# Correlation between redox pseudocapacitance, surface area, and cell potential with metal oxide electrodes

I'm having difficulty intuitively reconciling two different ideas concerning metal oxide thin films in contact with aqueous solutions:

1) that charge storage by both redox pseudocapacitance and electrical double layer formation depend on the thin film's surface area (i.e. more surface area = more active sites and interstitials = more redox reactions at surface groups & more counterion adsorption at interstitials = more adsorbed charge),

2) but that interfacial potential due to the presence of the same protons / hydroxyls / H2O does not depend on surface area (i.e. hydrolysis of H2O and adsorption of, or reaction with, H+ and OH- just changes the electrochemical potential of the surface groups, leading to an interfacial potential, measured as a cell voltage between two electrodes, that doesn't depend on surface area).

For the sake of clarity, the specific relation I'm describing in 2) is the pH-dependent voltage of certain metal oxide films. There is a lot of literature for, for example, iridium oxide showing that pH sensitivity of these electrodes is independent of surface area (from 10 square microns up to several square millimeters).

I can clearly see that, for example, the Nernst equation for electrodes at equilibrium features no term for amount of charge present; in this case, cell potential differences are altered only by the number of electrons transferred in the dominant redox couple(s) and their relative activities. This makes sense from a potential energy standpoint. However, from an electromagnetism perspective, additional point charges (assuming for a second that orbital electrons not used in bonds or that free protons can be described only as point charges) generate a linear, additive electrostatic potential; of course, solution counterions and/or free charge carrier redistribution in the metal oxide balance to negate an electric field outside of this localized area, but this doesn't change the fact that the potential energy in this regions is higher.

Of course, voltage is an integral over a path (in this case, between the ohmic contacts of a working electrode and reference electrode), and the electric field should be symmetric about the solution interface, but if all charges are balanced at equilibrium then how do we measure a cell potential at all? Is there just some scaling law at play here, where the charge magnitude inherent to each point charge multiplies, when redox reactions take place, at a faster rate than the number of point charges (active sites) per surface area?

Perhaps someone with a good intuitive understanding of cell potentials and redox potentials could shed some light on where my thinking is incorrect here.

Side note: I have viewed this, this, this, and this but they didn't have the answer I was looking for. I also thought about writing this question a little more generically than what I have here but it's long enough as it is.