Where do the electrons in lead–acid battery come from? [closed]








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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?




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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