Equivalent weight of H_3PO_2 when it dispropo | vedclass

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(D) The disproportionation reaction is: $3H_3PO_2 \rightarrow 2H_3PO_3 + PH_3$. In this reaction,the oxidation state of $P$ in $H_3PO_2$ is $+1$. In $

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$3M/4$

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(D) The disproportionation reaction is: $3H_3PO_2 ightarrow 2H_3PO_3 + PH_3$. In this reaction,the oxidation state of $P$ in $H_3PO_2$ is $+1$. In $H_3PO_3$,the oxidation state of $P$ is $+3$ (oxidation). In $PH_3$,the oxidation state of $P$ is $-3$ (reduction). For $H_3PO_2 ightarrow H_3PO_3$,the change in oxidation state is $3 - 1 = 2$. Since there are $2$ molecules of $H_3PO_3$,the total change is $2 	imes 2 = 4$. For $H_3PO_2 ightarrow PH_3$,the change in oxidation state is $1 - (-3) = 4$. The total number of electrons transferred is $4$. The $n$-factor for the disproportionation reaction is calculated as the total change in oxidation state per mole of reactant: $n = 4/3$. Equivalent weight $= M / n = M / (4/3) = 3M/4$.

Equivalent weight of $H_3PO_2$ when it disproportionates into $PH_3$ and $H_3PO_3$ is:

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