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. Unfortunately the English version doesn't contain the calculation of the voltage. I took the standard potentials from the book Elektrochemie by Hamann.


2 Answers 2


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


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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