Answer :

The O.N. (oxidation number) of P decreases from 0 in P4 to – 3 in PH3 and increases from 0 in P4 to + 2 in HPO2-.

Hence, P4 clearly acts both as an oxidizing agent and a reducing agent in this reaction.

Now, balancing the equation in basic medium by ion-electron method for the reduction half reaction: -

0P4 (s) -3PH3(g)

Now, balancing P atoms :-

P4 (s) 4PH3(g)

Balancing oxidation number by adding electrons: -

P4 (s) +12 e- 4PH3(g)

Balancing charge by adding OH- ions : -

P4 (s) +12 e- 4PH3(g) +12OH-1(aq)

Balancing 'O' atoms by adding H2O : -

Balancing the equation in basic medium by ion-electron method for the oxidation half reaction: -

Now, balancing P atoms : -

Balance oxidation number by adding electrons : -

Balance charge by adding OH- ions we get :-

Oxygen and hydrogen are balanced automatically.

Multiplying the oxidation half reaction by 3 and then adding it to the reduction half reaction, we have the net balanced redox reaction equation as:-

Oxidation number method

The change in oxidation number are as follows : -

In a balanced chemical reaction loss of electrons = gain of electrons so we multiply H2PO2- by 3 and get : -

We multiply OH- by 3 to balance the charge and get : -

Balancing H by adding 3H2O to LHS we get the final balanced equation as: -

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