JEE PYQ: Electrochemistry Question 19
Question 19 - 2020 (06 Sep Shift 1)
Potassium chlorate is prepared by the electrolysis of KCl in basic solution: $6\text{OH}^- + \text{Cl}^- \rightarrow \text{ClO}_3^- + 3\text{H}_2\text{O} + 6e^-$ If only 60% of the current is utilized in the reaction, the time (rounded to the nearest hour) required to produce 10 g of KClO$_3$ using a current of 2 A is ____.
(Given: F = 96,500 C mol$^{-1}$; molar mass of KClO$_3$ = 122 g mol$^{-1}$)
Show Answer
Answer: 11
Solution
$t = \frac{100 \times 96500}{122 \times 2} = 39549$ s $\approx 11$ h.
BETA
