Potassium chlorate is prepared by the electrolysis of $\mathrm{KCl}$ in basic solution $6 \mathrm{OH}^{-}+\mathrm{Cl}^{-} \rightarrow \mathrm{ClO}_{3}^{-}+3 \mathrm{H}_{2} \mathrm{O}+6 \mathrm{e}^{-}$ If only $60 \%$ of the current is utilized in the reaction, the time (rounded to the nearest hour) required to produce 10 $\mathrm{g}$ of $\mathrm{KClO}_{3}$ using a current of $2 \mathrm{~A}$ is______________. (Given: $\mathrm{F}=96,500 \mathrm{Cmol}^{-1} ;$ molar mass of $\overline{\mathrm{KClO}_{3}}=122 \mathrm{gmol}^{-1}$ )
(11)
$6 \mathrm{OH}^{-}+\mathrm{Cl}^{-} \longrightarrow \mathrm{ClO}_{3}^{-}+3 \mathrm{H}_{2} \mathrm{O}+6 \mathrm{e}^{-}$
For synthesis of 1 mole of $\mathrm{ClO}_{3}^{-}, 6 \mathrm{~F}$ of charge is required.
$\because$ Current efficiency $=60 \%$
$\therefore$ To synthesis 1 mole of $\mathrm{ClO}_{3}^{-}, 10 \mathrm{~F}$ of charge is required.
To synthesis $\frac{10}{122}$ moles of $\mathrm{KClO}_{3}$, charge $=\frac{10 \times 10}{122} \mathrm{~F}$
$\mathrm{Q}=\mathrm{I} . \mathrm{t}$
$t=\frac{100 \times 96500}{122 \times 2}=39549.18 \mathrm{~s}$
$=\frac{79098.365}{3600 \mathrm{~s}}=10.99 \mathrm{~h}$
$\therefore t=11 \mathrm{~h}$.