For the cell reaction
`Cu^(2+)(C_(1),aq.)+Zn(s)hArrZn^(2+)(C_(2),aq)+Cu(s)`
of an electrochemical cell, the change in free energy `(DeltaG)` of a given temperature is a function of

Options

A. `lnC_(1)`
B. `lnC_(2)//C_(1)`
C. `ln (C_(1) + C_(2))`
D. `ln C_(2)`

Answer

B

Explanation

`E_("cell") = E_("cell")^(@) +(RT)/(nF)"ln"([Cu^(2+)])/([Zn^(2+)])`
`= E_("cell")^(@) +(RT)/(nF)"ln"(C_(1))/(C_(2))`
Also `DeltaG = -nEF`
Thus `DeltaG` is function of `"ln"(C_(2))/(C_(1))`.