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STD 12 - 2. Electrochemistry — NEET STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceNEETSTD 12 - 2. Electrochemistry4 Marks
MCQ
Consider the change in oxidation state of Bromine corresponding to different emf values as shown in the diagram below

$\mathrm{BrO}_{4}^{-} \stackrel{1.82 \mathrm{V}}{\longrightarrow} \mathrm{BrO}_{3}^{-} \stackrel{1.5 \mathrm{V}}{\longrightarrow} \mathrm{HBrO}$$\stackrel{1.0652 \mathrm{V}}{\longrightarrow} \mathrm{Br}_{2} \stackrel{1.595 \mathrm{V}}{\longrightarrow} \mathrm{Br}^{-}$

Then the species undergoing disproportionation is

  • A
    $\mathrm{BrO}_{3}^{-} $
  • B
    $\mathrm{BrO}_{4}^{-} $
  • C
    $Br_2$
  • ✓
    $HBrO$
✓

Answer

Correct option: D.
$HBrO$
d
Calculate $E_{cell}^o$ corresponding to each compound under golng disproportlonation reactlon. The reaction for which $\mathrm{E}_{\text {cell }}^{\circ}$ comes out $+ve$ is spontaneous.

$\mathrm{HBrO} \longrightarrow \mathrm{Br}_{2} \quad \mathrm{E}^{\circ}=1.595,$ SRP (cathode)

$\mathrm{HBrO} \longrightarrow \mathrm{BrO}_{3}^{-} \quad \mathrm{E}^{\circ}=-1.5 \mathrm{V},$ SOP (Anode)

$2 \mathrm{HBrO} \longrightarrow \mathrm{Br}_{2}+\mathrm{BrO}_{3}^{-}$

$\mathrm{E}_{\text {cell }}^{\circ}=\mathrm{SRP}(\text { cathode })-\mathrm{SRP}(\text { Anode })$

$=1.595-1.5$

$=0.095 \mathrm{V}$

$\mathrm{E}_{\text {cell }}^{\circ}>0 \Rightarrow \Delta \mathrm{G}^{\circ}<0$ [spontaneous]

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