6.16 Chem­i­cal Po­ten­tial and Dif­fu­sion

The chem­i­cal po­ten­tial, or Fermi level, that ap­pears in the Fermi-Dirac dis­tri­b­u­tion is very im­por­tant for solids in con­tact. If two solids are put in elec­tri­cal con­tact, at first elec­trons will dif­fuse to the solid with the lower chem­i­cal po­ten­tial. It is an­other il­lus­tra­tion that dif­fer­ences in chem­i­cal po­ten­tial cause par­ti­cle dif­fu­sion.

Of course the dif­fu­sion can­not go on for­ever. The elec­trons that trans­fer to the solid with the lower chem­i­cal po­ten­tial will give it a neg­a­tive charge. They will also leave a net pos­i­tive charge be­hind on the solid with the higher chem­i­cal po­ten­tial. There­fore, even­tu­ally an elec­tro­sta­tic force builds up that ter­mi­nates the fur­ther trans­fer of elec­trons. With the ad­di­tional elec­tro­sta­tic con­tri­bu­tion, the chem­i­cal po­ten­tials of the two solids have then be­come equal. As it should. If elec­trons can trans­fer from one solid to the other, the two solids have be­come a sin­gle sys­tem. In ther­mal equi­lib­rium, a sin­gle sys­tem should have a sin­gle Fermi-Dirac dis­tri­b­u­tion with a sin­gle chem­i­cal po­ten­tial.

The trans­ferred net charges will col­lect at the sur­faces of the two solids, mostly where the two meet. Con­sider in par­tic­u­lar the con­tact sur­face of two met­als. The in­te­ri­ors of the met­als have to re­main com­pletely free of net charge, or there would be a vari­a­tion in elec­tric po­ten­tial and a cur­rent would flow to elim­i­nate it. The metal that ini­tially has the lower Fermi en­ergy re­ceives ad­di­tional elec­trons, but these stay within an ex­tremely thin layer at its sur­face. Sim­i­larly, the lo­ca­tions of miss­ing elec­trons in the other metal stay within a thin layer at its sur­face. Where the two met­als meet, a “dou­ble layer” ex­ists; it con­sists of a very thin layer of highly con­cen­trated neg­a­tive net charges next to a sim­i­lar layer of highly con­cen­trated pos­i­tive net charges. Across this dou­ble layer, the mean elec­tro­sta­tic po­ten­tial changes al­most dis­con­tin­u­ously from its value in the first metal to that in the sec­ond. The step in elec­tro­sta­tic po­ten­tial is called the “Gal­vani po­ten­tial.”

Gal­vani po­ten­tials are not di­rectly mea­sur­able; at­tach­ing volt­meter leads to the two solids adds two new con­tact sur­faces whose po­ten­tials will change the mea­sured po­ten­tial dif­fer­ence. More specif­i­cally, they will make the mea­sured po­ten­tial dif­fer­ence ex­actly zero. To see why, as­sume for sim­plic­ity that the two leads of the volt­meter are made of the same ma­te­r­ial, say cop­per. All chem­i­cal po­ten­tials will level up, in­clud­ing those in the two cop­per leads of the me­ter. But then there is no way for the ac­tual volt­meter to see any dif­fer­ence be­tween its two leads.

Of course, it would have to be so. If there re­ally was a net volt­age in ther­mal equi­lib­rium that could move a volt­meter nee­dle, it would vi­o­late the sec­ond law of ther­mo­dy­nam­ics. You can­not get work for noth­ing.

Note how­ever that if some con­tact sur­faces are at dif­fer­ent tem­per­a­tures than oth­ers, then a volt­age can in fact be mea­sured. But the phys­i­cal rea­son for that volt­age is not the Gal­vani po­ten­tials at the con­tact sur­faces. In­stead dif­fu­sive processes in the bulk of the ma­te­ri­als cause it. See sec­tion 6.28.2 for more de­tails. Here it must suf­fice to note that the us­able volt­age is pow­ered by tem­per­a­ture dif­fer­ences. That does not vi­o­late the sec­ond law; you are de­plet­ing tem­per­a­ture dif­fer­ences to get what­ever work you ex­tract from the volt­age.

Sim­i­larly, chem­i­cal re­ac­tions can pro­duce us­able elec­tric power. That is the prin­ci­ple of the bat­tery. It too does not vi­o­late the sec­ond law; you are us­ing up chem­i­cal fuel. The chem­i­cal re­ac­tions do phys­i­cally oc­cur at con­tact sur­faces.

Some­what re­lated to Gal­vani po­ten­tials, there is an elec­tric field in the gap be­tween two dif­fer­ent met­als that are in elec­tri­cal con­tact else­where. The cor­re­spond­ing change in elec­tric po­ten­tial across the gap is called the “con­tact po­ten­tial” or “Volta po­ten­tial.”

As usual, the name is poorly cho­sen: the po­ten­tial does not oc­cur at the con­tact lo­ca­tion of the met­als. In fact, you could have a con­tact po­ten­tial be­tween dif­fer­ent sur­faces of the same metal, if the two sur­face prop­er­ties are dif­fer­ent. “Sur­face po­ten­tial dif­fer­ence” or gap po­ten­tial would have been a much more rea­son­able term. Only physi­cists would de­scribe what re­ally is a gap po­ten­tial as a “con­tact po­ten­tial.”

The con­tact po­ten­tial is equal to the dif­fer­ence in the work func­tions of the sur­faces of the met­als. As dis­cussed in the pre­vi­ous sec­tion, the work func­tion is the en­ergy needed to take a Fermi-level elec­tron out of the solid, per unit charge. To see why the con­tact po­ten­tial equals the dif­fer­ence in work func­tions, imag­ine tak­ing a Fermi-level elec­tron out of the first metal, mov­ing it through the gap, and putting it into the sec­ond metal. Since the elec­tron is back at the same Fermi level that it started out at, the net work in this process should be zero. But if the work func­tion of the sec­ond metal is dif­fer­ent from the first, putting the elec­tron back in the sec­ond metal does not re­cover the work needed to take it out of the first metal. Then elec­tric work in the gap must make up the dif­fer­ence.

Key Points

$\begin{picture}(15,5.5)(0,-3) \put(2,0){\makebox(0,0){\scriptsize\bf0}} \put(12... ...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}} \end{picture}$

When two solids are brought in con­tact, their chem­i­cal po­ten­tials, or Fermi lev­els, must line up. A dou­ble layer of pos­i­tive and neg­a­tive charges forms at the con­tact sur­face be­tween the solids. This dou­ble layer pro­duces a step in volt­age be­tween the in­te­ri­ors of the solids.

$\begin{picture}(15,5.5)(0,-3) \put(2,0){\makebox(0,0){\scriptsize\bf0}} \put(12... ...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}} \end{picture}$

There is a volt­age dif­fer­ence in the gap be­tween two met­als that are elec­tri­cally con­nected and have dif­fer­ent work func­tions. It is called the con­tact po­ten­tial.