Consider the change in oxidation state of bromine corresponding to different emf values as shown in the given diagram:

$B r O_{4}^{-} 1.82 V - -- \to B r O_{3}^{-} 1.50 V - -- \to H B r O 1.595 V - --- \to B r_{2} 1.065 V - --- \to B r^{-}$

Then the species undergoing disproportionation is

H B r O

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B r O_{4}^{-}

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B r O_{3}^{-}

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B r_{2}

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Solution

The correct option is A $H B r O$
For a reaction to be spontaneous, $E_{c e l l}^{o}$ should be positive as $Δ G = - n F E_{c e l l}^{o}$

$H B r O \to B r_{2}; E^{o} = 1.595 V,$ SRP (cathode)
$H B r O \to B r O_{3}^{-}; E^{o} = - 1.5 00 V,$ SOP (anode)
$2 H B r O \to B r_{2} + B r O_{3}^{-}$

$E_{c e l l}^{o}$ $= S R P (c a t h o d e) - S R P (a n o d e)$
$= 1.595 V - 1.500 V = 0.095 V$

$E_{c e l l}^{o} > 0 \Rightarrow Δ G < 0$ (spontaneous)

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Similar questions

B r O_{4}^{-} 1.82 V - -- \to B r O_{3}^{-} 1.5 V - -- \to H B r O 1.595 V - --- \to B r_{2} 1.0652 V - ---- \to B r^{-}

B r O_{4}^{-}

1.82 V - -- \to

B r O_{3}^{-} 1.50 V - -- \to

H B r O 1.595 V - --- \to

B r_{2} 1.06552 - --- \to B r^{-}

In the conservation $B r_{2} ⟶ B R O_{3}^{-}$ , the oxidation state of bromine changes from

B r_{2}

B r {O_{3}}^{-}

B r_{2}

N a O H

B r^{-}

B r {O_{3}}^{-}

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