Consider the following cell reaction Cd ( s ) + Hg 2 SO 4 (   s ) + 9 5 H 2 O ( l ) ⇌ CdSO 4…

Consider the following cell reaction Cd(s)+Hg2SO4( s)+95H2O(l)⇌CdSO4·95H2O(s)+2Hg(l).
The value of Ecell0 is 4.315 V at 25°C. If ΔH°=-825.2 kJ mol-1, the standard entropy change ΔS° in J K-1 is _________ . (Nearest integer)
[Given : Faraday constant =96487 C mol-1]

Solution

For given cell reaction n=2
and ΔG°=-nFE°,cell ΔG°=ΔH°-TΔS
Then ΔH°=825.2×103 J/mole
T=298 K
E°cell=4.315 V
F=96487 C
ΔS°=--nFE°cell-ΔHT
ΔS°=--2×96487×4.315--825.2×103298=832.682×103-825.2×103298=7482298×103
ΔS°=25.1 J/k

Asked in: JEE Main 2021 (31 Aug Shift 1)

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