Equilibrium potential value of an ion and how to apply it to action potential ?

The equilibrium potential value of an ion or the Em value.

I will try making some sense out of this confusing term.

1. Ions across a cell membrane have two forces acting on them, the concentration gradient and the electrical gradient. Since, Na,K and Cl are the most abundant of ions, it is only needed to know the Em of these three ions.

2. The Em is the value of the membrane potential at which both the concentration gradient force and the electrical gradient force are equal and opposite. This means that these two forces will cancel each other and there will be no net flow of ions across the membrane.

3. Also, remember whenever given the chance ( permeability ) the ion will try to reach a stability, that is it will try to achieve a membrane potential equal to its Equilibrium potential.

There is an equation called the Nernst equation to measure the Em value. When you solve this equation the values of Em are

Na- (+61Mv)
K   ( -94.1Mv)
Cl  (-70Mv)

Lets apply these terms, to the action potential

·         At the firing level of action potential, both Na and K channels open, remember a golden rule..Na channels are fast to open and fast to close, K channels are lazy..they are slow to open and slow to close

·         The sodium channels open fast..the permeability of sodium increases to upto 5000 times..and since the Em of sodium is +61mv, it will rapidly enter the cell, so that it can make the membrane potential reach its Em value. It almost succeeds too, coming as close to +10mv..but by the this time..many sodium channels start to close, plus the slow opening potassium channels are fully active.

·         The Em value of potassium is -94mv, hence now it will rapidly move out of the cell..so that it can make the membrane potential equal to that of its Em value.

·         Why hyperpolarisation? The potassium channels are slow to close..so they keep effluxing potassium out..but it is gradually overridden by the Na-K pump..which limits the hyperpolarisation and restores the membrane potential to its initial value.