What does the graph of cell potential versus temperature look like? Would that look same as the graph between cell potential and temperature difference?

(Please correct me if I am missing some concept!)

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    – Poutnik
    Commented Oct 7, 2022 at 9:58
  • $\begingroup$ By temperature difference presumably you mean how potential changes with temperature (the slope) as suggested in the answer? $\endgroup$
    – Buck Thorn
    Commented Oct 7, 2022 at 11:52
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    – Poutnik
    Commented Oct 7, 2022 at 17:10

1 Answer 1


Let assume the classical representative of cells – the Daniell's cell:


\begin{align} E_\mathrm{cell} &= \Delta E^{\circ} + \frac{RT}{nF} \ln{\frac{a(\ce{Cu^2+})}{a(\ce{Zn^2+})}}\\ &= - \frac{\Delta G^{\circ}_\mathrm{r}}{nF} + \frac{RT}{nF} \ln{\frac{a(\ce{Cu^2+})}{a(\ce{Zn^2+})}}\\ &= \frac{1}{nF}\left( - \Delta H^{\circ}_\mathrm{r} + T \Delta S^{\circ}_\mathrm{r} + RT \ln{\frac{a(\ce{Cu^2+})}{a(\ce{Zn^2+})}}\right)\\ &= \frac{1}{nF}\left( - \Delta H^{\circ}_\mathrm{r} + T\left( \Delta S^{\circ}_\mathrm{r} + R \ln{\frac{a(\ce{Cu^2+})}{a(\ce{Zn^2+})}}\right)\right) \end{align}

$$\frac{\mathrm{d}E_\mathrm{cell}}{\mathrm{d}T}=\frac{1}{nF}\left( \Delta S^{\circ}_\mathrm{r} + R \ln{\frac{a(\ce{Cu^2+})}{a(\ce{Zn^2+})}}\right)$$

Equations assume the electrode local equilibrium states and for simplicity the same reactant and product heat capacities, otherwise calculations would be complicated by temperature trends of the reaction enthalpy and entropy.


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