How does elemental phosphorus react to give #PH_3# and #HPO_2^(-)# under basic conditions?

1 Answer
Dec 19, 2016

Answer:

This is a disproportionation reaction. Elemental phosphorus is reduced and oxidized.

Explanation:

#1/4P_4+3H^+ + 3e^(-) rarrPH_3# #"(i) reduction"#

#1/4P_4 + 2HO^(-) rarr HPO_2^(-) +H^(+) + 2e^(-)# #" (ii) oxidation"#

Overall: #2xx(i) + 3xx(ii)#

#5/4P_4+6H_2O rarr2PH_3 +3HPO_2^(-) +3H^(+)#

I will add #3xxHO^-# to each side of the equation to represent that the reaction is performed in basic conditions:

#5/4P_4+3HO^(-) + 3H_2O rarr2PH_3 +3HPO_2^(-) #

Is this balanced with respect to mass and charge?