During the electrolysis of aqueous NaCl,if 3 | vedclass

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(C) The reduction reaction of water during electrolysis is: $2 H_2O + 2 e^- \longrightarrow H_2 + 2 OH^-$. From the stoichiometry,$2 \ mole$ of $H_2O$

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$4$

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(C) The reduction reaction of water during electrolysis is: $2 H_2O + 2 e^- \longrightarrow H_2 + 2 OH^-$. From the stoichiometry,$2 \ mole$ of $H_2O$ require $2 \ mole$ of electrons ($2 \ F$ of charge). Therefore,$1 \ mole$ of $H_2O$ requires $1 \ F$ of charge. For $3 \ mole$ of $H_2O$,the theoretical charge required is $3 \ F$. Given the current efficiency is $75 \ \%$,the actual charge required $(Q)$ is calculated as: $Q 	imes 0.75 = 3 \ F$ $Q = \frac{3}{0.75} = 4 \ F$.

During the electrolysis of aqueous $NaCl$,if $3 \ mole$ of $H_2O$ are electrolysed,then how much charge (in $F$) is required if the current efficiency is $75 \ \%$?

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