In the following reaction,4P + 3KOH + 3H_2O t | vedclass

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(A) In the reaction $4P + 3KOH + 3H_2O 	o 3KH_2PO_2 + PH_3$,the oxidation state of $P$ changes as follows:$1$. In elemental phosphorus $(P_4)$,the ox

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$P$ is oxidized as well as reduced

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(A) In the reaction $4P + 3KOH + 3H_2O 	o 3KH_2PO_2 + PH_3$,the oxidation state of $P$ changes as follows:$1$. In elemental phosphorus $(P_4)$,the oxidation state of $P$ is $0$.$2$. In $KH_2PO_2$,the oxidation state of $P$ is calculated as: $1 + 2(+1) + x + 2(-2) = 0 \implies 1 + 2 + x - 4 = 0 \implies x = +1$.$3$. In $PH_3$,the oxidation state of $P$ is calculated as: $x + 3(+1) = 0 \implies x = -3$.Since the oxidation state of $P$ increases from $0$ to $+1$ (oxidation) and decreases from $0$ to $-3$ (reduction),$P$ is both oxidized and reduced. This is a disproportionation reaction.

In the following reaction,$4P + 3KOH + 3H_2O \to 3KH_2PO_2 + PH_3$,what happens to phosphorus $(P)$?

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