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:


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


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- -&gt; 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 $\ce{Pb^2+ + 2 e- -&gt; Pb}$. 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. Unfortunately the English version doesn't contain the calculation of the voltage. I took the standard potentials from the book Elektrochemie by Hamann.
The potentials depend on the form of the compounds. 
It is true that in solution 

$\ce{Pb^{2+}  (aq) + 2e^- -&gt; Pb (s)} $ 

is -0.126 V. 

But in the case of a battery we have:

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

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