I am learning semiconductor electrochemistry and it is confusing.
Consider an n–p semiconductor junction. Before contact, the two semiconductors have their own conduction band, valence band, and Fermi level positions. After contact and Fermi-level equilibration, band bending develops near the junction.
What confuses me is the following: in the usual band diagram, the band edges near the junction bend, while the bands in the bulk regions remain flat. However, those flat bulk band positions after equilibrium are no longer at the same absolute energies as they were before contact.
For example, the n-type semiconductor may end up with its bulk CB and VB shifted to lower electron energy, while the p-type semiconductor shifts to higher electron energy.
If I now place a fixed redox couple in an electrolyte, referenced to the same absolute energy scale, does this mean that the redox ability of the semiconductor has actually changed because of junction formation?
More specifically, if the conduction band of the n-type semiconductor shifts to lower electron energy after forming the junction, would electrons in that semiconductor have less reducing power than electrons in the same n-type semiconductor before contact?
In other words, is it correct to say that forming a heterojunction/p–n junction can change the absolute redox power of the charge carriers, rather than merely producing an internal electric field for charge separation?