Balance the following redox reactions:

${B r}_{2} (a q) + {O H}^{-} (a q) ⟶ {B r}^{-} (a q) + B r O_{3}^{-} (a q)$

{3 B r}_{2} (a q) + {6 O H}^{-} (a q) ⟶ {5 B r}^{-} (a q) + B r O_{3}^{-} (a q) + {3 H}_{2} O (l)

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3 {3 B r}_{2} (a q) + {6 O H}^{-} (a q) ⟶ {2 B r}^{-} (a q) + 4 B r O_{3}^{-} (a q) + {3 H}_{2} O (l)

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5 {3 B r}_{2} (a q) + {7 O H}^{-} (a q) ⟶ {2 B r}^{-} (a q) + 3 B r O_{3}^{-} (a q) + {3 H}_{2} O (l)

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{3 B r}_{2} (a q) + {7 O H}^{-} (a q) ⟶ {5 B r}^{-} (a q) + 3 B r O_{3}^{-} (a q) + {3 H}_{2} O (l)

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Solution

The correct option is A ${3 B r}_{2} (a q) + {6 O H}^{-} (a q) ⟶ {5 B r}^{-} (a q) + B r O_{3}^{-} (a q) + {3 H}_{2} O (l)$
The unbalanced chemical equation is as follows:
${B r}_{2} (a q) + {O H}^{-} (a q) ⟶ {B r}^{-} (a q) + B r O_{3}^{-} (a q)$
All atoms other than H and O are balanced.
The oxidation number of Br changes from 0 to -1. The change in the oxidation number is 1.
The oxidation number of Br changes from 0 to +5. The change in the oxidation number is 5.
To balance the increase in the oxidation number with decrease in the oxidation number, multiply $B r_{2}$ with 3 and $B r^{-}$ with 5.
$3 {B r}_{2} (a q) + {O H}^{-} (a q) ⟶ 5 {B r}^{-} (a q) + B r O_{3}^{-} (a q)$
To balance O atoms, add, two water molecules on LHS.
$3 {B r}_{2} (a q) + {O H}^{-} (a q) + 2 H_{2} O ⟶ 5 {B r}^{-} (a q) + B r O_{3}^{-} (a q)$
To balance H atoms, add 5 $H^{+}$ on the RHS.
$3 {B r}_{2} (a q) + {O H}^{-} (a q) + 2 H_{2} O ⟶ 5 {B r}^{-} (a q) + B r O_{3}^{-} (a q) + 5 H^{+}$
Since the reaction occurs in basic medium, add $5 O H^{-}$ ions on both sides.
$3 {B r}_{2} (a q) + 6 {O H}^{-} (a q) + 2 H_{2} O ⟶ 5 {B r}^{-} (a q) + B r O_{3}^{-} (a q) + 5 H^{+} + 5 O H^{-}$
On RHS, five protons and five hydroxide ions forms five water molecules out of which two cancels out with two water molecules on RHS.
${3 B r}_{2} (a q) + {6 O H}^{-} (a q) ⟶ {5 B r}^{-} (a q) + B r O_{3}^{-} (a q) + {3 H}_{2} O (l)$
This is the balanced chemical equation.

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Similar questions

5 B r^{-} (a q) + 6 H^{+} (a q) + B r O_{3}^{-} (a q) \to 3 B r_{2} (a q) + 3 H_{2} O (l)

- \frac{△ [B r O_{3}^{-}]}{△ t} = 0.01 m o l L^{- 1} m i n^{- 1}

\frac{△ [B r_{2}]}{△ t}

m o l L^{- 1} m i n^{- 1}

5 B r^{- 1} (a q) + 6 H^{+} (a q) + B r O_{3}^{-} (a q) \to 3 B r_{2} (a q) + 3 H_{2} O (l)

- \frac{Δ [B r O_{3}^{-}]}{Δ t} = 0.01 m o l L^{- 1} m i n^{- 1}, \frac{Δ [B r_{2}]}{Δ t}

m o l L^{- 1} m i n^{- 1}

5 B r_{(a q)}^{-} + B r O_{3 (a q)}^{-} + 6 H_{(a q)}^{+} \to 3 B r_{2 (a q)} + 3 H_{2} O_{(l)}

{B r O_{3}}^{-} (a q .) + 5 {B r}^{-} (a q .) + 6 H^{+} (a q .) ⟶ 3 {B r}_{2} (l) + 3 H_{2} O (l)

({B r}_{2})

5 B r^{-} (a q) + B r O_{3}^{-} (a q) + 6 H^{+} (a q) \to 3 B r_{2} (a q) + 3 H_{2} O (l)

B r_{2}

M s^{- 1}

M s^{- 1}

B r^{-}

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