Potassium chlorate is prepared by the electrolysis of KCl in basic solution 6 OH^-+ Cl^- → ClO_3^-+3 H_2 O +6 e^- If only 60 % of the current is…

Potassium chlorate is prepared by the electrolysis of KCl in basic solution $$6 \mathrm{OH}^{-}+\mathrm{Cl}^{-} \rightarrow \mathrm{ClO}_3^{-}+3 \mathrm{H}_2 \mathrm{O}+6 \mathrm{e}^{-} $$ If only $\displaystyle 60 \%$ of the current is utilized in the reaction, the time (rounded to the nearest hour) required to produce $\displaystyle 10$ g of $\displaystyle \mathrm{KClO}_3$ using a current of $\displaystyle 2$ A is \_\_\_\_. (Given : $\displaystyle \mathrm{F}=96,500 \mathrm{C} \mathrm{mol}^{-1}$; molar mass of $\displaystyle \mathrm{KClO}_3=122 \mathrm{~g} \mathrm{~mol}^{-1}$ )