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Question
Balance the following equation in basic medium by ion-electron method and oxidation number methods and identify the oxidising agent and the reducing agent.
\[\ce{P4(s) + OH–(aq) —> PH3(g) + HPO^–_2(aq)}\]
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Solution
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:
\[\ce{P4s →HPO2- (aq)}\]
The P atom is balanced as:
\[\ce{P4s →4HPO2- (aq)}\]
The O atom is balanced by adding 8 H2O molecules:
\[\ce{P4s + 8H2O → 4HPO2- (aq)}\]
The H atom is balanced by adding 12 H+ ions:
\[\ce{P4s +8H2O →4HPO2- (aq) + 12H+}\]
The charge is balanced by adding e– as:
\[\ce{P4s +8H2O →4HPO2- (aq) + 12H+ + 8e-}\] ...(i)
The reduction half equation is:
\[\ce{P_{4(s)} -> PH_{3(g)}}\]
The P atom is balanced as:
\[\ce{P4 (s) → 4PH3(g)}\]
The H is balanced by adding 12 H+ as:
\[\ce{P4 (s) + 12H + → 4PH3(g)}\]
The charge is balanced by adding 12e– as:
\[\ce{P4 (s) + 12H+ +12e- → 4PH3(g)}\] ...(ii)
By multiplying equation (i) with 3 and (ii) with 2 and then adding them, the balanced chemical equation can be obtained as:
\[\ce{5P4 (s) +24H2O →12HPO2^- + 8PH3(g) +12H+}\]
As, the medium is basic, add 12OH– both sides as:
\[\ce{5P4 (s) +12H2O +12OH- →12HPO2^- +8PH3(g)}\]
This is the required balanced equation.
Oxidation number method:
Let, total no of P reduced = x
∴ Total no of P oxidised = 4 – x
\[\ce{P4 (s) + OH- → xPH3(g) + 4 - xHPO2-}\] ... (i)
Total decrease in oxidation number of P = x × 3 = 3x
Total increase in oxidation number of P
= (4 – x) × 2 = 8 – 2x
∵ 3x = 8 – 2x
x = 8/5
From (i),
\[\ce{5P4 (s) + 5OH- → 8PH3(g) + 12HPO2-}\]
Since, reaction occures in basic medium, the charge is balanced by adding 7OH– on LHS as:
\[\ce{5P4 (s) +12OH- → 8PH3(g) +12HPO2-}\]
The O atoms are balanced by adding 12H2O as:
\[\ce{5P4 (s) + 12H2O + 12OH- → +12HPO2- + 8PH3(g)}\]
This is the required balanced equation.
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