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JEE Main 2020 — Electrochemistry Question with Solution

JEE Main 2020 (06 Sep Shift 1)

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 produce10g of KClO3 using a current of 2A is …….. (Given :F=96, 500Cmol; molar mass of KClO3=122 g mol-1)

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Show full solutionCorrect answer: 11
Correct answer
11

Step-by-step explanation

Number of faradays = I×t96500

 =2×t×60×6096500F

Mass of KClO3 produce = 10 Gm

  {2×t×60×60/96500}×0.60×122×(1/6)=10
So t=10.98 hours

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About this question

This is a previous-year question from JEE Main 2020, covering the Electrochemistry chapter of Chemistry. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.