Chemistry - Electrochemistry Question with Solution | TestHub

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