First slide

Standard EMF of a cell

Question

Consider the change in oxidation state of Bromine corresponding to different emf values as shown in the diagram below
Then the species undergoing disproportionation is

Difficult

A

Br₂
B

${BrO}_{4}^{-}$
C

${BrO}_{3}^{-}$
D

HBrO

Solution

In
$\large Br{O_4}^ -$
the oxidation state of Br is in +7 which is maximum and not under go disproportianation.
In Br ^-the oxidation state of Br is -1 which is in lowest and can't undergo disproportianation.
Disproportionation of
$\large Br{O_3}^ -$
:
Oxidation:
$\large Br{O_3}^ - \to Br{O_4}^ - \;E^{o}\;=1.82V$
Reduction:
$\large Br{O_3}^ - \to HBrO\;E^{o}\;=1.5V$
$\large {E_{cell}} = E_c^o - E_a^o = 1.5 - 1.82 = - 0.32V$
E_cell is negative hence does not undergo disproportionation.
Disproportionation of Br_2:
Oxidation:
$\large B{r_2} \to HBrO\;E^{o}\;=1.595V$
Reduction:
$\large B{r_2} \to B{r^ - }\;E^{o}\;=1.0652V$
$\large {E_{cell}} = E_c^o - E_a^o = 1.0652 - 1.595 = - 0.53V$
Since E_cell is negative doesn't undergo disproportionation.
Disproportionation of HBrO:
Oxidation:
$\large HBrO \to Br{O_3}^ - \;E^{o}\;=1.5V$
Reduction:
$\large HBrO \to B{r_2} \;E^{o}\;=1.595V$
$\large {E_{cell}} = E_c^o - E_a^o = 1.595 - 1.5 = 0.095V$
Since E_cell is positive it undergoes disproportionation.

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