Question

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

Answer 1

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?