JEE Main 2019 — Electrochemistry Question with Solution
From: JEE Main 2019 (Online) 12th January Morning Slot
Question
The standard electrode potential and its temperature coefficient for a cell are 2V and 5 104 VK1 at 300 K respectively.
The cell reaction is
Zn(s) + Cu2+ (aq) Zn2+ (aq) + Cu(s)
The standard reaction enthalpy (rH) at 300 K in kJ mol–1 is, [Use R = 8 JK–1 mol–1 and F = 96,000C mol–1]
The cell reaction is
Zn(s) + Cu2+ (aq) Zn2+ (aq) + Cu(s)
The standard reaction enthalpy (rH) at 300 K in kJ mol–1 is, [Use R = 8 JK–1 mol–1 and F = 96,000C mol–1]
Choose an option
Show full solutionCorrect option: A
Correct answer
A 412.8
Step-by-step explanation
G = -nFEcell = -2965002 = -386 kJ
S = nF = 296500 ( 5 104) = -96.5 kJ
At 298 K
TS = 298 (–96.5 J) = – 28.8 kJ
at constant T (=248 K) and pressure
G = H – TS
H = G + TS
= -386 - 28.8 = -412.8 kJ
S = nF = 296500 ( 5 104) = -96.5 kJ
At 298 K
TS = 298 (–96.5 J) = – 28.8 kJ
at constant T (=248 K) and pressure
G = H – TS
H = G + TS
= -386 - 28.8 = -412.8 kJ
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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.