I'm doing an experiment where I'm investigating the effect of temperature of the galvanic cell on the cell potential. But during the experiment, I noticed that the two half cells were at different temperatures. Could this be a limitation to the data that I collected? If so, how would it impact the cell potential that was measured?
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$\begingroup$ It is important you include the compact summary of your current knowledge, involving also conclusions of search for existing related or duplicate info or answers. It would prevent others to tell you what you already know or what you can easily find yourself. It also serves as evidence of your mandatory effort. Effort not shown can be considered as effort not done. $\endgroup$– PoutnikFeb 22, 2022 at 9:12
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$\begingroup$ The standard electrode potentials are temperature dependent, because the standard reaction Gibbs energy is temperature dependent too. $\endgroup$– PoutnikFeb 22, 2022 at 14:58
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Good experimental design means ensuring that you have repeatable and well defined experimental conditions - by not adequately controlling the temperature of the two half-cells you've introduced the variable $\Delta T_\text{half-cell}$. Since you aren't varying $\Delta T_\text{half-cell}$ in a systematic way and don't know if it's constant, you now have an undefined aberration convoluted in with whatever effect you wanted to measure. Either ensure that $\Delta T_\text{half-cell}$ is small enough to ignore, change your experimental setup to eliminate the temperature difference and repeat your measurements, or preform a series of measurements where you alter $\Delta T_\text{half-cell}$ in a systematic way. (This latter option is a lot of work since you'd have to vary $\Delta T_\text{half-cell}$ at several different cell temperatures.)
With that out of the way, what effect would $\Delta T_\text{half-cell}$ have on the overall cell potential? Lets start with the Nernst equation for the half-cell reaction $\ce{pP +ne^- -> qQ}$, which is:
$$ E\ce{(P|Q)} = E°\ce{(P|Q)} - \frac{RT\ce{(P|Q)}}{nF}\ln{\frac{([\ce{Q}]/c°)^q}{([\ce{P}]/c°)^p}} $$
(from Chemical Structure an Reactivity 1e by Keeler and Worthers, p 842f). $E°$ is the cell potential at standard conditions, $R$ is the molar gas constant, $T\ce{(P|Q)}$ is the temperature of the $\ce{(P|Q)}$, $n$ is the number of electrons transferred, $F$ is the Faraday constant, and $[\ce{Q}]/c°$ etc. is the concentration of $\ce{Q}$ relative to the standard concentration.
For the other half-cell reaction $\ce{aA +ne^- -> bB}$ we can write:
$$ E\ce{(A|B)} = E°\ce{(A|B)} - \frac{RT\ce{(A|B)}}{nF}\ln{\frac{([\ce{B}]/c°)^b}{([\ce{A}]/c°)^a}} $$
Remembering that the overall cell potential is $E = E\ce{(P|Q)} - E\ce{(A|B)}$, we arrive at the following:
$$ E = E° - \frac{R}{nF}\left[ T\ce{(P|Q)}\ln{\frac{([\ce{Q}]/c°)^q}{([\ce{P}]/c°)^p}} - T\ce{(A|B)}\ln{\frac{([\ce{B}]/c°)^b}{([\ce{A}]/c°)^a}}\right] $$
So, is $\Delta T_\text{half-cell}$ an important factor? It depends on the magnitude of $\Delta T_\text{half-cell}$ relative to the temperature range studied, the precision with which you measure $E$, and whether it's a constant difference or changes with time and/or cell temperature. You'll have to figure this out on your own, but the equation above should give you some numbers to work with.
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$\begingroup$ Note that $E°(T)=E°(\pu{298.15 K})+f(T)$ as the standard reaction Gibbs energy for $\ce{ox(aq) + H2(g) -> red(aq) + 2 H+(aq)}$ is function of temperature as well. $\Delta G^{°}_\mathrm{r}(T)=\Delta G^{°}_\mathrm{r}(\pu{298.15 K})+g(T)$, and therefore $E^{°}(T)= \frac{\Delta G^{°}_\mathrm{r}(T)}{nF} $ $\endgroup$– PoutnikFeb 23, 2022 at 8:54
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$\begingroup$ Relating $E$ and $\Delta G(T)$ is important since it gives thermodynamic insight, but I think the Nernst equation alone is enough to quantify the aberration. $\endgroup$ Feb 23, 2022 at 11:40
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$\begingroup$ @The N.E. addresses E change with activity, with the rate T dependent, but it does not address E° change with T. By other words, using the unit activities of relevant components, considering just the N.E. the galvanic cell voltage would not depend on temperature. But that is not true, as $\Delta G = \Delta H - T\Delta S$. The different thing is, if we can or cannot afford to neglect it, what is case dependent. $\endgroup$– PoutnikFeb 23, 2022 at 11:49