In the electrolysis of an aqueous cupric bromide solution, how many grams of bromine are formed on passing a current of $1$ ampere for $16$ minutes and $5$ seconds? Write the anode and the cathode half reactions.

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Solution

Cell reactions:
at anode: $C u$ $\to$ $C u^{2 +}$ + $2 e^{-}$
at cathode: 2 $B r^{-}$ + 2 $e^{-}$ $\to$ $B r_{2}$
we have,
$w =$ $\frac{E * I * t}{96500}$

$w =$ $\frac{40 * 1 * 101}{96500}$

$w =$ 0.041g

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