It is often necessary to calculate cell potentials away from the standard state. For this purpose, we can use the general expression for the free energy derived in Chapter 9:
where Q is the reaction quotient. Note that the above expression implies, as expected, that redox reactions are not in equilibrium. Substituting in the expressions for the free energy in terms of cell potentials gives
or
This is known as the Nernst equation. The Nernst equation is sometimes
written in terms of 
instead of the natural logarithm
using a temperature of 25 
. Combining all the constants that
arise when this is done gives the more familiar form:
Note, this form can only be applied at 25 !!! In fact, it is a bad idea
to commit this to memory. Instead, remember where it comes from, i.e., the
free energy, and then the above equation can be rederived when needed.
The Nernst equation can be applied to half-cell potentials as well as to complete
cell potentials. Let 
be the reaction quotient for a
half-cell reaction. Then, the Nernst equation reads
Example: Suppose the cell Zn|Zn 
||MnO 
|Mn is operated
at pH = 2.00 with [MnO ] = 0.12 M, [Mn ] = 0.001 M and
[Zn ] = 0.015 M. Calculate the cell voltage at 25 and at 80 .
Solution: From the notation for the cell, we can read off the unbalanced reaction:
which can be balanced by either method to give
with the half cell reactions:
Thus, 10.0 moles of electrons flow through the circuit for every
2 moles of MnO reduced and every 5 moles of Zn oxidized.
The expression for the reaction quotient Q is
The concentrations of all species are given in the problem, except for the 
concentration which, from the given pH, is 10 
.
From the data in Appendix E, the standard cell potential is given by
Therefore, using the Nernst equation, the cell potential at 25 is given by
and at 80 , it is
Thus, at the higher temperature, the cell voltage is reduced somewhat over
what it is at 25 .
The Nernst equation can also be used to measure equilibrium constants.
In an electrochemical cell, equilibrium occurs when all the reactants
are used up and products are formed. As this occurs, the cell potential
will drop to 0. That is 
, which also corresponds to
the condition 
, which has been our previous thermodynamic
condition for equilibrium. From the Nernst equation, if , then
since, at equilibrium Q=K.
Example: For the cell in the previous example, what is the
equilibrium constant at 25 ?
Solution: Using the above formula, and the fact that 
V,
the equilibrium constant becomes
which is almost the fourth power of google. Such an overwhelmingly huge equilibrium constant indicates that the reaction can be regarded as going almost to absolute completion.
In the same way, solubility products (another kind of equilibrium constant) can
also be measured, by constructing an appropriate galvanic cell. Suppose
we wish to measure the solubility product 
.
We could set up a half cell with a silver-chloride solution and a silver
electrode together with a standard hydrogen half cell:
H 
(1 atm)| (1 M)||Cl 
+ Ag 
|Ag
and measuring the cell potential. Since silver plates out at the silver
electrode, the concentration of Ag in the solution will not be known
but can be calculated. The solution can be prepared with a known
concentration of Cl ions, however, since these are spectator ions.
The following example illustrates this:
Suppose the above cell is set up and a silver chloride solution with
an initial concentration of 0.001 M is prepared and used at the cathode.
The cell potential is measured and determined to be 0.397 V. What is the
silver ion concentration and 
value at 25 .
Solution: Two things need to be noted. The silver-chloride solution is not in its standard state, and the reaction is not in equilibrium. Thus, the Nernst equation can be used:
In order to use it, we need the reaction quotient, which can be computed from the initial concentrations given in the problem and the standard cell potential.
The standard state cell potential is given by
where the standard state half-cell potential for the half reaction:
occurring at the cathode has been taken from Appendix E. The half-cell reaction occurring at the anode is
The reaction quotient can be determined from the overall reaction:
Thus,
Since 
atm and 
M, the reaction quotient becomes
in terms of the unknown 
concentration. Solving the Nernst equation for Q
gives
so that
Thus, the solubility product becomes:
