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_{4}^{-}

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B r O_{3}^{-}

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H B r O

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Solution

The correct option is D $H B r O$
$+ 1 H B r O \to 0 B r_{2}, E_{\frac{H B r O}{B r_{2}}}^{0} = 1.595 V$
$+ 1 H B r O \to + 5 B r O_{3}^{-}, E_{\frac{B r O_{3}^{-}}{H B r O}}^{0} = 1.5 V$
$E_{c e l l}^{0}$ for the disproportionation of $H B r O$ .
$E_{c e l l}^{0} = E_{\frac{H B r O}{B r_{2}}}^{0} - E_{\frac{B r O_{3}^{-}}{H B r O}}^{0}$
=1.595 -1.5
=0.095 V = +ve

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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} 1.065 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^{-}

_{3}

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

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