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Equivalent Mass

In acidic med...

Question

In acidic medium, equivalent weight of $K_{2} C r_{2} O_{7}$ (Molecular weight = M) is -

M/3

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M/4

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M/6

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M/2

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Solution

The correct option is C M/6
Reaction of $K_{2} C r_{2} O_{7}$ in acidic condition,
$K_{2} C r_{2} O_{7} + 14 H^{+} + 6 e^{-} \to 2 K^{+} + 2 C r^{3 +} + 7 H_{2} O$
Here 2 atom of $C r$ reduced from +6 oxidation to +3 oxidation state. Hence 6 $e^{-}$ is involved in the redox reaction.
$E q u i v a l e n t m a s s o f K_{2} C r_{2} O_{7} = \frac{M o l e c u l a r m a s s}{n u m b e r o f e l e c t r o n s i n v o l v e d i n t h e r e a c t i o n}$
Here, Molecular mass = M
Hence,
$E q u i v a l e n t m a s s o f K_{2} C r_{2} O_{7} = \frac{M}{6}$

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Similar questions

K_{2} C r_{2} O_{7}

K_{2} C r_{2} O_{7}

(M)

K_{2} C r_{2} O_{7}

K_{2} C r_{2} O_{7}

C r_{2} O_{7}^{2 -} + 14 H^{+} + 6 e^{-} \to 2 C r^{3 +} + 7 H_{2} O

K_{2} C r_{2} O_{7}

= M

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