For the cell reaction, $\mathrm{Cu}^{2+}\left(\mathrm{C}_{1},…

For the cell reaction, $\mathrm{Cu}^{2+}\left(\mathrm{C}_{1}, \mathrm{aq}ight)+\mathrm{Zn}(\mathrm{s})=\mathrm{Zn}^{2+}\left(\mathrm{C}_{2}, \mathrm{aq}ight)+\mathrm{Cu}(\mathrm{s})$ of an
electrochemical cell, the change in free energy, $\Delta \mathrm{G}$, at a given temperature is a function of
  1. $\ln \left(\mathrm{C}_{1}ight)$
  2. $\ln \left(\mathrm{C}_{2} / \mathrm{C}_{1}ight)$
  3. $\ln \left(\mathrm{C}_{2}ight)$
  4. $\ln \left(\mathrm{C}_{1}+\mathrm{C}_{2}ight)$

Solution

$\Delta \mathrm{G}=-\mathrm{nE}^{\circ} \mathrm{F}$
For concentration cell, $\mathrm{E}=\frac{\mathrm{RT}}{\mathrm{nF}} \ln \frac{\mathrm{C}_{2}}{\mathrm{C}_{1}}$
In it $\mathrm{R}, \mathrm{T}, \mathrm{n}$ and $\mathrm{F}$ are constant
So $\mathrm{E}$ is based upon $\ln \mathrm{C}_{2} / \mathrm{C}_{1}$
Now $\Delta \mathrm{G}=-\mathrm{nEF}=-\mathrm{nF} \times \frac{\mathrm{RT}}{\mathrm{nF}} \ln \mathrm{C}_{2} / \mathrm{C}_{1}$
$=-\mathrm{RTln} \mathrm{C}_{2} / \mathrm{C}_{1}$
At constant temperature $\Delta \mathrm{G}$ is based upon $\ln \left(\mathrm{C}_{2} / \mathrm{C}_{1}ight)$.

Asked in: JEE-TOPICTESTS-CHEMISTRY

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