Electrolysis of a dilute aqueous NaCl solutio | vedclass

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(B) The reduction reaction at the cathode is: $2H_2O(l) + 2e^- \rightarrow H_2(g) + 2OH^-(aq)$.From the stoichiometry,$1 \, mol$ of $H_2$ requires $2

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$19.3 	imes 10^4 \, s$

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(B) The reduction reaction at the cathode is: $2H_2O(l) + 2e^- ightarrow H_2(g) + 2OH^-(aq)$.From the stoichiometry,$1 \, mol$ of $H_2$ requires $2 \, mol$ of electrons.Therefore,$0.01 \, mol$ of $H_2$ requires $0.01 	imes 2 = 0.02 \, mol$ of electrons.Total charge $Q = n 	imes F = 0.02 \, mol 	imes 96500 \, C \, mol^{-1} = 1930 \, C$.Given current $I = 10 \, mA = 10 	imes 10^{-3} \, A = 0.01 \, A$.Using the formula $Q = I 	imes t$,we get $t = \frac{Q}{I} = \frac{1930 \, C}{0.01 \, A} = 193000 \, s$.This can be written as $19.3 	imes 10^4 \, s$.

Electrolysis of a dilute aqueous $NaCl$ solution is carried out by passing a current of $10 \, mA$. How much time will it take to liberate $0.01 \, mol$ of $H_2$ gas at the cathode? $(1 \, F = 96500 \, C \, mol^{-1})$

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