I'm really confused by how the terms Hyperpolarization and Depolarization are used in Cell biology and hope somebody can enlighten me hopefully. Here's what they mean for me so far:

Depolarization means the inner and outer side of the cell membrane become less polarized (so the Nernst potential tends to 0 due to log(1) being 0)

Hyperpolarization that the inner and outer side become more polarized (so the Nernst potential increases).

Assuming I'm not completely wrong until here. Let me give some examples, so maybe you can see better where the problem lies:

  • We have a model cell so (K+ internal is 140 mmol/l, K+ external is 4.5 mmol/l) and the external side is increased to 8 mmol/l , then a DEPOLARIZATION occurs because

roughly : $-60 \cdot \log(140/4.5) = -90 mV < -60 \cdot \log(140/8) = -74 mV$ ; or in other words, "energy" was used so a depolarization occurred (so far this seems correct with the solutions I can find to this example)

\ \

  • Now comes the second example (famous depolarization of a cell with the graph and the peak and repolarization):

enter image description here

  1. Na+ enters the cell : the concentrations im gonna use for this example are inside:$12 mmol/l$ and outside $140 mmol/l$

so before the Na enters we have: $-60 \log(12/140)= 60 mV$

if Na+ enters the Cell, a depolarization should occur (because the difference inside and outside is decreased)

this seems fine with the graph...

  • the part of the graph which goes over 0 (also called "overshoot"), is it not technically a hyperpolarisation+depolarization?

  • repolarization: this occurs by K+ ions leaving the cell, but that doesn't make sense according to the Nernst equation?

The concentration inside before the cells leaving is 140, outside 4.5 ; if K+ leave the cell, its simply going to depolarize more... So why does repolarization occur and not depolarization?

  1. It's commonly stated that Hyperpolarization is caused by K+ LEAVING the cell (for example with ion channels) , how does this make sense if the inner concentration is higher than the outer? If anything, this again should lead to depolarization

  2. Same thing with K+ entering the cell, isn't that supposed to cause a hyperpolarization as well?

I appreciate any kind of insight into this matter, since it seems that I'm completely wrong.

  • 2
    $\begingroup$ Why do you think that K+ leaving the cell is going to depolarise more? When Na+ enters the cell, it affects the total ionic concentration inside the cell, so that Na+ entering the cell (depolarisation) is “balanced” by K+ leaving the cell (repolarization). You also have to consider that a lot of channels are voltage-gated, so they regulate ionic transport based on a change in total membrane potential, regardless of which ion concentrations caused the change in potential. $\endgroup$
    – P...
    Commented Jun 8, 2019 at 11:25
  • $\begingroup$ considering the Na+ entering is probably smart BUT for example; if you have a cell why is it said that K+ leaving the cell will cause hyperpolarization? if it is not true due to the Nernst equation ? Same with K+ entering the cell ; regardless of whether Na enters or not ; if K+ leaves the cell potential from nernst will go towards 0 ; yet it is still claimed that this will cause a hyperpolarization? Is my understanding of depolarization and hyperpolarization incorrect? $\endgroup$
    – schokakola
    Commented Jun 8, 2019 at 11:39
  • $\begingroup$ since cell membranes are normally polarised at a resting potential of -70 mV, hyperpolarisation refers to any event that polarises the membrane even more, making the cell interior more negative beyond -70 mV (so the opposite of depolarisation). Hyperpolarisation is not specific to K+, and K+ leaving the cell doesn’t necessarily cause hyperpolarisation: it only does so when potassium channels stay open for a longer duration, allowing more K+ to leave the cell than needed to repolarise the cell back to -70mV. $\endgroup$
    – P...
    Commented Jun 8, 2019 at 12:19
  • $\begingroup$ The extra efflux of K+ and hyperpolarisation are important concepts in neurons, because nerve cells can only be excited (depolarised) from their resting potential. Going from a hyperpolarised state to the resting potential of -70mV allows a gap in time between excitation events. Because you wouldn’t want neurons firing constantly! :) $\endgroup$
    – P...
    Commented Jun 8, 2019 at 12:22
  • $\begingroup$ How does that relate to the simplified Nernst equation? I don't really see how your comment answers my question: $\frac{-60}{k} \log (C_{in}/C_{out})$. If you go from 140 inner and increase the concentration outside, then according to the equation it's going to go towards 0 ; this by definition is a depolarization ?????? $\endgroup$
    – schokakola
    Commented Jun 8, 2019 at 12:34

3 Answers 3


Here is how I think of the issue. First, keep in mind over the course of the action potential, ion concentrations on both the outside and inside of the neuron remain relatively unchanged. You can think of the Nernst potential as a charged battery, and they keep their concentrations relatively constant. Currents will flow, and the voltage will change, but this effects very few ions at a time, and does not effect the bulk concentration (See section 2.6 here). This is because any small change in concentration near the membrane (where voltage is measured) will quickly equalize with the surrounding bulk solution via diffusion.

Second keep in mind that the Nernst potential is an electro-chemical potential. Thus for potassium in particular, the chemical potential will overpower the electric potential driving potassium out of the cell, making the driving voltage of potassium negative.

So, as you state, the Nernst potential of sodium is $60$ $mV$ and for potassium is $-90$ $mV$. In your example there is a rest potential of $-70$ $mV$ this is given by the GHK potential mentioned in the comments. That means at rest, the balance between the sodium and potassium potentials is $-70$ $mV$. Here is the equation for reference. Note that this equation reduces to the Nernst potential when we consider only one ion.

$$ V = \frac{RT}{F} \ln(\frac{g_{Na}[Na_{out}]+g_{K}[K_{out}]}{g_{Na}[Na_{in}]+g_{K}[K_{in}]})$$

When the action potential starts, the depolarization is driven by the opening of sodium channels, that is $g_{Na}$ gets larger. Thus, when these channels open sodium rushes out and the cell voltage attempts to become more like the sodium Nernst potential, not zero. You can see that in the equation for the GHK voltage equation as the sodium conductance dominates the equation.

Near the peak of the action potential, the sodium channels close, and the potassium channels open. That is $g_{Na}$ gets small, and $g_{K}$ gets large. This means that now the potassium conductance dominates the equation, and the neuron's voltage becomes more like the $-90$ $mV$ Nernst potential. Then the repolarization occurs when the potassium channels close, and $g_{Na}$ and $g_K$ are back to there intial closed values, bring the rest potential back to $-70$ $mV$.

**note I am oversimplifying how the gates change in time, but I think this will help clear up your confusions.


My answer: it's not the concentration change that causes repolarization, it's the permeability change. The relevant equation here is the Goldman-Hodgkin-Katz equation, which entails that transmembrane voltage V_m is proportional to the ratio of [K]_o/[K]_i times permeability P_K:

GHK equation for sodium and potassium

[K]_i is always higher than [K]_o, so the ratio of [K]_o/[K]_i is always providing a negative contribution to V_m. During repolarization, the contribution from this ratio becomes less negative as its electrochemical gradient evens out, and if this was all that was happening then V_m would increase. However, there is a second relevant term that changes upon repolarization: potassium's permeability. P_K increases dramatically during repolarization, providing a negative contribution to V_m.

Note 1: Because potassium is positive and the cytoplasm is negative, the electrochemical gradient favors [K]_i > [K]_o. This explains why the transmembrane voltage decreases to -90 mV instead of 0 mV.

Note 2: The only reason why Vm is positive during an action potential in the first place is because there exists a high resting [Na]_o, and so a step change increase in sodium's permeability P_Na is what's driving the AP Phase 0 increase. As soon as P_Na drops, we would expect V_m to drop too regardless of what's happening with potassium.

Note 3: While I think your definitions make more sense (i.e. they're more literal), this quote (from Raz 2013: "depolarization is a positive change from the resting membrane potential") implies to me that "depolarization" has shifted from its historical definition to now just mean "an increase in transmembrane voltage".

  • $\begingroup$ Welcome to the site! References would be appreciated. Thanks! $\endgroup$
    – rotaredom
    Commented Jun 11, 2021 at 2:52
  • $\begingroup$ With respect to "note 3", there's really nothing biologically meaningful or special about the potential "0 mV", hence the use for "depolarization" meaning "any positive change from rest", sometimes including overshoot. $\endgroup$
    – Bryan Krause
    Commented Feb 28, 2022 at 21:23

Adding K+ on the outside of membrane is not the same as transfering K+ from inside to outside. When K+ is added to outside(hyperkalemia) the RMP becomes less negative(depolarisation) When K+ is transfered from in to out RMP becomes more negative(re/hyperpolarisation) We should not forget that most of the intracellular anions are nonpermeable protiens, so when K+ is transfered to outside they pull harder, and when K+ added outside, the pull decreases

  • 2
    $\begingroup$ Can you expand your explanation a bit, as well as expanding the acronyms? $\endgroup$
    – jakebeal
    Commented May 18, 2021 at 15:42

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