Redox potential of a lead–acid battery

In the German Wikipedia there are two reactions on the poles of the battery shown with the following potentials:



> 
> $$
> \begin{align}
> \ce{Pb + SO4^2- &-> PbSO4 + 2 e-} &|\pu{-0.36 V}\\
> \ce{PbO2 + SO4^2- + 4 H+ + 2 e- &-> PbSO4 + 2 H2O} &|\pu{+1.68 V}
> \end{align}
> $$
> 
> 
> $$E\_\mathrm{Ges}^0 = \pu{1.68 V} - (\pu{-0.36 V}) = \pu{2.04 V}$$
> 
> 
> 


I do understand the potential for the second $\pu{1.68 V}$ since for the second reaction the underlying redox pair is $\ce{Pb^4+ + 2 e- -> Pb^2+} $. For this redox pair the electrochemical standard potential is $\pu{1.69 V}$.


But for the first reaction I think that the underlying redox pair has to be $\ce{Pb^2+ + 2 e- -> Pb}$. This redox pair has standard potential of $\pu{-0.1263 V}$.


This result in a voltage of $\approx\pu{1.55 V}$. But Wikipedia and a book of mine tell the the voltage of this battery type is $\pu{2.04 V}$. 


What the reason for the $\pu{-0.36 V}$?


**Source:** This is from the [German Wikipedia article on lead-acid batteries.](https://de.wikipedia.org/wiki/Bleiakkumulator) Unfortunately the English version doesn't contain the calculation of the voltage. I took the standard potentials from the book [Elektrochemie by Hamann](https://www.amazon.de/Elektrochemie-Carl-H-Hamann/dp/3527310681).




The potentials depend on the form of the compounds. 
It is true that in solution 


$\ce{Pb^{2+} (aq) + 2e^- -> Pb (s)} $ 


is -0.126 V. 


But in the case of a battery we have:


$\ce{PbSO4 (s) + 2e^- -> Pb (s) + SO4^{2-} (aq)}$


And in this case the $\ce{Pb^{2+}}$ is in solid form and the potential is -0.356 V. 


In a battery the sulphate is insoluble and it is required that it sticks to the electrode, otherwise the reverse reaction can not occur. 


A table of potentials can be found [here](http://www.chemeddl.org/services/moodle/media/QBank/GenChem/Tables/EStandardTable.htm)


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The underlying redox pair is, as you say, $\ce{Pb^2+ +2e− -> Pb}$ and has standard potential of $\pu{−0.1263 V}$.


But standard potential is for $\pu{1 M}$ concentration. If you look at the "underlying" reaction, you must correct for the reduced concentration of $\ce{Pb^2+}$ due to its insolubility. The correction is made by using the Nernst equation and the solubility (or solubility product) of $\ce{PbSO4}$, and the half-reaction potential increases because the solubility of $\ce{Pb^2+}$ is low.


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