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Choose the Best Answer · Q21

Q.Consider the change in oxidation state of bromine corresponding to different emf values as shown in the diagram below: BrO4−→1.82VBrO3−→1.5VHBrO→1.595VBr2→1.0652VBr−\text{BrO}_4^{-} \xrightarrow{1.82\text{V}} \text{BrO}_3^{-} \xrightarrow{1.5\text{V}} \text{HBrO} \xrightarrow{1.595\text{V}} \text{Br}_2 \xrightarrow{1.0652\text{V}} \text{Br}^{-} Then the species undergoing disproportionation is

(a) Br2\text{Br}_2
(b) BrO4−\text{BrO}_4^{-}
(c) BrO3−\text{BrO}_3^{-}
(d) HBrO
Tamil Nadu Board Class 12 Chemistry, Electrochemistry: Latimer diagram for bromine showing standard emf values — BrO4⁻ →(1.82 V)→ BrO3⁻ →(1.5 V)→ HBrO →(1.595 V)→ Br2 →(1.0652 V)→ Br⁻, used to identify which species disproportionates.
Figure
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Step 1. In a Latimer-style diagram, a species in the middle of the chain undergoes disproportionation exactly when the reduction potential to its RIGHT (converting it to the next lower oxidation state) is GREATER than the potential to its LEFT (the potential by which it itself was formed from the higher oxidation state) — because this means the species finds it thermodynamically favourable to be simultaneously oxidised (reversing the left arrow) and reduced (following the right arrow).

Step 2. Check BrO3−\text{BrO}_3^-: left potential (BrO4−→BrO3−\text{BrO}_4^-\to\text{BrO}_3^-) = 1.82 V; right potential (BrO3−→HBrO\text{BrO}_3^-\to\text{HBrO}) = 1.5 V. Since left (1.82) > right (1.5), BrO3−\text{BrO}_3^- does NOT disproportionate. …

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