Derivation of the Nernst Equation

9.09.J/7.29J-Cellular Neurobiology, Spring 2005 Massachusetts Institute of Technology Department of Brain and Cognitive Sciences Department of Biology Instructors: Professors William Quinn and Troy Littleton 9.09J Cellular Neurobiology HANDOUT Quickie derivation of the Nernst equation (Weird, but solid as a rock) First, let's consider conditions. We have two chambers separated by a membrane. The chambers have different concentrations of an ion (say, potassium, K'). The membrane is permeable only to K'ions. (It has special holes hat only K ions fit through.) 2 (1 potassium ion concentration in chamber 1 Under these conditions, there will be a voltage AV across the membrane. How does it arise? There is a net diffusion of K ions from high concentration (chamber 1) to low concentration (chamber 2). If no other force were present this would continue until the concentration in both chambers were equal (makes sense, huh?). However, this process never goes that far. When a few K ions diffuse from chamber 1 to chamber 2, they create an excess of (+) charge in chamber 2. This creates the voltage difference and an equilibrium will be reached when the tendency of the ions to diffuse down the concentration gradient is exactly counterbalanced by the tendency of the voltage gradient to push them back. Maybe a potential energy diagram will help: Lokal Energy Potential chctual frice Als Elbania) All (centric) Distance Chann Tanon apparet As equilibrium (rate of Lon flow from 1 to 2 rate of ion flow from 2 to 1), the potential energy per lon must be equal in both chambers (othervise tons would flow down the "energy hill" and there would be cet diffusion). You can see from the diagram that equilibrium occurs when AU (chemical) (electrical), t.a., when an ion going across the membrane from chamber 1 to chamber I Loses exactly as much energy going up the voltage "hill" as it gains going dove che concentration hili. So the big equation is AU (electric) AU cham (1). What we have to do now is calculate the AU's and plug them ia. Let's scart with the hardest, M (cham). It's very difficult to decide how much work it would take to concentrace a bunch of lons in solution: if you try to squash them together mechanically, the water gets in the vay. Let's try an easier problem (which will turn out to be equivalent) concentrating ions in a gas. Let us imagine a gas of K ions. Impossible, you grumble. No nacter, I wittily reply. A gas is a gas; they all behave the gas lav PV ORT, and impossible ones will obey it too. This gas is particularly strange, since all the particles are electric charged, but we will vorry about electric charges separately, in the AU (electric) part of the deviation. O.K., we have, soles of this gas in a piston, we're going to concen- trate it by squashing it from an initial volume 1 to a final volume 12. Fy 3 Repanded gan, Comprarsal gas Iz Son Force We want to find out the change in potential energy involved in the squashing. We know from early on char (2) W vork AU change in potential energy 7- force AS distance through which the force acts. This Like a step in the right direction, since we're applying a force through a distance. There are, however, rue complications. (a); Equation 2 is for a uniform force. Note that as ve squash che gas ics pressure will increase and it will be harder and harder to squash ic further, so F is not a constant, it's a function of distance. We 2 could get around this problem by chopping up as into a lot of Litwie Lengths, de, finding the force 7(s) at each of them, multiplying the force there by the little piace, is giving a small increment 4U, and adding up all the little du's to get the total change AU. Fig. 4 old rule du -۲.۵۵ Farce A Figs new rule: AU-du- -(3)-da-f(s) da We know that the sasy way to sum up such little pieces is to integrats, so the new rule (for non-constant forces) is AU-P(a) da (3) So far so good. The other complication is(b): We've talked in terms of force Fand distance. But the gas law is vritten in terms of pressures and volumes, L.s. I (4) P-Pressure - Volume number of nolas of gas a constant, the "gaa law constant" Temperature, in 'Zalvin How do we get from forces to pressures and so forth? By cleverly noting that pressure (in navtons/dn) Forca/Area and Volume (distmce) xarea (of piaton). If ve divide force by the area (of the piston) and multiply distance 3 by arsa (of the piston) che product remains the same! Similarly for our integral expression: 40 - da-/Ada (