I'm making a lead-acid battery for an investigation. Since I only had sulfuric acid and lead electrodes, I decided to firstly run 2 lead electrodes through electrolysis in a sulfuric acid electrolyte. This would create the following half equations:
$$\ce{4H+ + 4e- -> 2H2}$$
$$\ce{Pb +2H2O->PbO2 +4H+ +4e-}$$
At this stage, I still need to perform some theoretical calculations of the required potential to run this electrolysis. Obviously, the standard reduction potential of the hydrogen ion is 0V. But I can't find the potential for the 2nd reaction. Could someone provide me with a source for it?
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2 Answers
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Your reactions are wrong.
$$\ce{Pb(s) + HSO^−_4(aq) -> PbSO4(s) + H+(aq) + 2e−}$$ $$\ce{PbO2(s) + HSO^−4(aq) + 3H+(aq) + 2e− -> PbSO_4(s) + 2H2O(l)}$$
Lead acid batteries are fairly reversible, hence their use as rechargable batteries. Anything that produces a gas, like in your reduction reaction, is unlikely to be reversible.
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You can find the electrode potentials for a lot of reactions on Wikipedia:
https://en.wikipedia.org/wiki/Standard_electrode_potential_(data_page)
I also think your second reaction isn't entirely correct. Lead atoms don't react with water to create lead oxide. Instead it's lead and sulfate ions reacting with water to create lead oxide and sulfuric acid.
