Chemistry - Electrochemistry Question with Solution | TestHub

${\mathrm{E}}_{\mathrm{cell}}^{\circ }$ is dependent on the equilibrium constant, $\mathrm{K}$, which is not dependent on the pressure of any gas involved in the cell. Relation between Electrochemical Cell Potential and Gibbs free Energy change is given by; $\mathrm{\Delta } \mathrm{G} = -\mathrm{nFE}$ where Delta G is Gibbs Free energy change , n is the number of electrons involved in the reaction and F is Faraday and E is the Cell potential.

We know that,

$\mathrm{dG}=\mathrm{VdP}-\mathrm{SdT}$

At constant pressure $\mathrm{dP}=0$

$\Rightarrow \mathrm{dG}=-\mathrm{SdT}$

$$\begin{gathered} \mathrm{S}=-\frac{\mathrm{dG}}{\mathrm{dT}} \\ ∆\mathrm{S}=-{\left(\frac{∆\left(\mathrm{dG}\right)}{\mathrm{dT}}\right)}_{\mathrm{P}} \end{gathered}$$

$\Rightarrow \frac{{\mathrm{dE}}_{\mathrm{cell}}^{°}}{\mathrm{dT}}=\frac{∆\mathrm{S}}{\mathrm{nF}}$

A cell is operating reversibly if the cell potential is exactly balanced by an opposing source of potential difference.