If M represents molecular mass of $B r_{2}$ then what will be its equivalent mass if it undergoes disproportionation reaction as shown:

$3 B r_{2} + 6 O H^{-} \to 5 B r^{-} + B r O_{3}^{-} + 3 H_{2} O$

\frac{M}{2}

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

\frac{M}{10}

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

\frac{M}{2}

\frac{M}{10}

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

$\frac{M}{6}$

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

Open in App

Solution

The correct option is C $\frac{M}{2}$ + $\frac{M}{10}$
$B r_{2}$ dispropotionates, its equivalent weight is the sum of equivalent weights of the two half reactions.

$2 e^{-} + B r_{2} \to 2 B r^{-}$ (x=2) (reduction)

$B r_{2} \to 2 B r O_{3}^{-} + 10 e^{-}$ ('n'=10) (oxidation)
2x=0 2x-12=-2
2x=10

$E_{w}$ = $\frac{M}{2} + \frac{M}{10}$

Suggest Corrections

Similar questions

B r_{2}

B r^{-}

{B r O_{3}}^{-}

3 B r_{2} + 6 O H^{-} \to 5 B r^{-} + {B r O_{3}}^{-} + 3 H_{2} O

B r_{2}

= M

For the reaction 3 $B r_{2} + 6 O H^{-} \to 5 B r^{-} + B r O_{3}^{-} + 3 H_{2} O$
Equivalent weight of $B r_{2}$ is [M = mol wt of $B r_{2}$ ]

3 B r_{2} + 6 O H^{⊖} \to 5 B r^{⊖} + B r O_{3}^{⊖} + 3 H_{2} O

B r_{2}

M

5 B r^{-} + B r O_{3}^{-} + 6 H^{+} \to 3 B r_{2} + 3 H_{2} O

B r_{2}

M

For the reaction

$3 {B r}_{2}$ + $6 {O H}^{Θ}$ $\to$ $5 {B r}^{Θ}$ + $B r {O_{3}}^{Θ}$ + $3 H_{2} O$

Equivalent weight of ${B r}_{2}$ (Molecular weight M) is

Join BYJU'S Learning Program