JEE Main 2019ChemistryElectrochemistryMediumMCQ

JEE Main 2019 — Electrochemistry Question with Solution

JEE Main 2019 (09 Apr Shift 1)

Question

The standard Gibbs energy for the given cell reaction in kJ mol-1 at 298 K is:
Zns+Cu2+aq⟶Zn2+aq+Cu(s) ,
E0=2 V at 298 K
(Faraday's constant ,F=96000 C mol-1)

Choose an option

Show full solutionCorrect option: D
Correct answer
D-384

Step-by-step explanation

For the reaction,
Cu2+aq+Zns⟶Zn2+aq+Cus
n=2, two electrons are exchanged
∆Go=-nFEo
=-2×96000×2=-384000
=-384 kJ

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About this question

This is a previous-year question from JEE Main 2019, 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.