JEE Main 2022ChemistryElectrochemistryMediumNumerical

JEE Main 2022 — Electrochemistry Question with Solution

JEE Main 2022 (27 Jun Shift 2)

Question

For the reaction taking place in the cell:

PtsH2gH+aqAg+aqAgs

Ecell°=+0.5332 V.

The value of ΔfG⊖ is kJmol-1. (in nearest integer)

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Show full solutionCorrect answer: 51
Correct answer
51

Step-by-step explanation

Anode ⇒12H2g⟶H+aq+e-Cathode ⇒Ag++e-⟶Ag(s)12H2g+Ag+aq⟶n=1H+aq+Ags

Relation between Gibbs Free Energy and EMF of a Cell is given by

ΔG°=-nFEcell°=-1×96500×0.5332=51453.8 J mol-1=51.4538 kJ mol-1

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About this question

This is a previous-year question from JEE Main 2022, 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.