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Question

Calculate the useful work done during the reaction in kJ/mol.
Ag(S)+12 Cl2(g)AgCl(s)
Given that E0Cl2/Cl=+1.36V
and E0Ag+|AgCl|Cl=+0.220V
PCl2=1 atm and T = 298 K

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Solution

For the cell reaction:
Ag(s)+12 Cl2(g)AgCl(s)E0=1.360.22=1.14 V
or, E=E00.0591log 1[Cl2]12
In standard conditions, PCl2=1 atm
log [Cl2]12=0
Useful work=Wmax=G=nEF=(1)×(1.14)×96500×103=110 kJ/mol


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