0x62448568…9229sent to0x8011df58…516c·#24,270,249·view on Etherscan
Water adsorbs to surfaces it can hydrogen bond to, and in the adsorbed phase the mobility of the ions that form by auto-ionization is asymmetric, the hydroxide ion is stuck within the adsorbate while the hydrogen ion is free to move. Thus, the hydrogen ion spreads out by diffusion, in equilibrium with the drift current from the resulting electric field (backwards towards the then negatively charged adsorbate). If there is dioxygen (O2) in the surrounding water, hydroxide ions occasionally break down into electrons, dioxygen and water, and the electrons move to combine with the hydrogen ions and dioxygen to form water. This consumes one water and produces one water - thus there is no change in quantity within the container. But this same process can happen over a membrane, into another container. If the surface that water is adsorbed to is permeable to protons and electrons, and there is dioxygen on the other side of the surface, protons will tend to diffuse over the membrane and the dioxygen and protons will occasionally accept electrons from hydroxide ions that break down in the adsorbate on the other side, thus one water is produced in the new compartment and one water is lost in the previous compartment. The same reaction that spontaneously happens within the adsorbate and bulk water in a compartment - when you add a separate compartment, will tend to move water. Naturally, if you have adsorbate on both sides of the membrane, the side with more adsorbate will tend to more frequently spontaneously produce water on the side with less adsorbate, compared to the side with less adsorbate. Adsorbate, like ice, is impaired by salt, so if you have salt water in one compartment and fresh water in the other, you will tend to produce water in the salt water compartment and consume it in the fresh water compartment. Likewise, the adsorbate is denser than liquid water thus an increase in pressure will increase the amount. The tendency for protons to move across the membrane (and thus for the likelihood that the water production happens in the new compartment) can also be increased by reducing the size of the first container.
The electro-chemical reactions in osmosis (as described above) require dioxygen for the positive pole in the circuit (where the electrons flow to). It is also symmetric, and equal amount of dioxygen is consumed on one side and produced on the other. Thus, it can happen back and forth. But what if you were to invert the architecture, such that the dioxygen is released in between the adsorbates rather than away from them? Such architecture requires that the membrane in between the compartments is containing the hydrogen ions (as they have to be in proximity to the dioxygen to combine with electrons), and that it is capable of providing space for dioxygen. You would need a base that holds the hydrogen ions, and some form of medium for the membrane itself which lets dioxygen pass through it while water cannot. With such a membrane, it can be assumed (for electron release to hydrogen ions on the other side) that the release of dioxygen and electrons from hydroxide would happen inwards towards the membrane rather than on the outside, as the reaction would be in closer proximity to the hydrogen ions on the other side. With this inverted architecture, the dioxygen released at one side can be simultaneously consumed on the other (i.e., it can diffuse freely across the membrane). This makes the dioxygen supply at the positive pole infinite or not a factor, as long as the quantity is high enough to start the process. To be able to move the hydrogen ions on a base within the membrane itself, you need to substitute it on the outside with a positively charged particle in equal quantity charge-wise.
To harness the electricity within this inverted architecture, you would need the membrane to be selectively permeable to electrons only where you place the machines to which you want to provide electricity. Thus, the architecture needs a membrane that is an insulator within which you can place your machines so that they run from one side and to the other, so that the electrons are forced to pass through them. You then have a membrane that is an insulator, has a base in it that stores hydrogen ions (on either side) and has substituted with another cation on the outside of the adsorbate on either side, and is permeable to dioxygen and also a store for dioxygen, and throughout this membrane you have placed machines that pass through the entire width of the membrane and that serve as conductive paths for the electrons.
In osmosis, the transfer of positive charge across the membrane reorients the electrical field to favour a movement of electrons across the membrane. In a similar way, your inverted architecture could have a higher concentration of a cation on one side of the membrane (like hydrogen ions in osmosis) and this cation would be prone to move across the membrane and trigger the discharge of electricity. If your membrane is only selectively permeable to this cation, you can choose exactly when to release the electricity - you simply open the cation channel. To achieve the concentration gradient, you would need the cation to be at a lower concentration on the other side of the membrane (just like the hydrogen ions in the case of osmosis are in lower quantity on the side water moves to).
When you discharge your battery on the side water moves from, and power your machines within the membrane, you will end up producing water in the compartment water moves to. To move this water back to the original compartment, you could simply reverse the process. But to do so, you would need a cation concentration gradient (just like in the discharge case, or in the case of plain osmosis). This cation would have to be at a lower concentration on the side water moved from, i.e., it would have to be another cation than the one that caused the discharge. And likewise, you would then need a store of hydrogen ions on the outside of the membrane as well, thus a base there as well.
Then, once you have discharged your cell, and then moved the water back out again, you would need to also move the cations back to the compartment they came from, so that you regenerate the concentration gradients.