JEE Main 2023ChemistryElectrochemistryHardNumerical

JEE Main 2023 — Electrochemistry Question with Solution

JEE Main 2023 (13 Apr Shift 2)

Question

At 298 K, the standard reduction potential for Cu2+/Cu electrode is 0.34 V. Given : KspCu(OH)2=1×10-20 Take 2.303RTF=0.059 V The reduction potential at pH=14 for the above couple is (-)x×10-2 V. The value of x is

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Show full solutionCorrect answer: 25
Correct answer
25

Step-by-step explanation

For the reaction:
Cu2+aq+2e-→Cus

The reduction potential is

E=Eο-RT2Fln1Cu2+

From the given data pH = 14 and

Ksp=CuOH2=1.0×10-20, we get

H+=10-14M,OH-=KWH+=10-14M210-14M=1M

     Cu2+=KspOH-2=1.0×10-20M21M=1.0×10-20M

            E=0.34V-0.059V2log11.0×10-20=0.34V-0.059×20V2E=0.34-0.59 =-0.25 V

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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.