0
$\begingroup$

I have been reading a Daniel Schroeder's book called "An introduction to Thermal physics", and after having completed first couple of paragraphs of "Free energy as available work" (chapter 5) I got this question.

I the book a $Pb$ based battery is considered with the discharge reaction that goes as follows:

$$Pb + PbO_2 + 4H^+ + 2SO_4^{2-} \rightarrow 2PbSO_4 + 2H_2O.$$

Then it is stated that, according to the tables, the change of the Gibbs free energy $\Delta G$ is -390 kJ/mol. This is, by definition of the Gibbs free energy, is the energy that we can count on as electrical work that the battery is capable of (in these ideal conditions). But, out of these 390 kJ/mol of "produced" energy 78 kJ/mol flows into the system as heat because the entropy of the final product is higher. In other words, we gain additional 78 kJ/mol of the battery's performance simply getting some heat from the environment.

In theory, if you make the entropy difference between the initial and the final products larger, then the performance of a battery can be improved. Do people work in this direction? For example, to go from solid to liquid as initial/final states. Because liquid state has more entropy, but the volume doesn't change that much (so we don't waste energy on expanding of the working substance). Also a small additional question. The heat we get from the environment is proportional to the temperature of the environment. Is that the reason why at warmer temperatures batteries perform better?

$\endgroup$
0
$\begingroup$

Here is the answer to one of your questions.

Batteries perform better when warm because the chemistry part of electrochemistry contains temperature dependence i.e., the reactions go faster at higher temperatures than at lower temperatures. In addition, for batteries with liquid electrolytes there comes a point where the electrolyte freezes, at which point the chemistry comes to a halt because the frozen electrolyte can no longer support the free transport of ions.

$\endgroup$
1
  • $\begingroup$ You may want to clarify that only endothermic reactions will speed up with increase in temperature. $\endgroup$
    – Sam
    Jul 3 '20 at 11:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.