Phosphorus when heated with NaOH solution gives Phosphine `(PH_(3))andH_(2)PO_(2)^(-)`. Give balanced equation.

Explanation

`P_(4)+NaOH+H_(2)O to PH_(3)+NaH_(2)PO_(2)`
`P_(4)+OH-PH_(3)+H_(2)PO_(2)to `
(a) `P_(4)+OH^(-)to PH_(3)+HPO_(2)^(-)`
(Reduction half cell )
``P_(4) to PH_(3)`
`P_(4)to 4PH_(3)`
`P_(4)+12H_(2)O to 4PH_(3)+12OH^(-)`("H balanced")`
`P_(4)+12H_(2)O+12e^(-)to 4PH_(3)+12OH^(-)`
`overset(P_(4)+12H_(2)O+12e^(-)to 4PH_(3)+12OH^(-))(3P_(4)+24 OH^(-)to 12H_(2)PO_(2)^(-)+12e^(-))/underline(4P_(4)+12H_(2)O+12e^(-)to 12H_(2)PO_(2)^(-)+4 PH_(3))`
Here `P_(4)` is oxidising agent as well reducing agent.
Oxidation half cell
`P_(4)to H_(2)PO_(2)^(-)`
`P_(4)to 4H_(2)PO_(2)^(-)`(".P.balanced")
`P_(4)+8H_(2)O to 4H_(2)PO_(2)^(-)` (Oxygen balanced)
`P_(4)+8H_(2)O+8OH^(-)to 4H_(2)PO_(2)^(-)+8H_(2)O` (Hydrogen balanced)
`P_(4)+8OH^(-)to 4H_(2)PO_(2)^(-)+4e^(-)`