# [Physiology] Intuition behind the Nerst equation

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.

## Answers and Replies

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

Gold Member
@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.

@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