Balance the following equations in basic medium by ion-electron method and oxidation number methods and identify the oxidising agent and the reducing agent.

P4(s) + OH–(aq) → PH3(g) + HPO–2(aq)

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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