Electrolysis of dilute aqueous NaCl solution | vedclass

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Question

(B) The cathode reaction for the electrolysis of aqueous $NaCl$ is: $2H_2O(l) + 2e^{-} \longrightarrow H_2(g) + 2OH^{-}(aq)$.From the stoichiometry,$1

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$19.3 	imes 10^4 \ s$

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(B) The cathode reaction for the electrolysis of aqueous $NaCl$ is: $2H_2O(l) + 2e^{-} \longrightarrow H_2(g) + 2OH^{-}(aq)$.From the stoichiometry,$1 \ mol$ of $H_2$ requires $2 \ mol$ of electrons.Therefore,$0.01 \ mol$ of $H_2$ requires $0.02 \ mol$ of electrons.Total charge $Q = n 	imes F = 0.02 	imes 96500 \ C = 1930 \ C$.Given current $i = 10 \ mA = 10 	imes 10^{-3} \ A = 0.01 \ A$.Using the formula $Q = i 	imes t$:$1930 = 0.01 	imes t$.$t = \frac{1930}{0.01} = 193000 \ s = 19.3 	imes 10^4 \ s$.

Electrolysis of dilute aqueous $NaCl$ solution was carried out by passing $10 \ mA$ current. The time required to liberate $0.01 \ mol$ of $H_2$ gas at the cathode is $(1 \ F = 96500 \ C \ mol^{-1})$

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