How long should aqueous NaCl be electrolysed by passing 100 ampere current, so that 0.5 mol chlorine is released at anode?

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TestHub · Accepted Answer

The oxidation reaction at the anode is:

$2 C l^{-} \to C l_{2} + 2 e^{-}$

From the stoichiometry of the reaction, 1 mole of $C l_{2}$ gas is produced by the passage of 2 moles of electrons.

Therefore, 0.5 mole of $C l_{2}$ gas requires $0.5 \times 2 = 1$ mole of electrons.

Charge required, $Q = 1 F = 96500 C$

Using the formula $Q = I \times t$ :

$96500 = 100 \times t$

$t = \frac{96500}{100} = 965 seconds$

Answer: 965 seconds

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