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$$ \begin{align} \ce{PbO2 + H2SO4 &-> PbSO4 + H2O + O} &\quad &\text{(anode)}\\ \ce{Pb + H2SO4 &-> PbSO4 + H2} &\quad &\text{(cathode)} \end{align} $$

The cathode’s $\ce{H2}$ joins the anode’s $\ce{O}$ making $\ce{H2O}$ in solution. The book says the cathode gets four electrons, but from where? Two liberated hydrogens have one each, that’s two. Where’d the other two come from?

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  • $\begingroup$ That's lead storage battery. $\endgroup$ – Zenix May 15 '20 at 15:07
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    $\begingroup$ I adjusted the formatting of the reactions, but the first one looks wrong. It would be nice if you could add the source you've taken them from and cite the textbook. $\endgroup$ – andselisk♦ May 15 '20 at 15:16
  • $\begingroup$ These aren't equations for battery, but just for reactions with conc. H2SO4. @andselisk Question could use your closehammer, I think ;) $\endgroup$ – Mithoron May 15 '20 at 15:47
  • $\begingroup$ @Mithoron I'm not sure about single-handedly closing this one. I don't think this is necessarily a bad or a homework question; it's just somewhat hard to trace the OP's reasoning in the absence of the source they took the reactions from. $\endgroup$ – andselisk♦ May 15 '20 at 15:56
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    $\begingroup$ @andselisk Well, that would be like "trihandedly" now ;) and it seems real unclear. $\endgroup$ – Mithoron May 15 '20 at 15:59
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The half-equations are not correctly written. There is no Oxygen atom released, and no H atom emitted, as the author proposes. And the cathode does not get 4 electrons, as he or she states. The correct half-equations should be, first at the anode : $$\ce{Pb + SO_4^{2-} -> PbSO4 + 2 e^-}$$ And at the cathode it is : $$\ce{PbO_2 + 4H+ + SO_4^{2-} + 2 e^- -> PbSO_4 + 2 H2O}$$

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  • $\begingroup$ Can we consider $\ce{PbO2}$ as a positive electrode $\endgroup$ – Adnan AL-Amleh May 15 '20 at 20:45
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    $\begingroup$ Yes ! $\ce{PbO_2}$ is a positive electrode. $\endgroup$ – Maurice May 16 '20 at 10:26
  • $\begingroup$ To be fair, there is generally hydrogen release in a lead-acid battery, especially during formation, but that's not the desired reaction. $\endgroup$ – Mark Wolfman Jun 29 '20 at 2:12

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