JEE Main 2023ChemistryElectrochemistryMediumNumerical

JEE Main 2023 — Electrochemistry Question with Solution

JEE Main 2023 (30 Jan Shift 1)

Question

Consider the cell

PtsH2g,1atmH+aq,1M|Fe3+aq,Fe2+aqPts

When the potential of the cell is 0.712 V at 298 K, the ratio Fe2+/Fe3+ is
(Nearest integer)
Given: Fe3++e-=Fe2+,E°Fe3+,Fe2+Pt=0.771 2.303RTF=0.06 V

Enter your answer

Show full solutionCorrect answer: 10
Correct answer
10

Step-by-step explanation

Given cell reaction:
PtsH2g,1atmH+aq,1M||Fe3+aq,Fe2+aqPts

at anode(oxidation) H2⟶2H++2e-

At cathode(reduction) Feaq3++e-⟶Feaq2+

E°cell=Ecathodeo - Eanodeo=EoH2|H+ +EoFe3+∣Fe2+=0.771 V

Thus, using Nernst equation,
⇒E=E°-0·061logFe2+Fe3+

⇒0.712=0+0.771-0.061logFe2+Fe3+

⇒logFe2+Fe3+=0.0590.06≈1

Fe2+Fe3+=10

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About this question

This is a previous-year question from JEE Main 2023, 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.