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

May 25, 2017

This is disproportionation............elemental phosphorus is simultaneously reduced and oxidized.......

#### Explanation:

$\text{Reduction half equation}$

$\frac{1}{4} {P}_{4} + 3 {H}_{2} O \left(l\right) + 3 {e}^{-} \rightarrow P {H}_{3} \left(g\right) + 3 H {O}^{-}$ $\left(i\right)$

$\text{Oxidation half equation}$

1/4P_4 + 2HO^(-)rarr ""^(-)stackrel(+I)P(OH)_2+e^(-) $\left(i i\right)$

And we take $\left(i\right) + 3 \times \left(i i\right)$ to eliminate the electrons which are conceptual particles..........

${P}_{4} + 3 H {O}^{-} + 3 {H}_{2} O \left(l\right) \rightarrow P {H}_{3} \left(g\right) + 3 {\left(H O\right)}_{2} {P}^{-}$

Which (I think!) is balanced with respect to mass and charge, as indeed it must be if we purport to represent chemical reality.