JEE Main 2017ChemistryElectrochemistryMediumMCQ

JEE Main 2017 — Electrochemistry Question with Solution

JEE Main 2017 (09 Apr Online)

Question

To find the standard potential of M3+/M electrode, the following cell is constituted: Pt/M/M3+ 0.001 mol L-1/Ag+ 0.01 mol L-1/Ag

The emf of the cell is found to be 0.421 volt at 298 K. The standard potential of half-reaction M3++3e-→M at 298 K will be:

(Given: EAg+Ag⊝ at 298 K=0.80 volt)

Choose an option

Show full solutionCorrect option: C
Correct answer
C0.32 volt

Step-by-step explanation

Ecell=Eocell−0.0591nlogQ

∴ Q=M3+Ag+3 =0.001(0.01)3=103

Ecell=Eocell−0.05913log103

Ecell=Eocell−0.05913×3=0.421Eocell−0.0591=0.421Eocell=0.48 V =EoAg+/Ag-EoM3+/MEoM3+/M=0.80-0.48 = 0.32 V

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About this question

This is a previous-year question from JEE Main 2017, 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.