[Physiology] Intuition behind the Nerst equation

[Hammad Shahid]

TL;DR

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.

 

[jim mcnamara]

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

 

[TeethWhitener]

@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 $z$.

 

[atyy]

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.

 

[Hammad Shahid]

@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.

 

[TeethWhitener]

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.

 

[Hammad Shahid]

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.