I'm trying to understand the physical details of the mechanism, by which galvanic cells work, instead of more abstract descriptions of the half reactions that take place and I find it hard to piece together concrete information on this. Below is a description of my basic understanding of the process and I'd be grateful if anyonce could comment on it, either confirming the validity of my assumptions, correcting them, or filling in the blanks, as necessary. I understand that my description is simplified, restricted to the most important aspects of the process and that other details, such as the interaction of H+ and OH- ions from the dissociation of water, with the electrodes and electrolyte have been ignored, but, as I've said, I'm trying to get a grasp of the basics. Hopefully, if what follows is not entirely off the track, it'll also help others with similar questions. So, here goes: Let us consider an electrode made of Zn, suspended in a solution of ZnSO4. The Zn atoms in the electrode tend to break away from the surface, leaving electrons behind and becoming suspended in the solution as ions. This happens presumably due to their own thermal kinetic energy, facilitated by molecules from the electrolyte attracting them, or bumping into them and knocking them free and depends therefore, on the ionization energy of the metal and the temperature. This process then leaves an excess of free electrons in the cathode, charging it negatively, which in turn attracts the suspended Zn ions back onto it. Since the electrolyte has ZnSO4 molecules dissolved in it, which dissosiate into Zn++ and SO4-- ions, they too interact, either capturing Zn++ ions which ejected from the surface, or providing Zn++ ions to be attracted to and deposited onto the surface. In any case, a dynamic equilibrium develops, with an excess of Zn++ ions in the electrolyte and an excess of e- in the cathode, which depends mostly on the temperature I suspect, but perhaps also on the electrolyte conecentration, and the process stops there. In the above, I'm unsure about what determines whether the Zn ions will be Zn+, or Zn++ ions, but it seems that the Zn in the electrode, will more readily ionize to Zn++, which, I suspect, has to do with its electron configuration, and/or the details of the metallic bond between Zn ions. I'd be greatly interested in any more detail on this. Now the same process develops with a Cu electrode immersed in a CuSO4 solution, but here, since Cu has a higher sum of first and second ionization energies, than Zn, the process will achieve equilibrium with fewer Cu++ ions suspended in the electrolyte and fewer excess e- in the electrode, hence a smaller negative charge (assuming the same temperature as before). If we then proceed to connect these two electrodes electrically, via a wire, electrons will flow from the more negatively charged Zn electrode to the less negatively charged Cu electrode until the charge is equalized, shifting the equilibrium at both electrodes. In the case of the Zn electrode, the excess of free e- will have been reduced, which will reduce the rate at which Zn++ ions from the solution deposit onto it and allow more Zn++ ions to escape. Conversly, at the Cu electrode, more excess e- will arrive, attracting more suspended Cu++ ions out of the solution. This will lead to a new equilibrium being established, with equal charge at both electrodes, somewhere inbetween their former charges, and the process will stop once more. It should be noted, that the shift is such at both sites, that the relative difference in excess metal ions suspended at each electrolytes has increased, further increasing their positive charge difference. If we now connect the two electrolytes in such a way, that ions can migrate between them, then the negative SO4-- ions in the less positively charged Cu half-cell are attracted to the more positvely charged Zn half-cell. This creates an excess of SO4-- ions in the Zn half-cell's electrolyte, binding more freely suspended Zn++ ions and allowing yet more to escape. Similarly, positve Zn++ ions will be attracted to the less positive Cu half-cell, where they will, presumably, either bind with the SO4-- ions floating around, thus forming ZnSO4, or be plated onto the negatively charged Cu electrode. In any case, the additional Zn++ ions escaping the Zn electrode and leaving more excess e- behind, coupled with the Zn++ ions, either depositing onto the Cu electrode, or binding SO4-- ions around it and causing Cu++ to deposit onto it, in any case, reducing excess e- there, will once again create charge imbalance and a flow of e- from the former to the latter. Thus the reaction will be sustained, until the Zn electrode has fully dissolved. I'm sorry in advance for the verbosity of the above, but, as I've said, it's the details of the physical mechanism, that drives the whole process, that I'm unclear about, so I wanted to make these details explicit. Any comments are welcome.