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Question

Calculate the equilibrium constant and free energy change of given following reaction for Daniell cell at 298 K temperature. Zn(s)+Cu2+(aq)Zn2+(aq)+Cu(s) Cell potential =1.1 volt (F=96500 coulomb)

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Solution

Ecell=E0cellRTnFlnQ
At equilibrium, Q=K and Ecell=0
So, E0cell=RTnFlnK

E0cell=2.303RTnFlogK
In the given reaction, exchange of two electrons is taking place. Therefore, n=2.
1.1=2.303×8.314×2982×96500logK
logK=37.207
K=1.62×1037
Gibb's free energy change ΔG=nFE=2×96500×1.1
ΔG=212,300 J=212.3 kJ

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