# [Physiology] Intuition behind the Nerst equation

In summary, the equation V(x) = 61*mV*(1/z)*(log[X(o)/X(i)]) represents the Nernst equation, which calculates the voltage (charge separation) across a membrane based on the valence (charge of ion) and concentrations of species inside and outside the cell. The intuition behind why the voltage decreases as the charge increases is that a higher charged ion species will move more easily across the membrane, leading to equilibrium being reached before too many ions are displaced. The voltage is proportional to the difference in charge between the two sides of the membrane, so a higher charge will result in a lower voltage.
TL;DR Summary
Why does the mV decrease as the valence increases?
The equation:

V(x) = 61*mV*(1/z)*(log[X(o)/X(i)])

Where:
z = valence (charge of ion)
[X(o)] = reference concentration (outside the cell)
[X(i)] = concentration of species inside the cell

I want to understand the intuition behind why the mV decrease as the charge increases. From what I understand, as a higher charged ion species moves across the membrane, the electrical force will be increased by a greater amount for each permeant. This will lead to equilibrium being reached before too many ions are displaced.

From there, I do not get how it will lead to a greater voltage (charge separation). While yes, less ions need to be moved in order to counter the chemical (concentration) force, at the same time they carry a higher magnitude charge, so doesn't that balance out?

Maybe I don't understand how voltage works exactly. I haven't taken physics so I do not know for sure how voltage and currents work.
Thanks.

Do you mean the Nernst Equation? Your description seems a bit confusing.

@jim mcnamara I’m pretty sure it’s a misinterpretation of the Nernst equation:
$$E_{cell}=E^{\ominus}-\frac{RT}{zF}\ln\frac{[\text{Red}]}{[\text{Ox}]}$$
At STP, and converting natural log to base ten log (edit: and assuming ##E^{\ominus}=0## and flipping the fraction in the logarithm), we get
$$E_{cell}=\frac{61\text{mV}}{z}\log_{10}\frac{[\text{Ox}]}{[\text{Red}]}$$
@Hammad Shahid In this case, “mV” is a unit (millivolts), not a variable. It’s not going to change with ##z##.

Last edited:
jim mcnamara and etotheipi
Voltage is proportional to the difference in charge between the two sides of the membrane.

Suppose we have a membrane that is permeable to X+ only. Initially, there is X+A- on one side of the membrane, and Y+B- on the other side. The X+ will move across the membrane from a region of high X+ concentration to region of low X+ concentration, leave A- behind, so one side of the membrane will become increasingly negatively charged relative to the other side. As more X+ moves across the membrane, the A- side becomes more negative, and will try to pull the X+ back. So the electric force will try to keep the X+ with the A-, while the concentration difference will try to drive the X+ to the other side of the membrane. The X+ will stop moving down its concentration gradient when the the opposing tendencies due to the electric force and concentration gradient match each other.

TeethWhitener said:
@jim mcnamara I’m pretty sure it’s a misinterpretation of the Nernst equation:
$$E_{cell}=E^{\ominus}-\frac{RT}{zF}\ln\frac{[\text{Red}]}{[\text{Ox}]}$$
At STP, and converting natural log to base ten log (edit: and assuming ##E^{\ominus}=0## and flipping the fraction in the logarithm), we get
$$E_{cell}=\frac{61\text{mV}}{z}\log_{10}\frac{[\text{Ox}]}{[\text{Red}]}$$
@Hammad Shahid In this case, “mV” is a unit (millivolts), not a variable. It’s not going to change with ##z##.
I think I wasn't clear in the question. I meant that having a higher charge will lower the answer in end. Since the answer is in mV, if let's sat having a +1 charge gives -50 mV as the answer, a +2 charge will give -25 mV.
I didn't understand why having a higher charge in a species would lead to a lower charge separation.

atyy
I think I wasn't clear in the question. I meant that having a higher charge will lower the answer in end. Since the answer is in mV, if let's sat having a +1 charge gives -50 mV as the answer, a +2 charge will give -25 mV.
I didn't understand why having a higher charge in a species would lead to a lower charge separation.
Because volts are joules per coulomb. In other words, the voltage In your equation represents the energy per unit charge to move an ion. If an ion has two units of charge, the voltage must be halved if the same amount of energy is expended.

atyy, jim mcnamara and Hammad Shahid
TeethWhitener said:
Because volts are joules per coulomb. In other words, the voltage In your equation represents the energy per unit charge to move an ion. If an ion has two units of charge, the voltage must be halved if the same amount of energy is expended.
Thanks. That makes sense. This is exactly what I was looking for.

atyy and jim mcnamara

## 1. What is the Nernst equation?

The Nernst equation is a mathematical equation that relates the equilibrium potential of an ion across a membrane to the concentration gradient of that ion. It is often used to calculate the resting membrane potential of a cell.

## 2. What is the significance of the Nernst equation?

The Nernst equation is significant because it helps us understand how ions move across a membrane and how this affects the electrical potential of a cell. It is also important in understanding the mechanisms of action potential generation and propagation.

## 3. How is the Nernst equation derived?

The Nernst equation is derived from the laws of thermodynamics and electrochemistry. It takes into account the concentration gradient and the charge of the ion to calculate the equilibrium potential across a membrane.

## 4. What factors influence the Nernst equation?

The Nernst equation is influenced by the temperature, the valence of the ion, and the ratio of the ion's concentrations on either side of the membrane. Changes in any of these factors can alter the equilibrium potential of the ion.

## 5. How is the Nernst equation used in research and medicine?

The Nernst equation is used in various research and medical applications, such as studying the electrical properties of cells, understanding the effects of drugs on ion channels, and diagnosing and treating disorders related to ion imbalances, such as electrolyte disturbances and cardiac arrhythmias.

• Biology and Medical
Replies
10
Views
2K
• Chemistry
Replies
1
Views
709
• Quantum Physics
Replies
1
Views
516
• Biology and Medical
Replies
2
Views
3K
• Atomic and Condensed Matter
Replies
1
Views
883
• Chemistry
Replies
8
Views
967
• Biology and Medical
Replies
1
Views
2K
• Biology and Medical
Replies
16
Views
6K
• Introductory Physics Homework Help
Replies
1
Views
764
• Biology and Medical
Replies
10
Views
3K