Potassium chlorate is prepared by the electrolysis of KCl in basic solution
\(6\mathrm{OH^-} + \mathrm{Cl^-} \to \mathrm{ClO_3^-} + 3\mathrm{H_2O} + 6e^-\)
If only 60% of the current is utilized in the reaction, the time (rounded to the nearest hour) required to produce 10 g of \(\mathrm{KClO_3}\) using a current of 2 A is ______.
(Given : F = 96,500 C \(\mathrm{mol^{-1}}\); molar mass of \(\mathrm{KClO_3}\) = 122 g \(\mathrm{mol^{-1}}\))
\(6\mathrm{OH^-} + \mathrm{Cl^-} \to \mathrm{ClO_3^-} + 3\mathrm{H_2O} + 6e^-\)
If only 60% of the current is utilized in the reaction, the time (rounded to the nearest hour) required to produce 10 g of \(\mathrm{KClO_3}\) using a current of 2 A is ______.
(Given : F = 96,500 C \(\mathrm{mol^{-1}}\); molar mass of \(\mathrm{KClO_3}\) = 122 g \(\mathrm{mol^{-1}}\))
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Correct Answer: 11
Solution and Explanation
Charge needed (100% efficiency) = 6 \(\times\) 96500 \(\times\) 0.08197 = 47459 C.
Only 60% of the current is utilised, so total charge to be passed = 47459/0.6 = 79098 C.
Time t = Q/I = 79098/2 = 39549 s = 39549/3600 = 10.99 h, which rounds to 11 h.
Answer: 11
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