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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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

$x = y$

Vedclass · Answer

(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 $y$.

Similar Questions

The number of electrons involved in the reduction of permanganate to manganese dioxide in acidic medium is $.......$.

The equivalent weight of $K_2Cr_2O_7$ in an acidic medium is...... (Molecular weight $= M$)

One of the products formed due to the reaction between $KMnO_4$ and $HCl$ is

$6 \times 10^{-3}$ mole $K_2Cr_2O_7$ reacts completely with $9 \times 10^{-3}$ mole $X^{n+}$ to give $XO_3^-$ and $Cr^{3+}$. The value of $n$ is

When $SO_2$ gas is passed through an acidic solution of $K_2Cr_2O_7$,what happens?

Vedclass Test Series

Exam Paper Generator

Online Exam Module