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

Answer 1

Serenity Allen

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 believe, is balanced in terms of mass and charge, as it really needs to be if we're going to claim to be depicting the reality of chemistry.

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Answer 2

Owen Ambrose

This is the formula for the following reaction: [ P_4 + 3NaOH + 3H_2O \rightarrow PH_3 + 3Na^+[H_2P(OH)_2]^- ] elemental phosphorus reacts with a strong base, like sodium hydroxide (NaOH), and water (H2O) undergoing basic hydrolysis to produce phosphine gas (PH3) and the sodium salt of hypophosphorous acid ([Na^+[H_2P(OH)_2]^-]), which is commonly referred to as sodium hypophosphite.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.