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

Dec 19, 2016

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

#### Explanation:

$\frac{1}{4} {P}_{4} + 3 {H}^{+} + 3 {e}^{-} \rightarrow P {H}_{3}$ $\text{(i) reduction}$

$\frac{1}{4} {P}_{4} + 2 H {O}^{-} \rightarrow H P {O}_{2}^{-} + {H}^{+} + 2 {e}^{-}$ $\text{ (ii) oxidation}$

Overall: $2 \times \left(i\right) + 3 \times \left(i i\right)$

$\frac{5}{4} {P}_{4} + 6 {H}_{2} O \rightarrow 2 P {H}_{3} + 3 H P {O}_{2}^{-} + 3 {H}^{+}$

I will add $3 \times H {O}^{-}$ to each side of the equation to represent that the reaction is performed in basic conditions:

$\frac{5}{4} {P}_{4} + 3 H {O}^{-} + 3 {H}_{2} O \rightarrow 2 P {H}_{3} + 3 H P {O}_{2}^{-}$

Is this balanced with respect to mass and charge?