 # Calculations on equilibria

Electrochemistry

## The Nernst equation at equilibrium

A cell producing a current, e.g. a 1.5 volt AA cell, is due to the reaction not being at equilibrium. This, on the other hand, also means that when there is no measurable current, an equilibrium has occurred, i.e. Q = K, and this can be used to calculate equilibrium constant for the reaction you are working with.

For the cells, the Nernst equation usually looks like this:

 E = E° - R · T · ln(Q) z · F

 E = EMF for the cell E° = EMF for the cell at standard conditions, i.e. E at 1 M and 25 °C R = the gas constant T = temperature in degrees Kelvin z = the number of electrons to be transferred F = Faraday's constant Q = the reaction quotient

which we can now adjust to:

 0 = E° - R · T · ln(K) z · F

 E = EMF for the cell, which is 0 at equilibrium E° = EMF for the cell at standard conditions, i.e. E at 1 M and 25 °C R = the gas constant T = temperature in degrees Kelvin z = the number of electrons to be transferred F = Faraday's constant K = the equilibrium constant

We are interested in finding K, so this is what we isolate in the equation:

 ln(K) = z · F · E° R · T

After this, it is really just a matter of drawing up the cell diagram and the reaction equation, so you know what you are working on.

## An example

To get a better sense of the formula, let us take a look at an example:

If we look at the cell

Zn(s) │ Zn2+ (aq) ║ Ag+ (aq) │ Ag(s)

we normally write the reaction like this:

Zn(s) + 2 Ag+(aq) 2 Ag(s) + Zn2+(aq)

But, when working with reactions at equilibrium, you have to remember that the reaction equation looks like this:

Zn(s) + 2 Ag+(aq) 2 Ag(s) + Zn2+(aq)

If we want to determine the equilibrium constant at 25 °C (298 K), we use the formula

 ln(K) = z · F · (E°(Ag/Ag+) - E°(Zn/Zn2+)) R · T

Looking at lists, we find that

E°(Ag/Ag+) = 0.800 V
E°(Zn/Zn2+) = -0.762 V
F = 9.65 · 104 C·mol-1
R = 8.31451 J·mol-1·K-1

and 2 electrons are transferred in reaction (i.e. z = 2), so

 ln(K) = 2 · 9.65 · 104 C·mol-1 · (0.800 V - (-0.762 V)) 8.31451 J·mol-1·K-1 · 298 K
ln(K) = 1.22 · 102
K = 6.93 · 1052 M-1

When calculating K like this, you cannot tell what the unit for the constant is from the equation. This you have to deduct separately from the reaction equation.