Question

Balance the chemical equation p4 + oh- gives ph3 + h2po2- + h+

Answer 1

first of all your equation is wrong

Answer 2

It is a 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 -