7
$\begingroup$

The Nernst equation can be applied to the whole cell reaction or a half-cell reaction. Consider the half cell reaction: $$\ce{Cu^{2+} + 2e^- -> Cu(s) \quad \quad E^0 = +0.337~\rm{V}}$$

The Nernst equation expressed in base 10 log at 25 °C is:

$$E = E^0 - \frac{0.1596}{2}\log_{10} \frac{a_{\ce{Cu^2+}}}{a_{\ce{Cu}}}$$

But $a_{\ce{Cu}} = 1$ by definition, and for low concentrations of $\ce{Cu^2+}$ where
$a_{\ce{Cu^{2+}}} \ll 1$ then: $\ce{[Cu^2+]} \approx a_{\ce{Cu^2+}}$.

But when $\ce{[Cu^2+]} < 1$ then the log will be negative so:

$$\lim_{\ce{Cu^2+}\to 0} (0.337 - 0.0798 \times \log_{10} [\ce{Cu^2+}]) = \infty $$

This doesn't seem realistic. For instance $\ce{[Cu^2+] = 1.0\times 10^{-1,000,000}}$ is fine mathematically, but that sort of concentration doesn't make any sense chemically.

Thus what is the limit of the half cell potential as the $\ce{[Cu^2+] -> 0}$?


Edit-1 - My gut reaction to this is that you can't really measure the activity of a single ion. If the ion of interest is low in concentration compared to the ionic strength in general, then it would seem that the activity of that ion should go to zero as its effect on the total overall ionic strength goes to zero.

$\endgroup$
  • $\begingroup$ A low concentration say $\ce{10^{-10} }$ $\pu { mol dm^{-3}}$ still contains a staggeringly huge number of molecules The formula is ok as its not possible to have negative concentrations and only $6.02x 10^{23}$ molecules $ \pu {mol^{-1}}$ $\endgroup$ – porphyrin Oct 1 '16 at 8:15
4
$\begingroup$

The Nernst equation does not need a replacement for low concentrations. Yes the limit is $\infty$, which essentially means you can't realistically have zero concentration of anything. We may think of absolutely pure water as an idealized situation. Well, as soon as it gets in touch with copper, some $\ce{Cu^2+}$ ions instantly jump into the solution, and from that point on, you have a certain non-zero concentration of them (though probably pretty low), and the equation works just fine.

$\endgroup$
  • 2
    $\begingroup$ Note also that the formula actually works best low concentrations. You actually use activities instead of concentrations, and activity is approximately equal to concentration at low concentrations. Not so at higher concentrations. $\endgroup$ – Zhe Oct 4 '16 at 1:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.