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The differential capacitance of ideal electrochemical double layers is constant, as the surface charge $\sigma$ linearily increases with the applied surface potential $\Psi$ relative to the point of zero charge

$$ C = \frac{\mathrm{d}\sigma}{\mathrm{d}\Psi} \rightarrow \sigma = C (\Psi - \Psi_\mathrm{ZPC}) \tag{1} $$

So, in a cyclic voltammetry experiment in which there is only capacitive current and no faradaic current due to charge transfer, the current will be constant

$$ i = C\frac{\mathrm{d}\Psi}{\mathrm{d}t } \tag{2} $$

This is because in a CV experiment the scan rate $\mathrm{d}\Psi/\mathrm{d} t$ stays constant (e.g. $\pu{10 mV/s}$). Experimentally it looks like this (not ideal, but close enough):

enter image description here

But, how is this consistent with the Helmholtz idea of the electrochemical double layer? Physically, there needs to be a linear increase in surface charge for higher potentials on the electrode. But, the surface-active Helmholtz plane is limited to the inner Helmholtz layer, with the rough diameter of one solvent molecule and one charged ion:

enter image description here

In my imagination, at low potentials, the whole electrode is already covered in positive excess charge, and the anions of the electrolyte already adhere to the surface.

As the inner Helmholtz plane is already filled up by anions, and the potential further away has already dropped to zero, I cannot understand how the surface charge can still increase to higher potentials as is needed for Eq. (1) to be valid.

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    $\begingroup$ I have to flip this around. At low voltages, why can't the surface charge density in the conductor change with voltage? Why would incomplete surface coverage be incompatible with a Helmholtz model? $\endgroup$
    – Buck Thorn
    Commented Jan 20 at 8:37
  • $\begingroup$ Maybe you can find further information in in A. J. Bard, L. R. Faulkner, H. S. White, Electrochemical Methods: Fundamentals and Applications, 3rd Edition, Wiley (2022) $\endgroup$
    – PAEP
    Commented Jan 21 at 18:29

1 Answer 1

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To start forget for a second about double layer models and particularly details of the electrolyte or other medium adjacent the electrode. Assume you have a perfect conductor which you hooked up to a voltage source and has a flat surface in contact with a vacuum. Classical electromagnetic theory (specifically Gauss' law) tells you that the electric field inside the perfect conductor is zero (there is no current) but that there is a field outside due to charge at the surface. If the surface is large and flat the surface charge generates a uniform field perpendicular to the surface. The field is related to the surface charge as $$\sigma = \epsilon_0 E$$ It is this field of course that attracts countercharges from whatever medium you place next to the surface. If you substitute the vacuum with a dielectric* this just becomes $$\sigma = \epsilon E$$ In this case a layer of counter-charges forms adjacent the surface due to polarization (reorientation) of molecules or distortion of the charge distribution including concentration of counterions. This polarization extends all the way to the opposing pole due to the voltage drop in the medium. A similar analysis leads to similar equations for a parallel plate capacitor.

The takeaway is that an electrolyte can be considered a dielectric* that is polarized with a net countercharge next to the surface of a charged conductor. That countercharge can in effect be zero, as it is in the case of vacuum. The extent of polarization depends on the surface charge of the conductor and the dielectric constant. The surface charge of the conductor depends on the voltage $\Delta \phi$ applied. It also depends on the geometry and the nature of the dielectric; for a parallel plate capacitor $\Delta \phi=Ed$ so charge depends on the plate separation $d$:

$$\sigma = \epsilon \Delta \phi/d \tag{1}$$

For an ideal conductor you can alter the surface charge by tuning the applied voltage, meaning it can be arbitrarily small. For small surface charges the counter-charge density will also be small and a linear variation with the applied voltage is not inconceivable. This linear relationship is expressed in terms of an electric susceptibility $\chi_{el}$ (1), in the present case of one component of the full tensor: $$P = \chi_{el} \epsilon_0 E$$ where $P$ is the polarization of the dielectric, the density of induced dipoles per unit volume, which can be shown equal to the surface density of counter-charges adjacent the electrode: $P=\sigma_{pol}$. The counter-charges also generate a field that partially cancels the field due to the surface charges. Applying Gauss's law then leads to $$E = \frac{\Delta \phi}{d} = \frac{\sigma_{\textrm{free}}}{\epsilon_0(1+\chi)}$$

Turning back again to equation (1), Since the capacitance is $C=Q/\Delta \phi$ and $Q=A \sigma$, it follows that $$C=\frac{\epsilon A}{d}$$ where $\epsilon=\epsilon_0(1+\chi)$.

In the Helmholtz model the dielectric contains ions* which respond to the applied electric field by moving to the oppositely charged electrode. The field then drops to zero some distance from the electrode and, as described in the problem, a counterlayer with the same total charge Q forms next to the conductor, representing a virtual parallel plate a very short distance away. The basic assumptions still hold, however, in essence that a linear electric susceptibility apply. This assumption requires above all that the applied voltage be low.


*A dielectric does not have free charges and cannot carry a current beyond that sustained while it is being polarized. The analysis here applies to an electrostatic situation, for instance a capacitor that has achieved a fully charged state and for which the current is zero.


Reference

  1. Alberty and Silbey, Physical Chemistry, 3rd Edition, Wiley 2001.
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