# Derivation and visualization of cell potential dependency on temperature and entropy

Working from the thermodynamic identity

$$\Delta{S} = -\left(\frac{\partial{\Delta{G}}}{\partial{T}}\right)_p,$$

is it allowed to integrate both sides to obtain an expression derived from the Nernst equation

$$E_\mathrm{cell} = \frac{T\,\Delta S}{zF}?$$

Can one plot the results and obtain $$\Delta S$$ from the gradient, assuming $$\Delta S$$ is constant for the reaction and $$E_\mathrm{cell}$$ values at some temperatures are given?

• You need to decide how to integrate both sides of $\displaystyle \Delta S=\left( \frac{\partial E}{\partial T}\right)_pzF$. – porphyrin Apr 3 at 11:52

What you are suggesting in your equation is that $$\Delta G_m=-T\Delta S_m$$ but you know that in fact $$\Delta G_m=\Delta H_m-T\Delta S_m$$. Your math is incorrect because you are missing a constant of integration.
The expression $$\Delta G_m=\Delta H_m-T\Delta S_m$$ can be rewritten using the Nernst equation as
$$-nFE = \Delta H_m -nFT\left(\frac{\partial E}{\partial T} \right)_p$$
by using the definition you provide for $$\Delta S_m$$ as the temperature-derivative of $$\Delta G_m$$, and the Nernst equation $$\Delta G_m = -nFE$$. In fact
$$\Delta S_m = nF\left(\frac{\partial E}{\partial T} \right)_p$$
Therefore if you plot $$nFE$$ as a function of T the slope of the curve at a given temperature will provide the value of $$\Delta S_m$$ at that T.