Potassium chlorate is prepared by the electrolysis of \(\mathrm{KCl}\) in basic solution as: \(6 \mathrm{OH}^{-}+\mathrm…

Potassium chlorate is prepared by the electrolysis of \(\mathrm{KCl}\) in basic solution as:

\(6 \mathrm{OH}^{-}+\mathrm{Cl}^{-} \rightarrow \mathrm{ClO}_{3}^{-}+3 \mathrm{H}_{2} \mathrm{O}+6 \mathrm{e}^{-}\)

If only \(60 \%\) of the current is utilized in the reaction, the time (rounded off to the nearest hour) required to produce \(10 \mathrm{~g}\) of \(\mathrm{KClO}_{3}\) using a current of \(2 \mathrm{~A}\) is _____.

(Given: \(\mathrm{F}=96,500 \mathrm{C} \mathrm{mol}^{-1}\); molar mass of \(\mathrm{KClO}_{3}=122\) \(\mathrm{g} \mathrm{mol}^{-1}\) )

[JEE Main 2020, 6 Sep (Shift 1)]