Chapter 4, End of Chapter, Review Questions, Exercise 1

The resting and the stimulating phase of the cell membrane during the firing of action potential is responsible for calculating the potential difference across the membrane. The membrane potential of the resting neuronal cell is -65 mV, which is very necessary for the functioning of the nervous system and conduction of nerve impulses.

In the resting membrane state, sodium is concentrated outside the cell. Upon stimulation, the voltage-gated Na⁺ ion channels open up, causing an influx of Na⁺ ions. This entry of the positive sodium ions causes the inner surface of the plasma membrane to attain a positive charge. The positive charge inturn repels the inward movement of Na⁺ ions, and pushes them back through their channels. At one given potential difference, meaning, the equilibrium potential that causes efflux of Na⁺ balances the force causing their influx. The E_Na value at 37℃ is +62mV. 

During the resting state, assuming that the membrane is impermeable to both K⁺ and Na⁺, it can be said that the V_m=0. If this membrane were to now become permeable to K⁺ ions, there would be an efflux of potassium ions according to the concentration gradient, until V_m is equal to the equilibrium potential of potassium ions (E_K). Therefore, the potential difference across the membrane changes from 0 to -80mV (E_k). 

There is a significant movement of sodium ions inside the cell (V_m- E_Na= -80-62= -142mV), because the membrane potential becomes exceedingly negative in comparison to E_Na.

But this movement does not occur until a stimulus causes the voltage-gated Na⁺ channels to open up. Upon stimulation, sodium moves from the outside of the cell to the inside of the cell in a direction that takes V_m towards the equilibrium potential of sodium during depolarization. This inward movement of sodium ion depolarizes the nerve cell until V_m= E_Na=62mV. The equilibrium potential of potassium is closer to the resting membrane potential.