Electrochemistry - Result Question 51
####53. For the cell reaction,
[1998]
$Cu^{2+}(C_1, aq)+Zn(s)=Zn^{2+}(C_2, aq)+Cu(s)$ of an electrochemical cell, the change in free energy, $\Delta G$, at a given temperature is a function of
(a) $\ln (C_1)$
(b) $\ln (C_2 / C_1)$
(c) $\ln (C_2)$
(d) $\ln (C_1+C_2)$
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Solution:
- (b) $\Delta G=-n E^{\circ} F$
For concentration cell, $E=\frac{R T}{n F} \ln \frac{C_2}{C_1}$
In it $R, T, n$ and $F$ are constant
So $E$ is based upon $\ln C_2 / C_1$
Now $\Delta G=-n E F=-n F \times \frac{R T}{n F} \ln C_2 / C_1$
$=-R T \ln C_2 / C_1$
At constant temperature $\Delta G$ is based upon $\ln C_2 / C_1$.
