For the reaction P_4 + xOH^- + yH_2O to PH_3 | vedclass

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Question

(C) The given reaction is a disproportionation reaction: $P_4 + xOH^- + yH_2O 	o PH_3 + H_2PO_2^-$.Balancing the reaction using the oxidation number

Vedclass · Accepted Answer

$x = y$

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(C) The given reaction is a disproportionation reaction: $P_4 + xOH^- + yH_2O 	o PH_3 + H_2PO_2^-$.Balancing the reaction using the oxidation number method:Oxidation half-reaction: $P_4 	o 4H_2PO_2^-$. The oxidation state of $P$ changes from $0$ to $+1$. Total increase in oxidation number is $4 	imes 1 = 4$.Reduction half-reaction: $P_4 	o 4PH_3$. The oxidation state of $P$ changes from $0$ to $-3$. Total decrease in oxidation number is $4 	imes 3 = 12$.To balance the change in oxidation numbers,multiply the oxidation half-reaction by $3$ and the reduction half-reaction by $1$: $3P_4 + 12OH^- + 12H_2O 	o 3PH_3 + 9PH_3 + 12H_2PO_2^-$.Simplifying the equation: $4P_4 + 12OH^- + 12H_2O 	o 4PH_3 + 12H_2PO_2^-$.Dividing by $4$: $P_4 + 3OH^- + 3H_2O 	o PH_3 + 3H_2PO_2^-$.Comparing this with the given equation,$x = 3$ and $y = 3$.Therefore,$x = y$.

For the reaction $P_4 + xOH^- + yH_2O \to PH_3 + H_2PO_2^-$,determine the correct relationship between $x$ and $y$.

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