Redox and Equivalent ConceptHard

Question

In the reaction, P4 + NaOH + H2O → PH3 + NaH2PO2, the equivalent weight of P4 is

Options

A.M
B.M/3
C.M/6
D.2M/3

Solution

$P_{4} + 3NaOH + 3H_{2}O \rightarrow PH_{3} + 3NaH_{2}PO_{2}$

The reaction is balanced by the loss and gain of 3 moles of electron per mole of P4 and hence, $E_{P_{4}} = \frac{M}{3}$

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