Pls calculate the n- factor of this reactionP4 + OH- ----> PH3 + H2PO2

1.	Pls calculate the n- factor of this reactionP4 + OH- ----> PH3 + H2PO2THANK YOU☺️☺️
Answer» Explanation: elemental phosphorus is both OXIDIZED and reduced… 14P4+3H++3e−→PH3REDUCTION: P(0) → P(-III): But basic conditions were specified…so we add three equiv hydroxide to each side… 14P4+3H2O+3e−→PH3+3HO−Oxidation: P(0) → P(-III): 14P4+2HO−→P(OH)+2+3e−Oxidation: P(0) → P(+III): Note that you got the wrong charge in your question… We add the oxidation reaction to the reduction equation to get… 12P4+3H++2HO−→P(OH)+2+PH3 And we CLEAN this up to get… 12P4+H++2H2O→P(OH)+2+PH3 …which I THINK is balanced with respect to MASS and charge

Pls calculate the n- factor of this reactionP4 + OH- ----> PH3 + H2PO2THANK YOU☺️☺️

Answer»

Explanation:

elemental phosphorus is both OXIDIZED and reduced…

14P4+3H++3e−→PH3REDUCTION: P(0) → P(-III):

But basic conditions were specified…so we add three equiv hydroxide to each side…

14P4+3H2O+3e−→PH3+3HO−Oxidation: P(0) → P(-III):

14P4+2HO−→P(OH)+2+3e−Oxidation: P(0) → P(+III):

Note that you got the wrong charge in your question…

We add the oxidation reaction to the reduction equation to get…

12P4+3H++2HO−→P(OH)+2+PH3

And we CLEAN this up to get…

12P4+H++2H2O→P(OH)+2+PH3

…which I THINK is balanced with respect to MASS and charge