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I am stuck in an ambiguity about the equilibrium potentials of neuron cells. The following text is picked up from khanacademy website:

In one part it is said that:

We'll start out with K at a higher concentration inside the cell than in the surrounding fluid, just as for a regular neuron. (Other ions are also present, including anions that counterbalance the positive charge on K but they will not be able to cross the membrane in our example.)

enter image description here

If potassium channels in the membrane open, K will begin to move down its concentration gradient and out of the cell. Every time a K ion leaves the cell, the cell's interior loses a positive charge. Because of this, a slight excess of positive charge builds up on the outside of the cell membrane, and a slight excess of negative charge builds up on the inside. That is, the inside of the cell becomes negative relative to the outside, setting up a difference in electrical potential across the membrane.

enter image description here

As potassium continues to leave the cell, separating more charges, the membrane potential will continue to grow. Thus, the building membrane voltage is an increasing force that acts counter to the tendency for net movement of potassium ions down the potassium concentration gradient.

Eventually, the electrical potential difference across the cell membrane builds up to a high enough level that the electrical force driving K back into the cell is equal to the chemical force driving K out of the cell. When the potential difference across the cell membrane reaches this point, there is no net movement of K in either direction, and the system is considered to be in equilibrium. Every time one K leaves the cell, another K will enter it.

enter image description here

In another part it is said that:

The electrical potential difference across the cell membrane that exactly balances the concentration gradient for an ion is known as the equilibrium potential. Because the system is in equilibrium, the membrane potential will tend to stay at the equilibrium potential. For a cell where there is only one permeant ionic species (only one type of ion that can cross the membrane), the resting membrane potential will equal the equilibrium potential for that ion.

The steeper the concentration gradient is, the larger the electrical potential that balances it has to be. You can get an intuitive feeling for this by imagining the ion concentrations on either side of the membrane as hills of different sizes and thinking of the equilibrium potential as the force you'd need to exert to keep a boulder from rolling down the slopes between them.

enter image description here

I concluded the two following information from the above text:

  1. The closer we get to the equilibrium state, the membrane potential decreases because the concentration gradient decreases. (The steeper the concentration gradient is, the larger the electrical potential that balances it has to be)

  2. The closer we get to the equilibrium state, the membrane potential increases, because positive ions leave the cell.

How to solve this paradox?

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  • $\begingroup$ Can you clarify your question? I think one of the issues here is understanding the difference between the membrane potential and the equilibrium potential. If positive ions leave the cell, the membrane potential becomes less positive/more negative. That is the only way positive ions leaving the cell will impact the membrane potential. The closer we get to the equilibrium state, the smaller the difference between the membrane potential and the equilibrium potential will be. $\endgroup$
    – De Novo
    Commented Jul 20, 2018 at 21:18
  • $\begingroup$ The closer we get to the equilibrium state, the smaller concentration gradient will be? $\endgroup$
    – m.taheri
    Commented Jul 21, 2018 at 8:25

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I interpret your question to be roughly: "the equilibrium potential is based on a concentration difference, but as ions move to create the electric potential, the concentration difference is changing. Why doesn't this in turn change the equilibrium potential such that the concentrations become equal and the potential becomes zero?"

If I'm correct...

The answer to your paradox is that very few ions have to move.

The number of potassium ions that would have to move across the membrane to establish the resting membrane potential is tiny. Electrical forces are very very strong. The imbalance in ions doesn't affect the concentrations inside and out appreciably.

For a typical cell, 1 microcoulomb of charge (6 × 10^12 monovalent ions) per square centimeter of membrane, transferred from one side of the membrane to the other, changes the membrane potential by roughly 1 V. This means, for example, that in a spherical cell of diameter 10 μm, the number of K+ ions that have to flow out to alter the membrane potential by 100 mV is only about 1/100,000 of the total number of K+ ions in the cytosol.

from Alberts, B., Johnson, A., Lewis, J., Raff, M., Roberts, K., & Walter, P. (2002). Ion channels and the electrical properties of membranes.

However, because membranes are also a little permeant to sodium, and because action potentials and synaptic transmission allow some additional ions to flow across the membrane, if you gave infinite time, eventually the ion concentrations inside and out would equalize, and the membrane potential would eventually be zero. The sodium-potassium pump solves this problem by continuing to keep potassium high and sodium low inside the cell.

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  • $\begingroup$ Note that I also posted a very similar answer here: biology.stackexchange.com/a/57066/27148 but to a moderately different question that I don't think is really a dupe. I think that Q&A is good reading though for understanding how ions actually move in establishing resting potentials. $\endgroup$
    – Bryan Krause
    Commented Jul 20, 2018 at 19:40
  • $\begingroup$ I think this answers a common question, but not the actual question/confusion the OP has. Take a look at the way the apparent paradox was worded, at the very end of the OP. $\endgroup$
    – De Novo
    Commented Jul 20, 2018 at 21:14
  • $\begingroup$ @DanHall Indeed I may have misinterpreted what they were trying to ask since I'm used to the problem I answered in students. Care to expand on your reading of it? I should add that my assumption was that they meant membrane potential increasing/decreasing to mean absolute value. $\endgroup$
    – Bryan Krause
    Commented Jul 20, 2018 at 21:17
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    $\begingroup$ Take a look at my comment to the OP. It's hard to say for sure without some dialogue. I didn't phrase it as an answer. Since your the neurophysiologist here, I thought it best to leave it to you. $\endgroup$
    – De Novo
    Commented Jul 20, 2018 at 21:20
  • $\begingroup$ @DanHall Okay thanks. I've added an explanation of my reading of their question, so hopefully that will either help them realize I am answering their question or help them realize how to rephrase so I can answer their actual question :) $\endgroup$
    – Bryan Krause
    Commented Jul 20, 2018 at 21:22
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Your first deduction is wrong. The sentence actually means with a greater concentration difference at the initial condition, eventually there should be a higher electrical potential to satisfy opposite chemical potential. The dynamic of the system can be described by an exponential function. The closer to the equilibrium potential we get, the smaller the rate of approaching to the final state i.e. the rate of electric potential escalation will be. Note that in reality this does not really affect the concentrations since a tiny amount of ion exchange can generate a considerable amount of electric potential.

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  • $\begingroup$ You mean the last picture is showing the initial state? $\endgroup$
    – m.taheri
    Commented Jul 20, 2018 at 20:16
  • $\begingroup$ The last picture demonstrates the final states for two systems with distinct initial concentration difference; Smaller for the right one compare to the left system. $\endgroup$ Commented Jul 21, 2018 at 5:05

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