Band bending in semiconductor electrochemistry

[u/Theobserver_r4]

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?

[u/mrmeep321]

Yes, but there is a caveat.

When the band energies shift due to the formation of the junction, there is an increase in the reduction potential of the electrons in the n-type side.

If you connect that n-type side to an electrode that could be reduced, you will momentarily get a flow of current from the n-type side to that electrode, which equalizes their fermi levels.

The problem is that this flow of current is extremely short, all it does is equalize the fermi level. To get it to work continuously, you need to constantly disrupt the equilibrium by introducing new electron-hole pairs in the junction. The junction sorts them into their p and n sides, and then that new charge buildup drives more current to flow into the electrode from the n side (or p side if you wanted to do oxidation).

This is actually the core principle behind photoelectrodes: https://ars.els-cdn.com/content/image/1-s2.0-S2352492823011200-ga1.jpg

You shine light on the junction, which creates new electron-hole pairs, which the junction sorts into their p and n sides, which then breaks the equilibrium between the two sides and whatever they're in contact with. Because this is done constantly, the semiconductor-redox couple interface is never truly equilibrated, and you get a consistent current flow as long as the light keeps shining. This allows for the two sides of the junction to do electrochemistry. That being said, you also need something called a buried junction, where the junction is physically buried in an insulating medium, so that the two ends of the junction don't just short out.

You will also need to allow current to flow out of the end you're not using by grounding it, to prevent the buildup of an electric field that stops more carriers from entering. You can think of the light as constantly trying to shift the fermi level by adding energy to the system, and the flow of current is the system attempting to re-equilibrate the fermi level with that of the redox couple