JEE Main 2018ChemistryElectrochemistryMediumMCQ

JEE Main 2018 — Electrochemistry Question with Solution

JEE Main 2018 (08 Apr)

Question

How long (approximate) should water be electrolysed by passing through 100 amperes current so that the oxygen released can completely burn 27.66 g of diborane? (Atomic weight of B=10.8 u)

Choose an option

Show full solutionCorrect option: D
Correct answer
D3.2 hours

Step-by-step explanation

B2H6+3O2→B2O3+3H2O
According to balanced equation
27.66 g B2H6 i.e. 1 mole B2H6 requires 3 mole of O2 . Now this oxygen is produced by electrolysis of water as follows
2H2O→4F2H2+O2
As 1 mole O2 is produced by 4 F charge
So 3 mole O2 will be produced by 12 F charge
So now on applying
Q = It
12×96500C=100×t(s)
t=12×96500100×3600 hours
t=3.2 hours

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About this question

This is a previous-year question from JEE Main 2018, 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.