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  • Potassium chlorate is prepared by the electrolysis of in basic solution 

Potassium chlorate is prepared by the electrolysis of KCl in basic solution \mathrm{Cl}^{-}+6 \mathrm{OH}^{-} \rightarrow \mathrm{ClO}_{3}^{-}+3 \mathrm{H}_{2} \mathrm{O}+6 \mathrm{e}^{-} If only 60% of current utilized in the reaction, the time (hr) required to produce 10 g of KClO_{3} using a current of 2A is ___ \left(F=96500 \mathrm{C} \mathrm{mol}^{-},\: \mathrm{KClO}_{3}=122 \mathrm{g} \cdot \mathrm{mol}^{-}\right)  
 

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The reaction occurs as follows:

\mathrm{Cl^{-}\: +\: 6OH^{-}\: \rightarrow \: ClO_{3}^{-}\: +\: 3H_{2}O\: +\: 6e^{-}}

Current efficiency = 60%

Now, moles of KClO_{3}= \frac{10}{122.5}

Thus, moles of e^{-}= \frac{60}{122.5} (assuming 100% efficiency)

Thus, actual moles of e^{-}= \frac{100}{122.5}

\mathrm{\therefore \: \frac{2\: x\: t}{96500}\: =\: \frac{100}{122.5}}

Thus, t = 39387.76 sec. = 10.94 hours.

Thus, time required is approximately 11 hours.

Posted by

Kuldeep Maurya

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