Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

Can somebody explain me the disproportionation of Phosphorus (P4) into Phosphine (PH3) and dihydro hypophosphite ion ??? Also I need balanced reaction for it.

Can somebody explain me the disproportionation of Phosphorus (P4) into Phosphine (PH3) and dihydro hypophosphite ion ??? Also I need balanced reaction for it.

Grade:11

1 Answers

Arun
25763 Points
2 years ago

disproportionation reaction of P4, i.e., P4 is oxidised as well as reduced in the reaction. P4 is the oxidising as well as the reducing agent. P4 is oxidised to H2PO2 - and reduced to PH3. So the(unbalanced) half-equations are as follows.

P4 => 4PH3, (1) and

P4 => 4 H2PO2- (2)

Following the general procedure, we can write

P4 + 12 H2O + 12 e- => 4 PH3 + 12 OH- (1a)

P4 + 8 OH- => 4 H2PO2 - + 4 e- (2a)

Eq(1a) is the reduction half-equation and Eq(2a) is the oxidation half equation.

Multiplying Eq (2a) by 3 so that the two half equations have equal number of electrons,

we get

3 P4 + 24 OH- => 12 H2PO2 - + 12 e- (2b)

Adding Eq (1a) and Eq(2b) and cancelling the common terms , we get the balanced ionic equation

4 P4 + 12 H2O + 12 OH- => 4 PH3 + 12 H2PO2 -

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free