The electrode potential for a half cell is the half cell's electromotive force (EMF). The electric potential difference measured on a battery, e.g. a 1.5 volt AA cell, is the entire cell's EMF, and is the result of the difference in contribution from the two half cells, i.e. the difference in EMF for the two half cells.

When calculating the electrode potential for a half cell, you use the Nernst equation, which looks like this for half cells (be aware that there is also a version for the entire cells):

E = E° -	R · T	· ln(Q)
	z · F

E	= the electrode potential for the half cell
E°	= the standard electrode potential for the half cell, 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

When working with the Nernst equation, it is important to be aware that we are working with the the reaction quotient Q, not the equilibrium constant K, unless we are at the point where the cell's EMF is 0. This will be shown in details on the page 'Calculations on equilibria'.

In some cultures, e.g. Denmark, it has been tradition to use e instead of E for notation of the electrode potential, but the correct notation is E according to IUPAC, so that is the notation being used here.

For calculations in the educational system, a simplified version of the Nernst equation is often used, working from the assumption that the temperature is 25 °C. Therefore you can encounter this version of the Nernst equation:

E = E° -	0.0592 V	· log(Q)
	z

E	= the electrode potential for the half cell
E°	= the standard electrode potential for the half cell, i.e. E at 1 M and 25 °C
z	= the number of electrons to be transferred
Q	= the reaction quotient

This simplified version of the Nernst equation made sense in the old days, when you didn't have the tools for computers we have today. Today it makes more sense to use the Nernst equation in its correct form, e.g. using a spreadsheet or some other software for math.

When doing calculations on electrode potentials, you start by writing the cell diagram for the half cell. This could be something like a silver electrode with 0.5 M AgNO₃ at 25 °C. That would look like this:

Ag(s) │ Ag⁺(aq, 0.50 M)

The cell reaction that goes with it is:

Ag⁺(aq) + e⁻ Ag(s)

From the information at hand, we can now list that

E°	= 0.800 V (found by looking it up)
R	= 8.31451 J/(mol · K)
T	= 298 K
z	= 1
F	= 9.65 · 104 C/mol
Q	= [Ag+]-1 = 2 M-1

We can now determine the electrode potential E(Ag/Ag⁺)

E(Ag/Ag+) = 0.800 V -	8.31451 J/(mol · K) · 298 K	· ln(2 M-1) = 0.78 V
	1 · 9.65 · 104 C/mol