PN junction diode's free electrons
I don't think your premise is correct.  When a p-type material (excess holes) joins an n-type material (excess electrons), some of these excess holes and electrons travel to the other side via diffusion and recombine with electrons and holes, respectively, on the opposite side.  These electrons and holes that diffuse to the other side are then gone via recombination, depleting them.  But, the p-type material has some electrons to begin with (it has excess holes), and the N-type material had some holes ( it has excess electrons), so  electons become  loose  on the P-side (when holes diffuse the N -side) and holes become loose on the N-side (when electrons diffuse to the p-side) and they line up at the junction boundary where the charged carriers were depleted.  This process achieves equilibrium when no more carriers travel across the junction.
When the junction is reverse biased, the holes in the p-type material are attracted to "negative polarity" of the voltage acrosss the junction ,  leaving more electrons behind at the junction.  Similarly, the positive polarity on the n-side attracts the electrons on that side, leaving the holes behind at the junction.  This increases  the opposing electric field at the junction which further opposes the exchange of carriers between the p and n materials thus expanding this depletion region further.   The holes and electrons are already there in the materials in different densities (p-type is mostly holes, but has electrons too, n-type is mostly  electrons, but also has holes.