I understand why changing the concentrations of the electrolytes in a voltaic cell affects the voltage. Given a zinc-copper cell for example, if I were to increase the concentration of the zinc sulphate electrolyte, the equilibrium between zinc ions and zinc metal would shift towards the side of the zinc metal. This would cause the reduction potential for zinc ions to increase, decreasing the overall cell potential. Please correct me if I’m wrong as this is how I understand it!

However, I don’t quite understand why temperature affects the voltage the way it does. According to the Nernst equation, if the reaction quotient is greater than one (meaning that the zinc sulphate concentration is greater than the copper sulphate concentration in this example), increasing temperature decreases voltage. Conversely, if the reaction quotient is less than one, increasing temperature increases voltage.

I am familiar with Le Chatlier's principle, but if I were to apply it to this scenario, given that the redox reaction taking place is exothermic, it would suggest that increasing temperature would always decrease the voltage produced since it would cause a shift towards the reactant side. This, of course, does not correspond with the relationship predicted by the Nernst equation.

I have looked all over the internet but I could not find any answer to this seemingly straightforward question. Any help is greatly appreciated!

  • 2
    $\begingroup$ Did you notice the temperature $T$ term in the Nernst equation? Try to derive the equation and what is the origin of $T$ there. Once you have done the mathematical derivation of the equation, one can think about temperature qualitatively. $\endgroup$
    – ACR
    Commented Feb 1, 2020 at 23:07
  • 4
    $\begingroup$ Your mistake is assuming that the equilibrium constant K and the voltage are linearly related, so that a shift in K towards reactants will reduce voltage. The potential is a function of TlnK, so if T increases more than lnK decreases, the potential will actually increase. $\endgroup$
    – Andrew
    Commented Feb 2, 2020 at 2:16

3 Answers 3


Both terms of the difference in the Nernst equation depend on the temperature.

$$E = E^\circ - \frac{R T}{z F} \ln Q$$

The second term obviously depends on temperature because it contains the temperature (and the other quantities are either constants or not temperature-dependent). The first term, however, is temperature-dependent as well. If you consider the relationship to the standard Gibbs energy of reaction, this becomes clear:

$$Δ_rG^\circ=−z F E^\circ $$

Rather than figure out the temperature-dependence by looking at the two terms, it is easier to forget the Nernst equation and directly relate the cell potential at non-standard conditions to the Gibbs energy of reaction under the same conditions:

$$Δ_rG=−z F E $$

At constant pressure, the partial derivative of the the Gibbs energy of reaction $Δ_rG$ with respect to the temperature is $-Δ_rS$. (Perhaps surprisingly, the temperature-dependence of the equilibrium constant depends on the sign of $Δ_rH$ instead, i.e. whether the reaction is endothermic or exothermic, see Van't Hoff equation).


You are correct in understanding that changing the concentration of electrolytes in a voltaic cell affects the voltage through the Nernst equation. As for the effect of temperature, it is important to remember that the Nernst equation takes into account the effect of temperature on the equilibrium constant (K) of the reaction, not just the effect of temperature on the reaction itself.

While an exothermic reaction would suggest that increasing temperature would shift the equilibrium towards the reactant side, it is not always the case. In fact, the direction of the shift depends on the overall enthalpy change (ΔH) of the reaction and the entropy change (ΔS). If the reaction is exothermic and the entropy change is positive, increasing temperature would actually shift the equilibrium towards the product side, resulting in an increase in voltage.

In the case of a zinc-copper cell, the Nernst equation predicts that increasing temperature would decrease the voltage if the zinc sulphate concentration is greater than the copper sulphate concentration (Q>1). This is because the reaction would be more product-favored at higher temperatures, and the excess of zinc ions would cause a decrease in the cell potential. On the other hand, if the copper sulphate concentration is greater than the zinc sulphate concentration (Q<1), increasing temperature would increase the voltage as the reaction would be more reactant-favored at higher temperatures.

Overall, the effect of temperature on the voltage of a voltaic cell depends on the specific reaction involved, and cannot be predicted solely based on Le Chatelier's principle.


Consider that even when exist a "driving force" to move ions to electrodes, it is necesary that particles stay not static, they need some capacity to move to change direction and take way to electrodes, and what gives this capacity to move, Energy, and in any particle system this energy distribution follows Maxwell-Bolzmann and this means that temperature counts.. e^-(1/kT)


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