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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2$\begingroup$ Welcome to CH SE site! Note that using photos/screenshots of text instead of typing text itself is highly discouraged. The image text content cannot be indexed nor searched for, nor can be reused in answers. Specifically handwritten scripts can be difficult to decipher. Consider copy/pasting or rewriting of at least essential parts. Suitable formatting can be done according to formatting math/chem expressions/equations. $\endgroup$– PoutnikOct 7, 2022 at 9:58
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$\begingroup$ By temperature difference presumably you mean how potential changes with temperature (the slope) as suggested in the answer? $\endgroup$– Buck Thorn ♦Oct 7, 2022 at 11:52
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1$\begingroup$ Note that having just the photo is still better than not having even just the photo. $\endgroup$– PoutnikOct 7, 2022 at 17:10
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1 Answer
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Let assume the classical representative of cells – the Daniell's cell:
$$\ce{Zn(s)}|\ce{ZnSO4(aq)}||\ce{CuSO4(aq)}|\ce{Cu(s)}$$
\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.
