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)

Question

Philoid · Accepted 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: -

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)