There is, however, only one pe for the solution if all the redox species are in equilibrium. Because the redox species are often out of equilibrium one cannot really speak of the pe of a solution. For example, the reduction of sulfate is often a non-equilibrium reaction. The figure below illustrates the problem with field E H determinations. Very often the field E H values are not consistent with the calculated E H based on various redox couples.
Many redox are notoriously slow to equilibrate.
Using thermodynamics one can show that this must represent a non-equilibrium solution because it is impossible to have measurable amounts of both present. The answer to this question lies in the mechanism of redox reactions. In redox reactions electrons must be transferred from a reductant to an oxidant. The ease with which an electron is transferred between reductant and oxidant depends on the energies and symmetries of the two orbitals involved. The transfer is impeded if the receiving orbital of the oxidant is higher in energy than the energy of the donating orbital of the reductant.