JEE Main 2019 — Electrochemistry Question with Solution
From: JEE Main 2019 (Online) 10th January Evening Slot
Question
In the cell
PtH2(g, 1 bar)AgClAg(s)|Pt(s)
the cell potential is 0.92 V when a 10–6 molal HCl solution is used. The standard electrode potential of (AgCl/ AgCl– ) electrode is :
Given, at
PtH2(g, 1 bar)AgClAg(s)|Pt(s)
the cell potential is 0.92 V when a 10–6 molal HCl solution is used. The standard electrode potential of (AgCl/ AgCl– ) electrode is :
Given, at
Choose an option
Show full solutionCorrect option: D
Correct answer
D0.20 V
Step-by-step explanation
Anode : H2(g) 2H+(aq) + 2e-
Cathode : AgCl(s) + e- Ag(s) + Cl-(aq)
---------------------------------------------------------------
H2(g) + 2AgCl(s) 2Ag(s) + 2H+(aq) + 2Cl-(aq)
From Nernst equation we know,
Ecell = E0cell -
Here,
Ecell = E0cell -
0.92 = E0AgCl/AgCl - -
E0AgCl/AgCl = 0.92 - 0.72 = 0.2 V
Cathode : AgCl(s) + e- Ag(s) + Cl-(aq)
---------------------------------------------------------------
H2(g) + 2AgCl(s) 2Ag(s) + 2H+(aq) + 2Cl-(aq)
From Nernst equation we know,
Ecell = E0cell -
Here,
Ecell = E0cell -
0.92 = E0AgCl/AgCl - -
E0AgCl/AgCl = 0.92 - 0.72 = 0.2 V
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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.