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JEE Main 20199 AprilMorning Shift
Exam
JEE Main
Year
2019
Date
9 April
Shift
Morning Shift
Subject
Chemistry
Chapter
Electrochemistry

JEE Main 2019 · 9 April · Morning Shift · Chemistry · Electrochemistry

The standard Gibbs energy for the given cell reaction in kJmo l ^ -...

The standard Gibbs energy for the given cell reaction in kJmol−1kJmo{l}^{- 1} at 298K298K is:
Zn(s)+Cu2+(aq)⟶Zn2+(aq)+Cu(s)Zn\left(s\right) + C{u}^{2 +}\left(aq\right) ⟶ Z{n}^{2 +}\left(aq\right) + Cu ( s ) ,
E0=2Vat298K{E}^{0} = 2Vat298K
(Faraday′sconstant,F=96000Cmol−1)( Faraday'sconstant, F = 96000Cmo{l}^{- 1} )

Options

  1. A−192- 192
  2. B192192
  3. C384384
  4. D−384- 384

Correct answer

D. −384- 384

Step-by-step solution

For the reaction,
Cu2+(aq)+Zn(s)⟶Zn2+(aq)+Cu(s)C{u}^{2 +}\left(aq\right) + Zn\left(s\right) ⟶ Z{n}^{2 +}\left(aq\right) + Cu\left(s\right)
n=2n = 2, two electrons are exchanged
∆Go=−nFEo∆ {G}^{o} = - nF{E}^{o}
=−2×96000×2=−384000= - 2 \times 96000 \times 2 = - 384000
=−384kJ= - 384kJ
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