For the reaction, $P b S + 4 H_{2} O_{2} ⟶ P b S O_{4} + 4 H_{2} O$

Which statement(s) is/are correct about the equivalent mass of respective species?

[ $M$ represents molar mass.]

E_{P b S} = \frac{M}{8}

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E_{H_{2} O_{2}} = \frac{M}{2}

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E_{P b S O_{4}} = \frac{M}{2}

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E_{P b S O_{4}} = \frac{M}{8}

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Solution

The correct options are
A $E_{P b S} = \frac{M}{8}$
B $E_{H_{2} O_{2}} = \frac{M}{2}$
D $E_{P b S O_{4}} = \frac{M}{8}$
$P b S + 4 H_{2} O_{2} \to P b S O_{4} + 4 H_{2} O$

+2 -2 +1 -1 +2 +6 -2 +1 -2

$P b S + 4 H_{2} O \to P b S O_{4} + 8 H^{+} + 8 e^{-}$ (oxidation)

$H_{2} O_{2} + 2 H^{+} + 2 e^{-} \to 2 H_{2} O$ (Reduction)

So, we can write,

$E_{P b S} = \frac{M}{8}$

$E_{P b S O_{4}} = \frac{M}{8}$

and $E_{H_{2} O_{2}} = \frac{M}{2}$

So, the correct ans is are $A, B$ and $D$ .

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