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

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Answer 1 · 2017-05-25T15:55:52

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

$"Reduction half equation"$

$1/4P_4 +3H_2O(l) + 3e^(-) rarr PH_3(g)+3HO^(-)$ $(i)$

$"Oxidation half equation"$

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

And we take $(i) + 3xx(ii)$ to eliminate the electrons which are conceptual particles..........

$P_4 + 3HO^(-)+3H_2O(l) rarr PH_3(g) + 3(HO)_2P^(-)$

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

Elemental phosphorus gives PH_3PH_3 and Na^+""^(-)P(OH)_2Na^+""^(-)P(OH)_2 upon basic hydrolysis. How is the reaction formulated?