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STD 12 - 2. Electrochemistry — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 2. Electrochemistry4 Marks
Question
Potassium chlorate is prepared by electrolysis of $\mathrm{KCl}$ 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 $x A$ has to be passed for $10 h$ to produce $10.0 \mathrm{~g}$ of potassium chlorate. The value of $\mathrm{x}$ is $.......$ (Nearest integer)

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

✓

Answer

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

$\rightarrow 10\, \mathrm{~g}\, \mathrm{KClO}_{3} \Rightarrow \frac{10}{122.6} \,\mathrm{~mol} \,\mathrm{KCO} 3$ 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 $\frac{x \times 10 \times 60 \times 60}{96500}\, \mathrm{~F}$ 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

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

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

$X=1.311$

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