I am having a real hard time understanding why the side with a high concentration of positive ions becomes the negative side of the membrane (in a neuron). This website http://www.st-andrews.ac.uk/~wjh/neurotut/mempot.html says its because "...the positive ions flow away from the high concentration side". But I still don't get it. Wouldn't it make more sense to then say that the high concentration site (more positive ions) has more positive voltage?

  • $\begingroup$ Hmm. look that presentation once again slowly. The membrane does not allow Cl- to pass, but it allows K+ to pass (perhaps due to their ionic radius) (en.wikipedia.org/wiki/Ionic_radius, see table). Now on the other chamber, K+ can diffuse-out, but Cl- not. So only K+ will come out; and at the moment a K+ come out to dilute chamber; the concentrated side would become unbalanced. Though there is + chrg enough; - charge is slight more. But that will create a back-pull on k+. The more diffusion proceed, the back pull go up and diffusion decrease. At end, diffusion stops at a max back pull $\endgroup$ Oct 3, 2016 at 18:46
  • $\begingroup$ Note that your salt soln on concentrated side, at start was neutral. There were too many positive charge, but as well the same number of negative charge. If you buy 1 packet of NaCl (edible salt) from market, it is much much more enriched in Na+ (than a soln. compatible to biological system). So it should be more positive, but it is neutral, because at the same time there are too many Cl-, too, and in equal number as Cl-, so no overall charge. $\endgroup$ Oct 3, 2016 at 18:53
  • $\begingroup$ If you take some + charge apart from some negative charges; the system should become more-negatively charged... whatever the total number-density of positive and negative charges at start. At end, you're getting More Cl- ions THAN the K+ ion in the same side. Count them. $\endgroup$ Oct 3, 2016 at 19:01
  • $\begingroup$ The eqlb at end (due to back-pull generated from that few extra Cl-, that try to take back the K+, against diffusion), is known in the name Gibbs Donnan Equilibrium. en.wikipedia.org/wiki/Gibbs%E2%80%93Donnan_effect $\endgroup$ Oct 3, 2016 at 19:07
  • $\begingroup$ Oh no on my 2nd comment's last line there's typing error... read "many Cl-, too, and in equal Number of Na+" (for table salt example). I had typed Cl- twice. $\endgroup$ Oct 3, 2016 at 19:15

2 Answers 2


Q. “Wouldn’t it make more sense to then say that the high concentration site (more positive ions) has more positive voltage?”

A. No.

I know little about electrophysiology, but looking at the graphics and explanation that precede the screen from which the sentence you quote is taken the statement is justified. Where are you going wrong?

Your argument implies that the side with the high concentration should be positive because it has more positive (K+) ions. But it has (initially) an equal number of negative ions (Cl) so it is initially electrically neutral. (The graphic indicates this by having no charge on the K — not a very satisfactory idea in my opinion.)

The important thing to note is the premise — that the membrane is permeable to K+ ions, but not to Cl ions. So, even though there is a concentration gradient, only the K+ ions can move across the membrane (the graphic shows those moving with a +ve charge). As they do so each leaves its ‘partner’ Cl ion behind, so a negative charge builds up at the high concentration side. At equilibrium there may be a high concentration of K+ ions at the ‘left’ of the diagram, but there is an even higher concentration of Cl ions!

Good luck with your further reading of these pages! It looks as if it gets much more complex as it goes on.


David's answer is good.

I would also add that the resting membrane potential is not strictly because of moving K+ ions - if you were "teleporting" ions out of the cell, you would need to move a lot more ions to get a potential of ~-70mV. Instead, the resting potential is essentially given by "how much electrical force would I need to put on this membrane to get those potassium ions to disobey their tendency to flow down their concentration gradient." You should always be thinking about the Goldman equation and the relationship between concentration gradients and permeabilities, or neurophysiology will be very confusing.


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