Balance the following equations in basic medium by ion-electron method and oxidation number methods and identify the oxidising agent and the reducing agent.
\( P_{ 4 \; (s) } \; + \; OH^{ – }_{ (aq) } \; \rightarrow \; PH_{ 3 \; (g) } \; + \; HPO^{ – }_{ 2 \; (aq) }\)
1 Answer
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:
_1638330046.bmp)
The charge is balanced by the addition of \( 12OH^{-} \)as follows:
_1638330280.bmp)
By adding \( 4H_2O\), the H and O atoms are balanced.
_1638330364.bmp)
The half-reduction equation is as follows:
_1638330391.bmp)
The P atom is in a state of equilibrium. By adding 12 electrons to the equation., it is balanced:
_1638330422.bmp)
The charge is balanced by the addition of \( 12OH^{-}\) as follows:
_1638330471.bmp)
12H2O is used to balance the 0 and H atoms as follows:
_1638330531.bmp)
The balanced chemical equation can be found by multiplying equations ∣ and (ii) by 3 and then adding them.
_1638330577.bmp)
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