In the reaction:
$3 B r_{2} + 6 C O_{3}^{2 -} + 3 H_{2} O \to 5 B r^{-} + B r O_{3}^{-} + 6 H C O_{3}^{-}$

Bromine is reduced and carbonate ion is oxidised.

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Bromine undergoes disproportionation.

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Bromine is reduced and water is oxidised.

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Only water is oxidised to carbonic acid.

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Solution

The correct option is C Bromine undergoes disproportionation.
Separating the oxidation and reduction reaction from the redox reaction:
$2 B r_{2} + 6 C O_{3}^{2 -} + 3 H_{2} O \to 5 B r^{-} + B r O_{3}^{-} + 6 H C O_{3}^{-}$
assigning the oxidation number on central atom ( $B r$ and $C$ ) in each molecules by considering oxidation number of $H = + 1, O = - 2$ we get oxidation state as:
$3 0 B r_{2} + 6 + 4 C O_{3}^{2 -} + 3 H_{2} O \to 5 - B r + + 5 B r O_{3}^{-} + 6 H + 4 C O_{3}^{-}$
as the oxidation number of $C$ does not change and for $B r$ it changes from 0 to -1 and 0 to +5 as:
$B r_{2} \to B r^{-}$ -reduction
$2 B r_{2} \to B r O_{3}^{-}$ -oxidation
Since $B r$ gets oxidized and reduced in the reaction it undergoes disproportionation.

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Similar questions

3 B r_{2} + 6 C O_{3}^{2 -} + 3 H_{2} O \to 5 B r^{-} + B r O_{3}^{-} + 6 H C O_{3}^{-}

In the following reaction

3Br₂ + 6CO₃^2- + 3H₂O → 5Br ^-+ BrO₃^- + 6HCO₃^-

3 B r_{2} + 6 C O_{3}^{2 -} + 3 H_{2} O \to 5 B r^{⊖} + B r O_{3}^{⊖} + 6 H C O_{3}^{⊖}

3 {B r}_{2} + 6 C O_{3}^{2 -} + 3 H_{2} O ⟶ 5 {B r}^{⊖} + B r O_{3}^{⊖} + 6 H C O_{3}^{⊖}

3 B r_{2} + 6 C O_{3}^{2 -} + 3 H_{2} O \to 5 B r^{-} + B r O_{3}^{-} + 6 H C O_{3}^{-}

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