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

1 Answer
May 25, 2017

Answer:

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.