Potassium chlorate is prepared by electrolysis of KCl
Question:

Potassium chlorate is prepared by electrolysis of KCl in basic solution as shown by following equation.

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

A current of xA has to be passed for 10h to produce 10.0g of potassium chlorate. the value of x is _______. (Nearest integer)

(Molar mass of $\mathrm{KClO}_{3}=122.6 \mathrm{~g} \mathrm{~mol}^{-1}$, $\mathrm{F}=96500 \mathrm{C}$ )

Solution:

Given balanced equation is

$\stackrel{\ominus}{60 \mathrm{H}}+\stackrel{\ominus}{\mathrm{Cl}} \longrightarrow \mathrm{ClO}_{3}^{-}+3 \mathrm{H}_{2} \mathrm{O}+6 \mathrm{e}^{-}$

$\rightarrow 10 \mathrm{~g} \mathrm{KClO}_{3} \Rightarrow \frac{10}{122.6} \mathrm{~mol} \mathrm{KCO}_{3}$ in obtained

$\rightarrow$ from the above reaction, it is concluded that by

$6 \mathrm{~F}$ charge $1 \mathrm{~mol} \mathrm{KClO}_{3}$ is obtained.

$\rightarrow$ By the passage of $6 \mathrm{~F}$ charge $=1 \mathrm{~mol} \mathrm{KClO}_{3}$

$\therefore$ By the passage of $\frac{x \times 10 \times 60 \times 60}{96500}$ F charge

$=\frac{1}{6} \times \frac{x \times 10 \times 60 \times 60}{96500}$

Now $\frac{x \times 10 \times 60 \times 60}{6 \times 96500}=\frac{10}{122.6}$

$\Rightarrow x=\frac{10 \times 965}{60 \times 122.6}=\frac{965}{735.6}=1.311 \simeq 1$

OR

$\mathrm{W}=\frac{\mathrm{E}}{\mathrm{F}} \times \mathrm{I} \times \mathrm{t}$

$10=\frac{122.6}{96500 \times 6} \times x \times 10 \times 3600$

X = 1.311

Ans.(1) 

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