-one else provides the answers, I will try to do my best next week. $\endgroup$ – Jaroslav Kotowski Apr 9 '14 at 8:38
  • $\begingroup$ Thanks for your efforts in advance! But I must add that I would even not understand the situation if there were only ONE electroactive species (for example only $H^+$ coming from H2SO4, for which the missing partner is $H_2$; no $Cu^{2+}$, no $Cd^{2+}$) in solution. $\endgroup$ – Sebi Apr 9 '14 at 11:39
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Nernst's equation

The Nernst Equation can be derived from the Gibbs energy $\Delta G = nFE $. The free energy for a cell reaction:

$$\Delta G = \Delta G^\circ - RT \ln Q$$

can be written as:

$$E = E^\circ - \frac {RT}{nF} \ln Q$$

for reaction

$$\ce{O + $n$e -> R}$$

$$E = E^\circ - \frac{RT}{nF} \ln\frac{a_\mathrm R}{a_\mathrm O}$$

where $a_i$ is activity of species $i$.

Activity In a solution, the activity of $i$ is defined: $a_i=\gamma_i \frac {C_i}{C^0}$ where $C_i$ is the concentration of solute, $C^\circ$ is standard concentration ($1\ \mathrm M$), and $\gamma_i$ is unitless activity coefficient.

The equation can be simplified in:

$$E = E^{\circ'} - \frac{RT}{nF} \ln\frac{[\ce{R}]}{[\ce{O}]}$$

But for the reaction

$$\ce{Cu^2+ + 2e- -> Cu}$$

the Nernst's equation will be:

$$ E = E^{\circ'} - \frac{RT}{nF} \ln\frac{[\ce{R}]=1}{[\ce{O}]} $$

because activity of solid (deposited copper) is 1.

If you try to calculate the potential of $0.01\ \mathrm M$ solution you will obtain the difference between standard concentration and the potential at $0.01\ \mathrm M$.

$$\begin{align} E &= E^{\circ'} - \frac{0.02559\ \mathrm V}{2} \ln\frac{1}{0.63}\\[6pt] &= 0.34\ \mathrm V - 0.0059\ \mathrm V \end{align}$$

The computed potential ($0.0059\ \mathrm V$) is lower than the standard potential and they are probably neglecting it.

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