<p>The <a href="https://en.wikipedia.org/wiki/Nernst_equation" rel="nofollow noreferrer">Nernst equation</a> gives the reversal potential/equilibrium potential, which is the voltage for which net flow <strong>for that specific ion</strong> equals zero. Net flow of zero for some ion means that the number of those ions moving out of a cell is equal to the number of those ions moving in.</p>
<p>The further the voltage is from the Nernst potential, the larger the 'effective' voltage is for that ion; that's the electrochemical driving force.</p>
<p>There is likely to be net current for other ions at the Nernst potential for a single ion; the purpose of the Goldman equation is to determine the voltage at which <em>overall</em> net current is zero. You can think of it like a weighted average of all the Nernst potentials, where each type of ion contributes more to the Goldman equation when there is more conductance across the membrane for that ion.</p>
<p>At any given moment of time, there is both the <em>actual</em> voltage, determined by the positions of charges and history, and the <em>equilibrium potential</em>, which you can calculate from the Goldman equation, the concentrations of all permeable ions, and the permeability of each ion. The actual voltage will get closer and closer to the equilibrium potential unless conductances change; the reason for this change in voltage is a net movement of ions (current), so the rate of change of voltage is given by the current.</p>
<p>Your formulation:</p>
<blockquote>
<p>if there is a concentration gradient which generates the potential of 80mV, the Nernst equilibrium (reversal potential) would have a value of -80mV. Is this correct?</p>
</blockquote>
<p>does not really make sense, I'd try to leave this thinking behind. Instead, start with:</p>
<ol>
<li><p>Concentration gradient and charge alone determine the reversal/equilibrium potential for each individual ion. Use Nernst to calculate it. No other ions matter.</p>
</li>
<li><p>In an imagined world with just one ion, membrane voltage will change to the reversal potential for that ion (assuming any conductance greater than zero). The potential &quot;generated&quot; by that ion's concentration gradient is exactly the reversal potential; that's what equilibrium means. If the voltage is any greater than this potential, there will be a net flow of this ion to make the voltage less; if the voltage is any less than this potential, there will be a net flow of this ion to make the voltage more.</p>
</li>
<li><p>In a more realistic world of multiple ions, the individual contribution of any one ion is always to influence the membrane potential towards its reversal potential. The &quot;weight&quot; or relative effect of this contribution depends on conductance: ions that can move more freely across the membrane have a bigger effect on voltage.</p>
</li>
</ol>

CC BY-SA 4.0
2023-05-05T11:21:36.513Z
Hey Bryan, thank you, once again, for the answer and for giving me pointers in how to think about this problem. This is the best explanation I have come across. 

Are you able to clarify this following segment further for me please? 

"At any given moment of time, there is both the actual voltage, determined by the positions of charges and history,"

What are you referring to with the term *history*?

note: I also wish to apologise once again for the comment I made last week, and thank you for the help. I appreciate it.
TheMatureNeuro
CC BY-SA 4.0
2023-05-05T13:30:32.960Z
@TheMatureNeuro No worries. Let's say we have a system with two ions, A and B, with Nernst potential for A=-50mV and Nernst for B=+50mV. If I tell you the membrane is right now permeable to A and impermeable to B, you know (or can calculate with Goldman) that the equilibrium potential is going to be about the reversal for A. The membrane potential itself doesn't change instantaneously, though. If I tell you the membrane has been permeable to A and not B for a long time, though, you can then assume the actual membrane has had time to change to that equilibrium potential.
Bryan Krause
CC BY-SA 4.0
2023-05-05T13:32:12.640Z
If suddenly you open a new conductance to B, the equilibrium potential instantly will shift to be less negative/more positive, but the actual membrane potential takes awhile to change. That rate of change is given by the overall permeability and the *capacitance* of the membrane.
Bryan Krause
CC BY-SA 4.0
2023-05-04T16:01:53.137Z
2023-05-04T16:01:53.137Z
Bryan Krause
<p>I am learning about the Nernst equilibrium in a neuron (along with the Goldman equation), and while I understand certain concepts individually, I struggle to bring it all together. The Nernst equilibrium is defined in terms of the reversal potential. That is, the membrane potential value needed to bring the net to 0. So, if there is a concentration gradient which generates the potential of 80mV, the Nernst equilibrium (<em>reversal potential</em>) would have a value of -80mV. Is this correct?</p>
<p>Now, the deviation from this equilibrium results in the <em>electrochemical driving force</em>. that is the difference between the membrane potential (which is the actual electrical potential) and the reversal potential. <em>how is the membrane potential interrelated to the concentration gradient?</em> Is the concentration gradient facilitated by the permeability of ions creating the membrane potential as a result?</p>
<p>Lastly, is the conductance of ions simply the degree to which the permeability to particular ion is allowed, and is modulated by the state of the membrane potential - Hyperpolarisation causes one particular ion to enter, depolarisation another.</p>
<p>I know there are several points and questions made here. I hope that it is relatively structured. As I have said, I myself struggle to put it all together in a coherent manner and make it more intuitive.</p>

<neuroscience><theoretical-neuroscience><neurophysiology>
Does the Nernst equilibrium in a neuron for a specific ion represent the potential voltage that would be needed to bring the net to 0?
TheMatureNeuro
# Does the Nernst equilibrium in a neuron for a specific ion represent the potential voltage that would be needed to bring the net to 0?

(Asked by: username_0 on 2023-05-04)

I am learning about the Nernst equilibrium in a neuron (along with the Goldman equation), and while I understand certain concepts individually, I struggle to bring it all together. The Nernst equilibrium is defined in terms of the reversal potential. That is, the membrane potential value needed to bring the net to 0. So, if there is a concentration gradient which generates the potential of 80mV, the Nernst equilibrium (*reversal potential*) would have a value of -80mV. Is this correct?

Now, the deviation from this equilibrium results in the *electrochemical driving force*. that is the difference between the membrane potential (which is the actual electrical potential) and the reversal potential. *how is the membrane potential interrelated to the concentration gradient?* Is the concentration gradient facilitated by the permeability of ions creating the membrane potential as a result?

Lastly, is the conductance of ions simply the degree to which the permeability to particular ion is allowed, and is modulated by the state of the membrane potential - Hyperpolarisation causes one particular ion to enter, depolarisation another.

I know there are several points and questions made here. I hope that it is relatively structured. As I have said, I myself struggle to put it all together in a coherent manner and make it more intuitive.

**Tags:** neuroscience, theoretical-neuroscience, neurophysiology


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## 1 answers

### Answer by username_1 on 2023-05-04. Score: 0

The [Nernst equation](https://en.wikipedia.org/wiki/Nernst_equation) gives the reversal potential/equilibrium potential, which is the voltage for which net flow **for that specific ion** equals zero. Net flow of zero for some ion means that the number of those ions moving out of a cell is equal to the number of those ions moving in.

The further the voltage is from the Nernst potential, the larger the 'effective' voltage is for that ion; that's the electrochemical driving force.

There is likely to be net current for other ions at the Nernst potential for a single ion; the purpose of the Goldman equation is to determine the voltage at which *overall* net current is zero. You can think of it like a weighted average of all the Nernst potentials, where each type of ion contributes more to the Goldman equation when there is more conductance across the membrane for that ion.

At any given moment of time, there is both the *actual* voltage, determined by the positions of charges and history, and the *equilibrium potential*, which you can calculate from the Goldman equation, the concentrations of all permeable ions, and the permeability of each ion. The actual voltage will get closer and closer to the equilibrium potential unless conductances change; the reason for this change in voltage is a net movement of ions (current), so the rate of change of voltage is given by the current.

Your formulation:

> if there is a concentration gradient which generates the potential of 80mV, the Nernst equilibrium (reversal potential) would have a value of -80mV. Is this correct?

does not really make sense, I'd try to leave this thinking behind. Instead, start with:

1. Concentration gradient and charge alone determine the reversal/equilibrium potential for each individual ion. Use Nernst to calculate it. No other ions matter.
2. In an imagined world with just one ion, membrane voltage will change to the reversal potential for that ion (assuming any conductance greater than zero). The potential "generated" by that ion's concentration gradient is exactly the reversal potential; that's what equilibrium means. If the voltage is any greater than this potential, there will be a net flow of this ion to make the voltage less; if the voltage is any less than this potential, there will be a net flow of this ion to make the voltage more.
3. In a more realistic world of multiple ions, the individual contribution of any one ion is always to influence the membrane potential towards its reversal potential. The "weight" or relative effect of this contribution depends on conductance: ions that can move more freely across the membrane have a bigger effect on voltage.

> **Comments:**
> **username_0**: Hey Bryan, thank you, once again, for the answer and for giving me pointers in how to think about this problem. This is the best explanation I have come across.   Are you able to clarify this following segment further for me please?   "At any given moment of time, there is both the actual voltage, determined by the positions of charges and history,"  What are you referring to with the term *history*?  note: I also wish to apologise once again for the comment I made last week, and thank you for the help. I appreciate it.  
> **username_1**: @username_0 No worries. Let's say we have a system with two ions, A and B, with Nernst potential for A=-50mV and Nernst for B=+50mV. If I tell you the membrane is right now permeable to A and impermeable to B, you know (or can calculate with Goldman) that the equilibrium potential is going to be about the reversal for A. The membrane potential itself doesn't change instantaneously, though. If I tell you the membrane has been permeable to A and not B for a long time, though, you can then assume the actual membrane has had time to change to that equilibrium potential.  
> **username_1**: If suddenly you open a new conductance to B, the equilibrium potential instantly will shift to be less negative/more positive, but the actual membrane potential takes awhile to change. That rate of change is given by the overall permeability and the *capacitance* of the membrane.  

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