Question

Elemental phosphorus gives `PH_3` and `Na^+""^(-)P(OH)_2` upon basic hydrolysis. How is the reaction formulated?

Answer 1

This is disproportionation............elemental phosphorus is simultaneously reduced and oxidized.......

Explanation:

`"Reduction half equation"`

`1/4P_4 +3H_2O(l) + 3e^(-) rarr PH_3(g)+3HO^(-)` `(i)`

`"Oxidation half equation"`

`1/4P_4 + 2HO^(-)rarr ""^(-)stackrel(+I)P(OH)_2+e^(-)` `(ii)`

And we take `(i) + 3xx(ii)` to eliminate the electrons which are conceptual particles..........

`P_4 + 3HO^(-)+3H_2O(l) rarr PH_3(g) + 3(HO)_2P^(-)`

Which (I think!) is balanced with respect to mass and charge, as indeed it must be if we purport to represent chemical reality.