I am using advanced lead-carbon batteries (these ones) in an off-grid solar system. The spec specifies equalizing voltage 14.1 V for 12 V block @ 25 °C (77 °F). I keep the batteries in an outdoor shed, and the temperature where I live tends to hang around 0 °C (32 °F) these days.

So, for lead acid chemistry, would the equalizing voltage go up or down with temperature going down? I feel like it would go up (more voltage needed to do the same job at lower temperatures). How does this reasoning stand? What are the underlying scientific considerations?

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    $\begingroup$ If the batteries are commercially available, I would ask the manufacturer. They surely have more information than what is in the specs. $\endgroup$ – Karsten Theis Jun 2 '19 at 10:15
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    $\begingroup$ @KarstenTheis I am interested in the underlying scientific considerations too, not just in the number to configure my solar system. $\endgroup$ – Greendrake Jun 2 '19 at 23:16
  • $\begingroup$ I cannot confirm the accuracy of this entry in the wikipedia but: "There are no known, independently verified ways to reverse sulfation. There are commercial products claiming to achieve desulfation through various techniques (such as pulse charging), but there are no peer-reviewed publications verifying their claims. Sulfation prevention remains the best course of action, by periodically fully charging the lead-acid batteries." en.wikipedia.org/wiki/… $\endgroup$ – Buck Thorn Jun 3 '19 at 13:19
  • $\begingroup$ @BuckThorn sulphate crystals may not be completely or substantially removable, but if they were not at all (i.e. reaction not reversible) there would have been no lead acid batteries. I think the question can be generalised as "how does temperature affect the voltage that is required to dissolve lead sulphate at the same speed". $\endgroup$ – Greendrake Jun 3 '19 at 19:31
  • $\begingroup$ The mechanism of reversing sulfation appears to be to exceed the nominal charging voltage by a small amount, in order to counter internal resistive losses. This strikes me as a problem that is specific to your batteries and their history and in a sense too broad to give you a detailed answer. As a starting point however you can take Loongs or similar answer which gives you the baseline voltage you need to exceed. as computed from thermodynamic principles. In practiceyou may want to measure the current during charging and see how that changes. $\endgroup$ – Buck Thorn Jun 8 '19 at 9:11

Generally, the electrode reactions are both based on lead in different oxidation states. At the negative electrode, $\ce{Pb}$ is oxidized to $\ce{Pb^2+}$ during discharge. $$\ce{Pb <=> Pb^2+ + 2 e-}$$ At the positive electrode, $\ce{Pb^4+}$ is reduced to $\ce{Pb^2+}$. $$\ce{Pb^4+ + 2 e+ <=> Pb^2+}$$ For a classical lead–acid battery, the overall cell reaction is approximately $$\ce{Pb + PbO2 + 2H+ + 2 HSO4- <=> 2 PbSO4 + 2 H2O}$$ As long as $\ce{Pb}$, $\ce{PbO2}$, and $\ce{PbSO4}$ are available at the electrodes, the equilibrium cell voltage depends only on the acid concentration (that's why equilibtrium cell voltage can be estimated based on measured acid density). The dependence of the equilibrium voltage on concentration is given by the Nernst equation: $$U_\text{cell}=\left(1.931+0.0592\log\frac{a_{\ce{H+}}\cdot a_{\ce{HSO4-}}}{a_{\ce{H2O}}}\right)\ \mathrm V$$ This equation applies to one cell; a battery, however, has six cells in a row. Thus the total voltage is $$U_\text{battery}=6\left(1.931+\frac{RT}{ZF}\log\frac{a_{\ce{H+}}\cdot a_{\ce{HSO4-}}}{a_{\ce{H2O}}}\right)\ \mathrm V$$ where
$R$ is the gas constant,
$T$ is temperature,
$z$ is the number of electrons transferred in the cell reaction, and
$F$ is the Faraday constant.

We are only interested in the dependence of voltage $U$ on temperature $T$. All the other parameters may be taken as constants. Thus, our equation can be simplified to $$U=\left(11.586+T\cdot k\right)\ \mathrm V$$ where $k$ is a constant.

The exact parameter values for a classical lead–acid battery might differ a bit from a modern battery with lead–carbon electrodes; however, we can calibrate our model using the data given by the manufacturer, i.e. $U=14.1\ \mathrm V$ at $T=25\ \mathrm{^\circ C}=298.15\ \mathrm K$ $$14.1\ \mathrm V=\left(11.586+298.15\ \mathrm K\cdot k\right)\ \mathrm V$$ and we find $$k=0.008432\ \mathrm{K^{-1}}$$ Therefore, our complete equation is $$U=\left(11.586+T\cdot 0.008432\ \mathrm{K^{-1}}\right)\ \mathrm V$$ For a new temperature of $T=0\ \mathrm{^\circ C}=273.15\ \mathrm K$, we get an estimate of $$\begin{align}U&=\left(11.586+273.15\ \mathrm K\cdot 0.008432\ \mathrm{K^{-1}}\right)\ \mathrm V\\ &\approx13.9\ \mathrm V\end{align}$$

  • $\begingroup$ So far this tells how battery's voltage output depends on temperature. But the question is about equalizing voltage which is voltage that needs to be applied to the battery in order to remove sulfate crystals that build up on the plates over time. At 25 ∘C 14.1 V do the job. Are you sure that lower voltage is needed at lower temperatures to remove the same sulfate crystals? $\endgroup$ – Greendrake Jun 2 '19 at 10:05
  • $\begingroup$ Wouldn't the standard cell potential also be temperature dependent? What is the reaction enthalpy of the process? Also, the conductivity and the kinetics of various steps in the process would change, potentially affecting the required equalizing voltage. $\endgroup$ – Karsten Theis Jun 2 '19 at 10:14
  • $\begingroup$ @Greendrake Whatever the value of 14.1 V given by the manufacturer is referring to – I only made an estimate for the temperature correction of this value. $\endgroup$ – user7951 Jun 2 '19 at 14:59
  • $\begingroup$ @Loong Sorry but this does not answer the question at all then. I did not ask how voltage output depends on temperature. $\endgroup$ – Greendrake Jun 2 '19 at 20:06
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    $\begingroup$ @Greendrake You asked how the equalizing voltage (14.1 V at 25 °C) depends on temperature. And I tried to answer that. $\endgroup$ – user7951 Jun 3 '19 at 4:23

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