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Oxidation Number Method

Find the equi...

Question

Find the equivalent weight of $K_{2} C r_{2} O_{7}^{2 -}$ in the given reaction, if the molar mass of $C r_{2} O_{7}^{2 -}$ is 'M' ?
$K_{2} C r_{2} O_{7}^{2 -}^{+} C r^{3 +}$

\frac{M}{6}

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\frac{M}{3}

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M

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2 M

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Solution

The correct option is A $\frac{M}{6}$
In acidic medium, the balanced reaction will be:
$6 e^{-} + + 6 C r_{2} O_{7}^{2 -} + 14 H^{+} \to 2 + 3 C r^{3 +} + 7 H_{2} O$
n -factor $n_{f} o f K_{2} C r_{2} O_{7} = | 3 - 6 | \times 2 = 6$
$Equivalent weight of K_{2} C r_{2} O_{7} = \frac{M o l . w t o f K_{2} C r_{2} O_{7}}{6} = \frac{M}{6}$

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Similar questions

C r_{2} O_{7}^{2 -}^{+} C r^{3 +}

C r_{2} O_{7}^{2 -}

C r_{2} O_{7}^{2 -}

M g / m o l

C r_{2} O_{7}^{2 -}^{+} C r^{3 +}

C r_{2} O_{7}^{2 -}

C r_{2} O_{7}^{2 -}

M g / m o l

C r_{2} O_{7}^{2 -} H^{+} - - \to C r^{3 +}

C r_{2} O_{7}^{2 -}

K M n O_{4}

M n O_{4}^{-}

K M n O_{4} \to M n^{2 +}

{C r_{2} O_{7}}^{2 -} \to C r^{3 +}

{C r_{2} O_{7}}^{2 -} = \frac{Molar Mass}{3}

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