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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 -
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