I understand the electrical conductivity of pure water is very low, but not zero, and is due to the slight number of H+ and OH- ions naturally present. I understand that they will move under the influence of a potential difference and be the charge carriers for the resulting current. What I don't understand is how the electrons are lost at one electrode, and gained at the other when the potential difference between the electrodes is rather small. When the potential difference is greater than about 1.23 Volts, there will be electrolysis with H2 and O2 bubbling off the electrodes, and the chemical equations for this process describe it fine, but what about 0.5 volts? What is the chemistry in this case? It can't be the same.
On the surfaces of the electrodes equilibrium is virtually never achieved. Differently from bulk voltage (which approaches zero faster than the interfaces'), interfaces have inherently a non-null resultant electromagnetic field due to polarization of electronic densities on the outer atoms and perturbations from external particles collisions, no matter what the electrodes are made of. In short terms: interfaces of phases are anisotropic.
Local imperfections in the bulk, such as holes (in solids) or the formation of ions (in fluids) also produce a potential difference that allows current flow, and eventually some electrons could cross the path between the electrodes transported by many ions (which are subject to other chemical equilibria), but the force that binds them to the ions are mostly strong enough to prevent this motion in the bulk.
Local imperfections are also the reason the concentration of molecules in a system is described by distribution models such as Maxwell-Boltzmann distribution for potential, energy, velocity or local temperature.