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I have a galvanic cell composed of a cathode with 1M copper sulphate solution, and an anode with 0.001M zinc sulphate solution. I am increasing the temperature and recording the resultant effect on the EMF. According to the Nernst equation, EMF should increase linearly with temperature in this scenario:

EMF = 0.0003T + 1.1 (I have rearranged the equation)

I don't know why this happens. I thought it was because of Ohm's law, which states voltage is directly proportional to resistance, and hence, by increasing the resistance of the galvanic cell by increasing temperature, the EMF also increased. But I read on other websites that increases in temperature can decreases the EMF too. So I am quite confused and don't know whether I made a mistake with the Nernst equation.

Is anyone able to explain why EMF, in this situation, increases alongside temperature? It would be lovely if there was some theory explaining why (like Ohm's law or something similar). Sorry if this question isn't formatted properly, this is my first one.

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    $\begingroup$ There is the explicit dependence in RT/nF .ln(a) terms and there is implicit T dependence of E°. And there is possible temperature dependent kinetic factor of deviation from expectation. $\endgroup$
    – Poutnik
    May 7, 2023 at 9:24
  • $\begingroup$ Exactly. RT/nF ln(a) should scale with temperature, but so does E0 aswell. Difficult to predict $\endgroup$
    – Mäßige
    May 7, 2023 at 9:29
  • $\begingroup$ How did you rearrange the equation to get EMF = 0.0003 T + 1.1 ? $\endgroup$
    – Maurice
    May 7, 2023 at 13:11
  • $\begingroup$ EMF is proportional to Gibbs free energy, whose partial derivative with respect to temperature is an entropy term. $\endgroup$
    – Karsten
    May 8, 2023 at 3:51
  • $\begingroup$ @Karsten That assumes dH/dT = dS/dT = 0 $\endgroup$
    – Poutnik
    May 9, 2023 at 4:06

1 Answer 1

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According to the Nernst equation, EMF should increase linearly with temperature in this scenario.

The statement and the equation you have written are fine. Considering the comments, I will try to extend the equation taking into account the change of the standard cell potential with temperature, in order to see if it has an impact or not. After this I will address your questions.

The system is the Daniel's cell, where the half-reactions and the global reaction, are \begin{align} \ce{Cu^2+ (aq)} + 2e^- &\rightarrow \ce{Cu(s)} \\ \ce{Zn^2+ (aq)} + 2e^- &\rightarrow \ce{Zn(s)} \\ \ce{Cu^2+ (aq)} + \ce{Zn(s)} &\rightarrow \ce{Cu(s)} + \ce{Zn^2+ (aq)} \tag{1} \end{align}

Eq. (1), written this way, is spontaneous.

As temperature changes, the standard Gibbs energy of reaction will be modified. We find its dependence with temperature integrating the Gibbs-Helmholtz equation, at constant pressure and concentration of species, and we put in parenthesis whether a property is evaluated in $T$ or $T^\Theta$ \begin{align} \frac{d(\Delta_r G^\Theta/T)}{dT} &= -\frac{\Delta_r H^\Theta}{T^2} \\ \frac{d(\Delta_r G^\Theta/T)}{dT} &= -\frac{\Delta_r H^\Theta(T^\Theta) + \Delta_r C_p(T - T^\Theta)}{T^2} \\ \int_{\Delta_r G^\Theta(T^\Theta)/T^\Theta}^{\Delta_r G^\Theta(T)/T} d(\Delta_r G^\Theta/T) &= \Delta_r H^\Theta(T^\Theta) \int_{T^\Theta}^T -\frac{dT}{T^2} - \Delta_r C_P \int_{T^\Theta}^T \left(\frac{1}{T} - \frac{T^\Theta}{T^2}\right) \; dT \\ \frac{\Delta_r G(T)}{T} - \frac{\Delta_r G^\Theta(T^\Theta)}{T^\Theta} &= \Delta_r H^\Theta(T^\Theta) \left(\frac{1}{T} - \frac{1}{T^\Theta}\right) \\ &-\Delta_r C_P \left[\ln\left(\frac{T}{T^\Theta}\right) + \left(\frac{T^\Theta}{T} - 1\right)\right] \\ \frac{\Delta_r G(T)}{T} &= \frac{\Delta_r G^\Theta(T^\Theta)}{T^\Theta} + \Delta_r H^\Theta(T^\Theta)\left(\frac{1}{T} - \frac{1}{T^\Theta}\right) \\ &- \Delta_r C_P \left[\ln\left(\frac{T}{T^\Theta}\right) + \left(\frac{T^\Theta}{T} - 1\right)\right] \\ \Delta_r G(T) &= \left(\frac{T}{T^\Theta}\right)\Delta_r G^\Theta(T^\Theta) + \left(1 - \frac{T}{T^\Theta}\right) \Delta_r H^\Theta(T^\Theta) \\ &- \Delta_r C_P \left[T\ln\left(\frac{T}{T^\Theta}\right) - (T - T^\Theta)\right] \hspace{3 cm} (2) \\ \end{align} The relation between the standard cell potential and standard Gibbs energy of reaction, in this case, is $ \Delta E^\Theta(T) = -\Delta_r G^\Theta(T) / 2F $. Dividing Eq. (2) on both sides by $-2F$ we get \begin{align} \Delta E^\Theta(T) &= \left(\frac{T}{T^\Theta}\right)\Delta E^\Theta(T^\Theta) - \left(1 - \frac{T}{T^\Theta}\right) \frac{\Delta_r H^\Theta(T^\Theta)}{2F} \\ &+ \frac{\Delta_r C_P}{2F} \left[T\ln\frac{T}{T^\Theta} - (T - T^\Theta)\right] \hspace{3 cm} (3) \\ \end{align} Eq. (3) is very general. The only restriction is that we assume the specific heat capacites independent of temperature. The following data (see references) is useful for evaluating Eq. (3): \begin{array}{c| c c c c} \text{Substance/Property} & \ce{Cu^2+ (aq)} & \ce{Zn (s)} & \ce{Cu (s)} & \ce{Zn^2+ (aq)} \\ \hline \Delta_f H^\Theta \; \pu{(kJ/mol)} & 64.77 & 0 & 0 & -153.89 \\ \Delta_f G^\Theta \; \pu{(kJ/mol)} & 65.49 & 0 & 0 & -147.06 \\ C_P \; \pu{(J/mol·K)} & - & 25.40 & 24.44 & - \\ \end{array} Thereby \begin{align} \Delta_r H^\Theta(T^\Theta) &= \sum_j \nu_j \Delta_f H^\Theta(T^\Theta) = -218660 \; \pu{\frac{J}{mol}} \\ \Delta_r G^\Theta(T^\Theta) &= \sum_j \nu_j \Delta_f H^\Theta(T^\Theta) = -212550 \; \pu{\frac{J}{mol}} \\ \Delta_r C_P &= \sum_j \nu_j C_{P,j}(T^\Theta) = -0.96 \; \pu{\frac{J}{mol·K}} \end{align}

Now we can evaluate Eqs. (2) and (3). The results are as follows:

enter image description here

As the reaction is exothermic, higher the temperature the less spontaneous it will be. As the Gibbs energy goes up, less will be the energy that it can be converted to electrical work to the surroundings. This is reflected in the decrease of the standard cell potential. However, to be honest, the changes are not that important considering this temperature range. As a conclusion, at least for the Daniell's cell, the change of $\Delta E^\Theta$ with temperature is not a determining factor.

Back to the question The Nernst equation is \begin{align} \Delta E(T) = \Delta E^\Theta(T) - \frac{RT}{2F} \ln\left(\frac{\ce{[Zn^2+]}}{\ce{[Cu^2+]}}\right) \tag{4} \end{align} At fixed concentrations, Eq. (4) is just a function of temperature, but now both members are a function of it. Lets consider your equation in the post (in red color), and the Nernst equation taking into account the change of $\Delta E^\Theta$ with temperature (in blue color):

enter image description here

Note also that the slope of your equation will overestimate the performance of the cell.

For your case, you have an excess of $\ce{Cu^2+}$ with respect to $\ce{Zn^2+}$. While increasing the temperature, you increase the electrochemical potential of both $\ce{Cu^2+}$ and $\ce{Zn^2+}$, remembering that $$ \tilde{\mu}_j = \mu_{j}^\Theta + RT\ln(C_j/C_j^\Theta) + z_jF\phi^S \tag{5} $$ where $\phi^S$ is the potentail of the liquid phase where the ions are. This translates into a higher tendency of $\ce{Cu^2+}$ of being reduced, and also of $\ce{Zn^2+}$ of being reduced. However, with the concentration difference, copper ions get the upperhand. This tendency translates into a higher potential difference between the terminals in the cell, so that the performance is increased at higher temperatures. In more simpler terms, you are making the job "easier" for $\ce{Cu^2+}$ to release two electrons in the electrolyte/solid interface by giving thermal energy. However, you need to give that energy from the surroundings.

I thought it was because of Ohm's law, which states voltage is directly proportional to resistance, and hence, by increasing the resistance of the galvanic cell by increasing temperature, the EMF also increased.

When considering the Nernst equation, the current is zero, because we are studying the electrochemical equilibrium. This is approximated when you measure the voltage with an "infinite" resistance in the voltmeter. If this wasn't the case, you would see the reaction take place, and the voltmeter reading changing over time. The resistance is always "infinite" in all your readings.

References

Atkins, P. and De Paula, J., "Physical Chemistry", 9th ed, W.H. Freeman and Company New York (2016).

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  • $\begingroup$ This is hands down an excellent answer. There are just a couple of things I'd correct with respect to formatting. First, we have mhchem for chemical expressions enabled (not just symbols of elements). Second, the standard state is never denoted with the capital letter Theta. Instead, it's either plimsoll (^⦵, Unicode, works with XeLaTeX) or a circle (^\circ, works for pdfLaTeX). Third, textual labels and subscripts are usually upright. And my personal recommendation is to use $p$ for pressure since $P$ is usually reserved for power in PC. $\endgroup$
    – andselisk
    May 9, 2023 at 17:53
  • $\begingroup$ @andselisk Hello andselisk. Thank you! I will make the changes afterwards, so it doesn't bother incoming questions in the feed, when night comes in South America. I didn't quite get the comment about mhchem, where should I employ it? $\endgroup$ May 10, 2023 at 12:22
  • $\begingroup$ For example, in place of \ce{Cu^2+ (aq)} + 2e^- &\rightarrow \ce{Cu(s)} just use the \ce{…} macro once: $\ce{Cu^2+(aq) + 2 e^- -> Cu(s)}$. $\endgroup$
    – andselisk
    May 10, 2023 at 12:58

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