1.

Calculate the equilibrium constant K for the reaction at 298 K Zn(s) + Cu^(2+)(aq)

Answer»

Solution :`DeltaG^(@) =-nFE^(@)`
`n=2, F = 96487 C MOL^(-1), E_("cell")^(@) = 1.1 V`
Therefore, `DeltaG^(@) =-2 xx 96487 xx 1.1 =-21227 J mol^(-1)`
Equilibrium constant K can be OBTAINED from the relation
`DeltaG^(@) =-RT ln K`
or `-21227 =-8.314 xx 298 xx ln K`
or ln `K = 21227/(8.314 xx 298) = 8.5677`
or `K = 3.696 xx 10^(8)`


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