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

$$P_{ 4 \; (s) } \; + \; OH^{ – }_{ (aq) } \; \rightarrow \; PH_{ 3 \; (g) } \; + \; HPO^{ – }_{ 2 \; (aq) }$$

Asked by Pragya Singh | 1 year ago |  74

##### Solution :-

The oxidation number of P drops from 0 to -3 in P4 and increases from 0 to +2 in HPO2 . As a result, P4 serves as both an oxidizing and reducing agent in
this process. Ion-electron method: The half-equation for oxidation is:

$$P_4\rightarrow HPO_2^{-}$$

The P atom is balanced in the following way:

The charge is balanced by the addition of $$12OH^{-}$$as follows:

By adding $$4H_2O$$, the H and O atoms are balanced.

The half-reduction equation is as follows:

The P atom is in a state of equilibrium. By adding 12 electrons to the equation., it is balanced:

The charge is balanced by the addition of $$12OH^{-}$$ as follows:

12H2O is used to balance the 0 and H atoms as follows:

The balanced chemical equation can be found by multiplying equations ∣ and (ii) by 3 and then adding them.

Answered by Abhisek | 1 year ago

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