NEET · Chemistry · STD 12 - 2. Electrochemistry
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 \(Br_2\)
- D \(HBrO\)
Answer & Solution
Correct Answer
(D) \(HBrO\)
Step-by-step Solution
Detailed explanation
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)
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