Membrane Potential Calculations
Bch 462, Arizona State University
Neal Woodbury
One of the most confusing aspects of membrane potential calculations is the fact that the
sign of the membrane potential is essentially arbitrary. So, I am going to use a
convention that is commonly used in reporting potentials in the literature. The
convention is this:
 Potentials will always be written as the voltage of the inside relative to the
outside. In other words when the membrane potential is negative, that means that
the potential inside is more negative than the potential outside.
Another confusing point is that the sign on the energy term dealing with voltage will
change if you change the direction of the transport.
 We will take care of this by always calculating the free energy change associated
with ions being transferred from the outside to the inside. Then, we will switch
the direction of the transfer (and the sign of the free energy change) if we need to
consider the free energy for transfer in the opposite direction.
First consider the concept of coupling chemical reactions. Lets assume that I am using an
ATP powered pump to drive the reaction:
H+in  H+out.
I am pumping H+ out in this case. The pump couples the proton transfer reaction to the
chemical hydrolysis of ATP:
ATP + H2O  ADP + Pi
We I say coupling, it means I am going to make one reaction happen if and only if the
other reaction happens. This means we are really talking about the overall reaction:
H+in +ATP + H2O  H+out + ADP + Pi
The free energy of this total reaction is, of course, the sum of the free energies of the two
individual reactions.
So lets try some calculations
Problem 1 If we convert the energy of ATP entirely into a membrane potential by
pumping protons, how large a membrane potential could we create? Assume that both
sides of the membrane are held at pH 7.0 by other cellular processes.
Well, if we converted all the energy available in ATP to a membrane potential, that
means that the ATP powered pump would pump until the driving force of the ATP
hydrolysis was equal to the energy of the membrane potential pushing back on the pump.
Hence, the free energy change associated with the membrane potential should exactly
balance the free energy change associated wit the hydrolysis of ATP. In other words the
free energy changes for the two reactions should be equal and opposite in sign and the
total free energy for the reaction will be zero.
The standard free energy change for ATP hydrolysis is -30.5 kJ/mole. In a typical cell,
the concentration of ATP is about 10 mM, that of ADP is about 10 mM and that of Pi is
about 1 mM (on an exam you would be given these numbers). Therefore the free energy
change is given by:
G = G0 + RTln([ADP][Pi]/[ATP])
Remember that while water appears in the chemical reaction for ATP hydrolysis, the
standard state of water is the concentration of the pure water and thus its relative
concentration (relative to the standard state) is 1.0 since the concentration of water in the
solution is essentially the same as in the standard state (the fact that there is a little bit of
ATP and ADP and H+ in the water does not significantly change its concentration).
Therefore water does not appear in thermodynamic equations for biochemical reactions
as either a reactant or a product.
RT at room temperature is 2.45 kJ/mole. Hence the free energy is
G = kJ/mole + (2.45 kJ/mole)ln(0.01*0.001/0.01) = - 47.4 kJ/mole
Notice two things. First, I did not put any units on the concentrations of ADP, Pi and
ATP. Why? Because these are relative concentrations (relative to the standard state of
1M) They are really ratios between the actual concentration and the standard state
concentration. This leads to the second point. The values of these numbers are always
the values one would have if they were in the units of Molar. You canNOT write
RTln(10*1/10)
As though the values were in mM. This would give a different (and wrong) answer.
Next we need to calculate the free energy change for transfer of the proton across the
membrane. Remember that we said that we would always calculate the free energy for
the reaction in the direction from out to in and then reverse it (and change the sign of the
free energy difference) if necessary. So, for the reaction:
H+out  H+in
We can write:
G = zF + RTln([H+in]/[H+out])
Because the system is buffered on both sides and remains pH 7.0, the ratio of the
concentrations inside and outside is 1.0 (in fact, since pH 7.0 is the standard biochemical
state, the relative values of both H+in and H+out are 1.0). Thus the second term in the free
energy change equation is zero. Giving us:
G = zF

We are transferring H+ and thus the charge (z) is +1. F is 96.5 kJ/(volt mole):
G = (96.5 kJ/volt mole)

We really want the free energy for the reaction:
H+in  H+out
Which is just the negative of the free energy for the reaction in the other direction, thus
G = -(96.5 kJ/volt mole)
Now to get the total free energy change for the coupled reaction, we add this to the free
energy change for the ATP hydrolysis reaction:
Gtot = -(96.5 kJ/volt mole)- 47.4 kJ/mole
Recall that the problem asks what is the maximum potential that can be formed by using
ATP to pump protons across the membrane. Recall also that this happens when the free
energy change for the coupled reaction is zero.
 = -(96.5 kJ/volt mole)- 47.4 kJ/mole
Solving for  gives:
 = -0.491 V
The minus comes about because we are measuring the voltage on the inside relative to
the outside (remember the convention we picked). Thus, by pumping protons from in to
out we have made the inside more negative than the outside.
Problem 2 What would happen if instead the outside H+ concentration was buffered at
pH 6.0 and the inside pH was at pH 8.0? Now what would be the maximum potential
that could be generated?
Well again for the reaction
H+out  H+in
We can write:
G = zF + RTln([H+in]/[H+out])
But now, [H+out] = 10-6M/10-7M = 10 (remember the standard state for H+ in biochemistry
is 10-7 M so [H+] is always relative to that). Likewise, [H+in] = 10-8M/10-7M = 0.1.
Plugging these numbers into the equation we have
G = (+1)(96.5 kJ/mole vole) + (2.45 kJ/mole)ln(0.1/10)
= (96.5 kJ/mole vole)11.3 kJ/mole
Note that the second term is negative (a spontaneous reaction) because of the way we
have written the reaction down, the protons are going in the direction of the gradient
(from high to low concentration). Again, what we really wanted was the opposite
reaction:
H+in  H+out
Which has a free energy that is just the negative of that for the reaction in the other
direction, thus
G = -(96.5 kJ/mole vole)11.3 kJ/mole
Now adding this to the free energy of the ATP hydrolysis reaction gives:
Gtot = -(96.5 kJ/mole vole)11.3 kJ/mole – 47.4 kJ/mole
Again, the highest membrane potential will be obtained when the pumps works until the
free energy change of the two reactions is equal and opposite and the total free energy
change is zero:
 = -(96.5 kJ/mole vole)– 36.1 kJ/mole
 = -0.374 V
This makes sense. The potential is still in the same direction because we are pumping
protons in the same direction. It is smaller because now we are pumping the protons
against a gradient so not only will the membrane potential oppose the action of the pump,
the concentration gradient will oppose it as well.
Your turn. Redo problem 2 for a chloride ion pump that uses ATP to pump chloride out
of the cell. Assuming that the chlorine ion concentration inside is held constant at 10
mM and that outside it is 100 mM. Note that it is quite reasonable that the chloride ion
concentration would not change. The amount of ion that has to be pumped across the
membrane in order to generate the membrane potential is very low compared to 10 mM.
I get  = +0.433 volts.
Problem 3: Now turn the problem around. In theory, how many moles of ATP could be
produced by transferring one mole of H+ from the outside to the inside of the cell under
conditions were there was a pH of 6.0 outside and 8.0 inside and there was a membrane
potential of –0.200 V?
From above, we know that each mole of ATP would need about 47.4 kJ in order to be
generated under normal cell conditions. So the question simply becomes, how much
energy is available by allowing the transfer of a mole of H+ under the stated conditions?
G = zF + RTln([H+in]/[H+out])
= (+1)(96.5 kJ/mole volt)(-0.200 V) + (2.45kJ/mole)ln(0.1/10) = -30.58 kJ/mole
The fact that it is negative means simply that this is energy available to couple to ATP
production. We can see however that the amount available from the proton transfer is
less than that required to make one mole of ATP. There is only enough energy to make
part of a mole (about 0.64 moles). In fact as we shall see in the chapter on oxidative
phosphorylation that more than one mole of H+ will be transferred for every mole of ATP
made.
Now lets attack the Sodium Potassium ATPase problem.
Problem 4. How large a membrane potential would you expect to generate using the
Na+/K+ atpase assuming that the concentrations of Na+ and K+ are 10 mM Na+ inside, 100
mM K+ inside and the reverse outside?
So now we have three reactions:
ATP  ADP + Pi
3Na+in  3Na+out
2K+out  2K+in
The total couple reaction is
ATP + 3Na+in + 2K+out  ADP + Pi + 3Na+out + 2K+in
Again we want the total free energy change here to be zero because the pump action is
just being balanced by the energy of the opposing membrane potential and the gradient.
So we need to add all the free energies up, set them to zero and then solve for . For
the first reaction, we already know that the free energy change is:
G1 = –47.4 kJ/mole
Let’s write the second reaction for one mole of Na+ and multiply the result by 3 in the
end. In addition, we need to turn it around (in the direction out to in so that it matches
our convention for membrane potential signs) for now in order to do the calculation and
then we will reverse it later:
Na+out  Na+in
We see that
G = zF + RTln([Na+in]/[Na+out])
= (+1)(96.5 kJ/mole volt) + (2.45 kJ/mole)ln(0.01/0.10)
= (96.5 kJ/mole volt) - 5.64 kJ/mole
Reversing this to the correct direction (in to out) and multiplying by 3 so that we have the
energy for 3 equivalents of Na+ as the equation requires we have:
G2 = -(289.5 kJ/mole volt) + 16.92 kJ/mole
For potassium, we do not have to turn the equation around, but we will first calculate the
free energy change for the reaction:
K+out  K+in
This is:
G = zF + RTln([K+in]/[K+out])
= (+1)(96.5 kJ/mole volt) + (2.45 kJ/mole)ln(0.1/0.010)
= (96.5 kJ/mole volt) + 5.64 kJ/mole
To get the actual free energy change for 2 K+ we just multiply by 2. This gives:
G3 = (193 kJ/mole volt) + 11.3 kJ/mole
Now we just add up the three free energy terms for the coupled reaction:
Gtot = -47.4 kJ/mole - (289.5 kJ/mole volt) + 16.9 kJ/mole + (193 kJ/mole volt) +
11.3 kJ/mole
= -19.2 kJ/mole – (96.5 kJ/mole volt)

Solving for  gives
 = - 0.199 V
Which is not far off from the membrane potential in your cells. Note that the potential is
negative. This means that the inside of the cell is negative compared to the outside as
expected if we are pumping 3 positive charges out for every 2 positive charges that come
in.
Your turn. Valinomycin is an antibiotic that allows transfer of K+ across the cell
membrane. Redo the last problem after addition of valinomycin assuming that this
caused the potassium concentration to become 10 mM on both sides of the membrane but
left the Na+ concentration the same. I get –0.316 V. It is higher because the pump does
not have to fight against the potassium gradient.