Diffusion current
When a p-n junction is formed, a diffusion phenomena causes electrons from the n-doped region to diffuse to the p-doped region. At the same time (even if it's an abstraction) holes diffuse from the p-type region to the n-type one. The atoms that lose a carrier (electron or hole) become ions, which means that instead of being neutral, they have a positive or negative net charge. This happens because the ideal equilibrium would have the same concentration of mobile carriers equal all over the region.
Ohmic current
However, this diffusion causes the growth of a region, populated by ions, called depletion region, because all atoms have lost their carrier. These ions, as we said, are electrically charged, and cause an electric field directed from the n-region to the p-region, pushing carriers in the opposite way than diffusion. Therefore an equilibrium is reached in which the current (movement of carriers) caused by diffusion is perfectly balanced by the current caused by the electric field (ohmic current).
Effect of biasing
Applying a potential to the junction causes a perturbation on this equilibrium, making one of the currents dominant on the other. Reverse biasing the junction causes the ohmic current to prevail, while forward biasing increases the diffusion current.
Now, the diffusion current is a much stronger phenomena, from which derives the exponential growth of the forward bias current with the bias voltage. Ohmic current, on the other side, is much weaker, and saturates quite soon (neglecting avalanche effect) because the width of the depletion region (which determines the resistivity) is proportional to the reverse bias voltage.
I think, the answer is relatively simple.
Do you know the working principle of a "Schottky diode", which is based on a semiconductor-metal junction?
Now - what happens if you connect a voltmeter (or any other load) across the diode? You create two Schottky junctions which exactly compensate the diffusion voltage inside the pn diode. Thus, no voltage can be measured.
With other words: You cannot use the diffusion voltage to drive any current through an external load.
Best Answer
In a cold semiconductor below the knee voltage electrons from the n side haven’t enough energy to enter the p side and, likewise, holes from the p side haven’t enough energy to enter the n side. When you heat it, a large fraction of charge carriers (on one of these sides, or both) acquire necessary energy. The Fermi–Dirac distribution can provide an approximation, although forward-biased junction is a strongly non-equilibrium system.
In terms of the Shockley diode equation, it is described as “the saturation current \$I_{\mathrm S}\$ increases with \$T\$ faster than \$\exp(V_{\mathrm D} / n V_T)\$ decreases”.