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Question

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

6OH + Cl -> ClO3 + 3H2O + 6e

If only 60% of the current is utilized in the reaction, the time (rounded to the nearest hour) required to produce 10g of KClO3 using a current of 2A is ………….

(Given: F = 96,500 C mol-1; molar mass of KClO3=122g mol-1)


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Solution

Step 1: To find the charge produce to make 10g KClO3

Numberofmoles=MassofthesubstanceMolarmass=10122moles6OH-+Cl-ClO3-+3H2O+6e-

Here, the total number o charge transfer is 6e-

It is conducted that by 6F charge 1 mole of KClO3 is obtained.

10gKClO3=10122×6F

Step 2 :- To calculate the time required to produce 10g KClO3.

By using the formula, Q=I×t

Where Q= charge transfer, I= current, and t= time

Q=I×t10122×6F=2×t10122×6=2×tF

In the above questions, it is given that 60% of current is utilized, that is 60100

10122×6=2×t×60100×360096500{1F=96500C}t=11hrs.

Final answer:- The time required to produce 10g of KClO3 using a current of 2A is 11hrs.


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