Redox Reactions Ques 7
7. Consider the change in oxidation state of bromine corresponding to different emf values as shown in the diagram below :
[2018]
$BrO_4^{-} \xrightarrow{1.82 V} BrO_3^{-} \xrightarrow{1.5 V} HBrO$
$ Br^{-} \longleftarrow _{1.0652 V} Br_2 \longleftarrow 1.595 V $
Then the species undergoing disproportionation is
(a) $BrO_3^{-}$
(b) $BrO_4^{-}$
(c) $HBrO$
(d) $Br_2$
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Answer:
Correct Answer: 7.(c)
Solution:
- (c) Calculate $E _{\text{cell }}^{\circ}$ corresponding to each compound undergoing disproportionation reaction. The reaction for which $E _{\text{cell }}^{\circ}$ comes out $+ve$ is spontaneous.
$ \begin{aligned} & HBrO \longrightarrow Br_2 E^{\circ}=1.595 V, SRP \text{ (cathode) } \\ & HBrO \longrightarrow BrO_3^{-} E^{\circ}=-1.5 V, SOP \text{ (anode) } \\ & 2 HBrO \longrightarrow Br_2+BrO_3^{-} \\ & E _{\text{cell }}^{\circ}=SRP \text{ (cathode) }-SRP \text{ (anode) } \\ & =1.595-1.5 \\ & =0.095 V \\ & E _{\text{cell }}^{\circ}>0 \Rightarrow \Delta G^{\circ}<0 \text{ [spontaneous] } \end{aligned} $
Reaction in (d) involves comproportionation or synproportionation. When two reactants, each containing the same element but with a different oxidation number, form a product in which the element involved reach the same oxidation number.
It is opposite to disproportionation.
BETA
