$. . . B r O_{3}^{-} (a q) + . . . B r^{-} (a q) + . . . H^{+} (a q) \to . . . B r_{2} (l) + . . . H_{2} O$
After balancing the above equation. What is the ratio of coefficient $B r O_{3}^{-}$ to $B r^{-}$ ?

1 : 5

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1 : 3

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1 : 2

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1 : 1

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2 : 3

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Solution

The correct option is A 1 : 5
The unbalanced chemical equation is
$B r O_{3}^{-} (a q) + B r^{-} (a q) + H^{+} (a q) \to B r_{2} (l) + H_{2} O$
The oxidation number of bromine increases from -1 (in $B r^{-}$ ion) to 0 (in $B r_{2}$ ).
Increase in the oxidation number is 1.
The oxidation number of bromine decreases from +5 (in $B r O_{3}^{-}$ ) to 0 (in $B r_{2}$ ).
Decrease in the oxidation number of bromine is 5.
Balance increase in oxidation number with decrease in oxidation number.
$B r O_{3}^{-} (a q) + 5 B r^{-} (a q) + H^{+} (a q) \to 3 B r_{2} (l) + H_{2} O$
Hence, the ratio of $B r O_{3}^{-}$ to $B r^{-}$ is 1: 5.

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