- #1
Hammad Shahid
- 64
- 3
- 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.
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.