How many Faradays are required to reduce $1 m o l$ $B r O_{3}^{-}$ to $B r^{-}$ ?

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Solution

The correct option is A 6
The reaction of the reduction is,
$B r O_{3}^{-} + 6 e^{-} + 3 H_{2} O \to B r^{-} + 6 O H^{-}$
The oxidation state of $B r$ in $B r O_{3}^{-}$ is +5 and reduced to $- 1$ . Hence, 6 electrons are involved in the reaction.

One faraday of charge is the magnitude of the charge of one mole of electrons.

Thus, 6 faradays are required for this reaction.
Hence, option (c) is correct.

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