Potassium chlorate is prepared by the electrolysis of KCl in basic solution

6OH^– + Cl^– -> ClO₃^– + 3H₂O + 6e^–

If only 60% of the current is utilized in the reaction, the time (rounded to the nearest hour) required to produce 10g of KClO₃ using a current of 2A is ………….

(Given: F = 96,500 C mol^-1; molar mass of KClO₃=122g mol^-1)

Step 1: To find the charge produce to make 10g ${\mathrm{KClO}}_3$

$$\begin{gathered} \mathrm{Number}\mathrm{of}\mathrm{moles}=\frac{\mathrm{Mass}\mathrm{of}\mathrm{the}\mathrm{substance}}{\mathrm{Molar}\mathrm{mass}}=\frac{10}{122}\mathrm{moles} \\ 6{\mathrm{OH}}^{-}+{\mathrm{Cl}}^{-}\to {\mathrm{ClO}}^{3-}+3{\mathrm{H}}_2\mathrm{O}+6{\mathrm{e}}^{-} \end{gathered}$$

Here, the total number o charge transfer is $6{\mathrm{e}}^{-}$

It is conducted that by 6F charge 1 mole of ${\mathrm{KClO}}_3$ is obtained.

$10\mathrm{g}{\mathrm{KClO}}_3=\frac{10}{122}\times 6\mathrm{F}$

Step 2 :- To calculate the time required to produce 10g ${\mathrm{KClO}}_3$_.

By using the formula, $\mathrm{Q}=\mathrm{I}\times \mathrm{t}$

Where Q= charge transfer, I= current, and t= time

$$\begin{gathered} \mathrm{Q}=\mathrm{I}\times \mathrm{t} \\ \frac{10}{122}\times 6\mathrm{F}=2\times \mathrm{t} \\ \frac{10}{122}\times 6=\frac{2\times \mathrm{t}}{\mathrm{F}} \end{gathered}$$

In the above questions, it is given that 60% of current is utilized, that is $\frac{60}{100}$

$$\begin{gathered} \frac{10}{122}\times 6=\frac{2\times \mathrm{t}\times \frac{60}{100}\times 3600}{96500}\{1\mathrm{F}=96500\mathrm{C}\} \\ \mathrm{t}=11\mathrm{hrs}. \end{gathered}$$

Final answer:- The time required to produce 10g of ${\mathrm{KClO}}_3$ using a current of 2A is 11hrs.