electric circuits - Pn junction voltage drop?

This image from wikipedia, explains that there occurs a potential drop across a pn semiconductor junction, and an electric field confined to the depletion region.

I already know the reason for the existence of this drop and the calculation of the difference but I have two questions regarding this drop.

1) If the n and p doped regions are externally connected using a perfectly conducting wire, why will not any current flow?
Connecting the two regions should equalize their potentials (since wire is a resistance less) conductor, and therefore the gradient at the junction is destroyed resulting in a diffusion current which is obviously against conservation of energy as the semiconductor has non-zero resistivity. Where will extra potential drops will be created so that Kirchoff's voltage rule holds without any current and the built-in potential difference $V_o$ of the juction persists?

2) (probably naive) If an external Bias is applied, of say $|V|<|V_o|$, then the pn potential difference across the junction will just reduce by that amount ($|V|$). Assuming that the external voltage source is ideal without any resistance, what will be the potential drops which would sum to zero in this case? Will drop due to the resistance of semiconductor play a par in it? If yes, then there cannot be any ideal resistance-less semiconductor junction, can there?

Answer

Calling it a built-in voltage is something of a misnomer. People usually think of "voltage" as "what you measure with a voltmeter". So "voltage" is normally synonymous with "electrochemical potential of electrons" (in stat mech terminology) and with "difference in fermi level" (in semiconductor terminology). Under this definition, the built-in "voltage" is not actually a voltage.

Then what is it? It's what chemists call "galvani potential", and some physicists call "electrostatic potential". It's the line-integral of electric field. (Maybe you should call it "built-in potential", not "built-in voltage".)

Voltage / fermi level measures the total "happiness" of electrons, the sum of all influences on the electron. The electric field (galvani potential) is just one of those many influences. Other influences include diffusion (entropy), the kinetic energy of the electron's wave function, etc. etc. But it's the sum of all influences that determines how the electron moves. That's why it is the voltage, not the galvani potential, that determines the most important things like current flow and energy dissipation.

So to summarize: The "voltage" across a p-n junction is zero, when the word "voltage" is defined in the most common and sensible and intuitive way. After all, the junction is in equilibrium; an electron is equally happy to be on either side.

For more details see my other answer: Fermi level alignment and electrochemical potential between two metals

Going around a loop, both the voltage differences and the galvani potential differences sum to zero. But only the former is really important. For the galvani potential differences, most of them are unobservable, like the volta potential at the junction when you solder an aluminum wire to a copper wire. It is possible to figure out the galvani potential differences everywhere in a p-n junction circuit, including at the wire contacts, at the voltmeter, and so on. If you do figure them out, and add them all up, you'll get zero! But since none of those parameters matter for the circuit behavior, people rarely think about them or try to figure them out.