JEE Main 2019 — Electrochemistry Question with Solution
From: JEE Main 2019 (Online) 9th April Morning Slot
Question
The standard Gibbs energy for the given cell
reaction in kJ mol–1 at 298 K is :
Zn(s) + Cu2+ (aq) Zn2+ (aq) + Cu (s),
E° = 2 V at 298 K
(Faraday's constant, F = 96000 C mol–1)
Zn(s) + Cu2+ (aq) Zn2+ (aq) + Cu (s),
E° = 2 V at 298 K
(Faraday's constant, F = 96000 C mol–1)
Choose an option
Show full solutionCorrect option: C
Correct answer
C–384
Step-by-step explanation
Here Zn is losing two electrons and Cu is gaining two electrons. So only two electrons are involved in the reaction.
n = 2
Go = - nFEo
= -2 96000 2
= -384 kJ/mol
n = 2
Go = - nFEo
= -2 96000 2
= -384 kJ/mol
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