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JEE Main 201912 JanuaryMorning Shift
Exam
JEE Main
Year
2019
Date
12 January
Shift
Morning Shift
Subject
Chemistry
Chapter
Electrochemistry

JEE Main 2019 · 12 January · Morning Shift · Chemistry · Electrochemistry

The standard electrode potential E ^ o and its temperature...

The standard electrode potential Eoanditstemperaturecoefficient(dEdT){E}^{o} and its temperature coefficient \left(\dfrac{dE}{dT}\right) for a cell are 2Vand−5×10−4VK−12V and - 5{\times 10}^{- 4}V{K}^{- 1} at 300K300K, respectively. The reaction is Zn(s)Zn\left(s\right) + {Cu}^{2 +}\left(aq\right)→Zn2+(aq)\to {Zn}^{2 +}\left(aq\right) + Cu\left(s\right).Thestandardreactionenthalpy(ΔrH−). The standard reaction enthalpy \left({\Delta}_{r}{H}^{-}\right) at 300Kinmol−1300K in mo{l}^{- 1} is

[[Use R=8JK−1mol−1andF=96,500Cmol−1]R = 8J{K}^{- 1}{mol}^{- 1} and F =96,500 C{mol}^{- 1} ]

Options

  1. A−412.8- 412.8
  2. B206.4206.4
  3. C−384.0- 384.0
  4. D192.0192.0

Correct answer

A. −412.8- 412.8

Step-by-step solution

ΔG=−nFEcell=−2×96500×2=−386kJ\Delta\text{G} = - {nFE}_{cell} = - 2 \times 96500 \times 2 = - 386kJ
ΔS=nFdEdT=2×96500×−5×10−4Jdegreecelcious−1\Delta\text{S} = nF\dfrac{dE}{dT} = 2 \times 96500 \times - 5 \times {10}^{- 4}Jdegree{celcious}^{- 1}
=−96.5J= - 96 .5 J

At 298K298K,

TΔS=298×(−96.5J)=−28.757kJT \Delta\text{S} = 298 \times \left(- 96 .5 J\right) = - 28 .757 kJ
At constant temperature \left(T = 248K\right)andpressure,<br/><br/>and pressure,<br /><br />\Delta G = \Delta H - T \Delta\text{S}<br/><br/><br /><br />\Delta H = \Delta G + T \Delta\text{S}<br/><br/><br /><br />= - 386 - 28.757 = -414.75 ≃ -412.8
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