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

Question

1690 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-19T15:29:49

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

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

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

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?

How does elemental phosphorus react to give PH_3PH_3 and HPO_2^(-)HPO_2^(-) under basic conditions?