Consider the change in oxidation state of Bromine corresponding to different emf values as shown in the diagram below.

Then the species undergoing disproportionation is __________.

B r_{2}

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B r O_{3}^{-}

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H B r O

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B r O_{4}^{-}

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Solution

The correct option is C $H B r O$
$H B r O$ will disproportionate to $B r O_{3}^{-}$ and $B r_{2}$

$B r O_{3}^{-} 1.5 V - -- \to H B r O$
Reverse the above equation.
$H B r O - 1.5 V - ---- \to B r O_{3}^{-}$ .....(1)
$H B r O 1.595 V - ---- \to B r_{2}$ .....(2)
Add equations (1) and (2)
$2 H B r O - 1.5 V + 1.595 V = 0.095 V - ---------------- \to B r O_{3}^{-} + B r_{2}$
The positive value of net cell reaction indicates that the overall cell reaction is spontaneous.

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Similar questions

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

_{3}

B r_{2}

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

B r_{2}

N a O H

B r^{-}

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

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