Question 19. Balance the following equations in basic medium by
ion-electron method and oxidation number methods and identify the
oxidising agent and the reducing agent.
(a)P4(s) + OH(aq) → PH3(g) + H2PO2(aq)
Answers
Here is your 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 HPO-2. Hence, P4 acts both as an oxidizing agent and a reducing agent in this reaction.
Ion-electron method:
The oxidation half equation is:
P4(s) → H2PO-(aq)
The P atom is balanced as:
P0 4(s) → 4H2P + 1O-(aq)
The O.N. is balanced by adding 4 electrons as:
P4(s) → 4H2PO-(aq) + 4e-
The charge is balanced by adding 8OH- as:
P4(s) + 8OH - (aq) → 4H2PO-2(aq)
The O and H atoms are already balanced. The reduction half equation is:
P4(s) → PH3(g)
The P atom is balanced as
P04(s) → 4 P-3H3(g)
The O.N. is balanced by adding 12 electrons as:
P4(s) + 12e- → 4 PH3(g)
The charge is balanced by adding 12OH- as:
P4(s) + 12e- → 4 PH3(g) + 12OH-(aq) .....(i)
The O and H atoms are balanced by adding 12H2O as:
P4(s) + 12H2O(l) + 12e- → 4 PH3(g) + 12OH-(aq) ....(ii)
By multiplying equation (i) with 3 and (ii) with 2 and then adding them, the balanced chemical equation can be obtained as:
P4(s) + 3OH-(aq) + 3H2O → PH3 + 3H2PO-2(aq)