JEE Mains · Chemistry · STD 12 - 2. Electrochemistry
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 \(\left.\mathrm{KClO}_{3}=122.6 \mathrm{~g} \mathrm{~mol}^{-1}, \mathrm{~F}=96500 \mathrm{C}\right)\)
- A \(1\)
- B \(2\)
- C \(3\)
- D \(4\)
Answer & Solution
Correct Answer
(A) \(1\)
Step-by-step Solution
Detailed explanation
Given balanced equation is \(6 \mathrm{OH}^{-}+\mathrm{Cl} ^{-}\rightarrow \mathrm{ClO}_{3}^{-}+3 \mathrm{H}_{2} \mathrm{O}+6 \mathrm{e}^{-}\) \(\rightarrow 10\, \mathrm{~g}\, \mathrm{KClO}_{3} \Rightarrow \frac{10}{122.6} \,\mathrm{~mol} \,\mathrm{KCO} 3\) in obtained…
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