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Question: Balance the following equation in the basic medium by ion-electron method and oxidation number metho...

Balance the following equation in the basic medium by ion-electron method and oxidation number method and identify the oxidising agent and the reducing agent.
P4(s)+OH−(aq)→PH3(g)+H2PO2−(aq){P_4}(s) + O{H^ - }(aq) \to P{H_3}(g) + {H_2}P{O_2}^ - (aq)

Explanation

Solution

To answer this question, you should recall the ion-electron method to balance equations. In the ion-electron method also known as the half-reaction method, we proceed by separation the reaction into two halves - one for oxidation and one for reduction. Now balance each of these reactions separately and combine them such that you eliminate the extra ions to arrive at the final balanced redox equation.

Complete step by step answer:
From the reaction,
P4(s)+OH−(aq)→PH3(g)+H2PO2−(aq){P_4}(s) + O{H^ - }(aq) \to P{H_3}(g) + {H_2}P{O_2}^ - (aq)
We can see that the oxidation number of P decreases from 0 to −3 - 3 in PH3P{H_3} and increases from 0 to +2 + 2 in H2PO2−{H_2}P{O_2}^ - . Hence, P4{P_4}is oxidizing as well as reducing agent. Now let’s proceed stepwise to balance this equation using the ion-electron method:
1.The oxidation half-reaction is:
P4(s)→H2PO2−(aq){P_{4}}(s) \to {H_2}P{O_2}^ - (aq)
To balance P{\text{P}} atoms multiply RHS of reaction with 4:
P4(s)→4H2PO2−(aq){P_4}(s) \to 4{H_2}P{O_2}^ - (aq)
The oxidation number is balanced by adding 4 electrons on RHS:
P4(s)→4H2PO2−(aq)+4e−{P_4}(s) \to 4{H_2}P{O_2}^ - (aq) + 4{e^ - }
The charge is balanced by adding 8 hydroxide ions on LHS:
P4(s)+8OH−(aq)→4H2PO2−(aq){P_4}(s) + 8O{H^ - }(aq) \to 4{H_2}P{O_2}^ - (aq)
The O{\text{O}} and H{\text{H}} atoms are also now balanced.

2. The reduction half-reaction is
P4(s)→PH3(g){P_4}(s) \to P{H_3}(g)
The oxidation number is balanced by adding 12 electrons on LHS.
P4(s)+12e−→PH3(g){P_4}(s) + 12{e^ - } \to P{H_3}(g)
The charge is balanced by adding 12 hydroxide ions on RHS
P4(s)+12e−→PH3(g)+12OH−{P_4}(s) + 12{e^ - } \to P{H_3}(g) + 12O{H^ - }
The oxidation half-reaction is multiplied by 3 and the reduction half-reaction is multiplied by 2 to cancel out the electrons.

The half-reactions are then added to obtain a balanced chemical equation.P4(s)+3OH−(aq)+3H2O(l)→PH3(g)+3H2PO2−(aq){P_{4}}(s) + 3O{H^ - }(aq) + 3{H_2}O(l) \to P{H_3}(g) + 3{H_2}P{O_2}^ - (aq)

Note: We can balance this reaction using another method known as the oxidation number method
During reduction, the total decrease in the oxidation number for 4 P{\text{P}} atoms is 12.
During oxidation, the total increase in the oxidation number for 4 P{\text{P}} atoms is 4.
The increase in the oxidation number is balanced with a decrease in the oxidation number by multiplying H2PO2−  {H_2}P{O_2}^ - \;with 3.
P4(s)+OH−(aq)→PH3(g)+3H2PO2−(aq){P_4}(s) + O{H^ - }(aq) \to P{H_3}(g) + 3{H_2}P{O_2}^{ - }(aq)
To balance O atoms, multiply OH−  O{H^{ - \;}}ions by 6.
P4(s)+6OH−(aq)→PH3(g)+3H2PO2−(aq){P_4}(s) + 6O{H^ - }(aq) \to P{H_3}(g) + 3{H_2}P{O_2}^ - (aq)
To balance H{\text{H}} atoms, 3 water molecules are added to L.H.S and 3 hydroxide ions on R.H.S.
P4(s)+6OH−(aq)+3H2O(l)→PH3(g)+3H2PO2−(aq)+3OH−(aq){P_4}(s) + 6O{H^ - }(aq) + 3{H_2}O(l) \to P{H_3}(g) + 3{H_2}P{O_2}^ - (aq) + 3O{H^ - }(aq)
Subtract 3 hydroxide ions from both sides:
P4(s)+3OH−(aq)+3H2O(l)→PH3(g)+3H2PO2−(aq){P_{4}}(s) + 3O{H^ - }(aq) + 3{H_{2}}O(l) \to P{H_{3}}(g) + 3{H_2}P{O_2}^ - (aq)