JEE Mains · Chemistry · STD 12 - 2. Electrochemistry
Potassium chlorate is prepared by the electrolysis of \(KCl\) in basic solution \(6 OH ^{-}+ Cl ^{-} \rightarrow ClO _{3}^{-}+3 H _{2} O +6 e ^{-}\) If only \(60 \%\) of the current is utilized in the reaction, the time (rounded to the nearest hour) required to produce \(10\, g\) of \(KCIO _{3}\) using a current of \(2\, A\) is.......... (Given : \(F =96,500\, C\, mol ^{-1}\) molar mass of \(\left. KClO _{3}=122\,gmol ^{-1}\right)\)
- A \(11\)
- B \(8\)
- C \(18\)
- D \(22\)
Answer & Solution
Correct Answer
(A) \(11\)
Step-by-step Solution
Detailed explanation
\(\frac{10}{122} \times 6=\frac{2 \times( hr ) \times 3600 \times 60 \%}{96500}\) \(+( hr )=\frac{96500}{122 \times 72}=10.98 hr\) \(=11\) hours
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