You visited us 1 times! Enjoying our articles? Unlock Full Access!

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}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{M}{3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

M

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2 M

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

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 $n_{f} o f K_{2} C r_{2} O_{7} = | 3 - 6 | \times 2 = 6$
$Equivalent weight of K_{2} C r_{2} O_{7} = \frac{M o l . w t o f K_{2} C r_{2} O_{7}}{6} = \frac{M}{6}$

Suggest Corrections

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}

Join BYJU'S Learning Program

Explore more

Join BYJU'S Learning Program