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?

  • 2
    $\begingroup$ You need to decide how to integrate both sides of $\displaystyle \Delta S=\left( \frac{\partial E}{\partial T}\right)_pzF$. $\endgroup$
    – porphyrin
    Apr 3 '21 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.


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