A recent United States National Chemistry Olympiad question asked:

A certain voltaic cell has a standard cell potential that increases with increasing temperature. Which best explains this observation?

A) ΔH° rxn>0

B) ΔS° rxn>0

C) ΔG° rxn>0

D) E°>0

The answer is B, but I can't quite figure out why the answer is B and not A.

I know that Le Chatelier's principle usually works in electrochemical cells. It's what makes the Nernst equation work: in a voltaic cell, if the equilibrium shifts left, the reaction will shift right to compensate, and the voltage will increase (and if the equilibrium shifts right the voltage will decrease). However, I'm a bit confused on whether this applies to heat (enthalpy) being considered as a reactant/product. If the temperature of a voltaic cell with an endothermic reaction is increased, will the voltage increase to compensate? The answer to the above problem would suggest it wouldn't. But let's make it more complicated: what if the same reaction also happens to have a negative change in entropy? The Gibbs free energy of the reaction would become less negative in accordance with ΔG=ΔH-TΔS, and ΔG=-nFE would suggest that the potential would decrease, not increase. This enthalpy/entropy conflict has been asked about before on CSE, but I think things are different in the context of electrochemistry. I've heard that if Gibbs free energy of a reaction with a positive change in entropy becomes more negative when temperature increases as a result of ΔG=ΔH-TΔS, the increased free energy is balanced out by ΔG=-RTln(K), since you would need to divide ΔG by temperature when calculating the equilibrium constant (so if you wanted to significantly increase the equilibrium constant by increasing the temperature, it wouldn't work since the two effects would offset each other to some degree). As a result, enthalpy rather than entropy determines how equilibrium shifts when temperature changes, in accordance with Le Chatelier's principle. But here's the thing: ΔG=-RTln(K) doesn't need to be used when considering electrochemical cells; you can go straight from ΔG=ΔH-TΔS to ΔG=-nFE. That means you don't have to divide by temperature. So do things work differently for electrochemical reactions vs. non-electrochemical ones? Does entropy rather than enthalpy determine how potential changes with temperature? Assume enthalpy and entropy change negligibly with temperature.

Can this be explained without complicated physical chemistry?

  • 1
    $\begingroup$ ΔG=ΔH-TΔS - exactly this. $\endgroup$
    – Mithoron
    Commented Apr 6 at 22:21
  • $\begingroup$ But for non-electrochemical reactions at least, I know you can't use that equation alone to predict shifts in equlibrium. $\endgroup$
    – unstable
    Commented Apr 7 at 2:44

1 Answer 1


The equation $\Delta G_r=-nFE$ helps as does $\displaystyle \Delta S_r=\left(\frac{\partial E}{\partial T}\right)_pnF$ where $\Delta G_r$ is the reaction free energy. The slope $\displaystyle \frac{\partial E}{\partial T}$ is positive as $E$ increases with temperature and so $\Delta S_r$ increases.

$E$ increases with temperature and so $\Delta G_r$ becomes more negative with temperature increase. As $\Delta G_r=\Delta H_r-T\Delta S_r$ if $\Delta H_r \gt 0$ increases $\Delta G_r$ becomes more positive, i.e. less negative, whereas $\Delta S_r$ is positive so $\Delta G_r$ becomes more negative i.e. less positive.

( The equation for the entropy change starts with $dG=Vdp-SdT$ at constant pressure $dp=0$ and so $ dG=-SdT$ or $\displaystyle S=-\left(\frac{\partial G}{\partial T}\right)p$ then substitute $G$ with the potential $E$.)

  • $\begingroup$ I'm not sure I quite understand this. Does entropy "win" over enthalpy when determining how the potential changes? $\endgroup$
    – unstable
    Commented Apr 12 at 23:36
  • $\begingroup$ Not always, the options in your question define what happens. Each case is its own, in some reactions it might be enthalpy in others entropy is most important $\endgroup$
    – porphyrin
    Commented Apr 13 at 10:38

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