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JEE Main 202131 AugustMorning Shift
Exam
JEE Main
Year
2021
Date
31 August
Shift
Morning Shift
Subject
Chemistry
Chapter
Electrochemistry

JEE Main 2021 · 31 August · Morning Shift · Chemistry · Electrochemistry

Consider the following cell reaction Cd _ ( s ) + Hg _ 2 SO _ 4 ( s...

Consider the following cell reaction Cd(s)+Hg2SO4(s)+95H2O(l)⇌CdSO4⋅95H2O(s)+2Hg(l).{Cd}_{( s )} + {Hg}_{2}{SO}_{4 ( s )} + \dfrac{9}{5}{H}_{2}{O}_{( l )} ⇌ {CdSO}_{4} \cdot \dfrac{9}{5}{H}_{2}{O}_{( s )} + 2{Hg}_{( l )} .
The value of Ecell0{E}_{cell}^{0} is 4.315V4 . 315V at 25°C.25 ° C . If ΔH°=−825.2kJmol−1,\Delta H ° = - 825 . 2kJ{mol}^{- 1}, the standard entropy change ΔS°\Delta S ° in JK−1J{K}^{- 1} is _________ . (Nearest integer)
[Given : Faraday constant =96487Cmol−1= 96487C{mol}^{- 1}]

Correct answer

0

Step-by-step solution

For given cell reaction n=2n = 2
and ΔG°=−nFE°,cellΔG°=ΔH°−TΔS\Delta G ° = - nFE ° , cell \Delta G ° = \Delta H ° - T \Delta S
Then ΔH°=825.2×103J/mole\Delta H ° = 825 . 2 \times {10}^{3}J / mole
T=298KT = 298K
E°cell=4.315VE{°}_{cell} = 4 . 315V
F=96487CF = 96487C
ΔS°=−(−nFE°cell−ΔH)T\Delta S ° = \dfrac{- \left(- nFE{°}_{cell} - \Delta H\right)}{T}
ΔS°=−(−2×96487×4.315−(−825.2×103))298=832.682×103−825.2×103298=7482298×103\Delta S ° = \dfrac{- \left(- 2 \times 96487 \times 4 . 315 - \left(- 825 . 2 \times {10}^{3}\right)\right)}{298} = \dfrac{832 . 682 \times {10}^{3} - 825 . 2 \times {10}^{3}}{298} = \dfrac{7482}{298} \times {10}^{3}
ΔS°=25.1J/k\Delta S ° = 25 . 1J / k
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