1
$\begingroup$

Double layer theories like that of Gouy-Chapman give the electrostatic potential $\ \phi\ $ in solution. For a symmetrical electrolyte and small potential drops $\ \phi\left(x=0\right)-\phi\left(x\rightarrow\infty\right)=\phi_0-0\ \text V=\phi_0$, the theory gives $$ \phi=\phi_0\exp\left(-\kappa x\right) $$ where $x$ is measured from the electrode/solution interface into the bulk solution. As $\kappa$ is very large $\left(\sim10^{-9}\frac{1}{\text m}\right)$, the electric field $$ E=-\frac{\text{d}\phi}{\text{d}x}=-\kappa\phi_0\exp\left(-\kappa x\right) $$ falls to a constant value (0 V in this example) very quickly.

In the equations for the transport of ions in diluted solutions, the migration flux of species $j$ is $$ J_{\text {j,migr}}=-z_ju_jFc_j\frac{\text{d}\phi}{\text{d}x} $$

As the flux is proportional to $E$ it must be zero everywhere in solution according to Gouy-Chapman. But in every book about electrochemistry migration is said to contribute to the flux $J_j$ of species $j$ except when $c_j$ is much more smaller than the concentrations of other species. But why should there be migration in a macroscopic cell at all? In double layer theories ions gather near the electrode surface and compensate the charge on the metal, so there should be no electric field in solution. Isn't this what the equation for $\phi$ above says?

Or are there two different types of "electrostatic potentials" in solution: One associated with the double layer theories and one associated with transport? $$ \phi_{\text{Gouy-Chapman}}\neq\phi_{\text{Migration}} $$

If this is the case: How are both potentials related to each other (and why do books always use the same symbol for both)?

$\endgroup$

Your Answer

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

Browse other questions tagged or ask your own question.