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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