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Is it possible that $\ce{2 OH-}$ from water could react to form $\ce{H2O2}$?

I mean most autoionization forms $\ce{H3O+}$ and $\ce{OH-}$.

However I think it is possible that $\ce{H-}$ and $\ce{OH+}$ forms, though the amount formed of these 2 is small. This is from water acting as an acid and H+ acting as a base.

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The only "hydroxyl"-type species whose recombination has been observed to yield hydrogen peroxide are hydroxyl radicals.

$$\ce{HO^. + ^.OH -> H2O2}$$

In order to generate these hydroxyl radicals from water, a lot of energy has to be provided, typically by radiolysis (short wave uv, $\gamma$, electron beam).

The conceivable steps yielding $\ce{HO^.}$ are either

$$\ce{H2O -> H^. + ^.OH}$$ or $$\ce{H2O -> H2O^{+.} + e-}$$

$$\ce{H2O^{+.} -> H+ + ^.OH}$$

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Is it possible that $\ce{2 OH-}$ from water could react to form $\ce{H2O2}$?

There would be two electrons left unaccounted for. There needs to be a balanced chemical equation, then we could consider the equillbrium constant for the reaction.

We could write $\ce{2H2O -> H2 + H2O2}$, but it is extremely unfavorable, like water decomposing to hydrogen and oxygen gas. You could calculate the Gibbs free energy of the reaction and an equilibrium constant, but the equilibrium constant will be extremely small.

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  • $\begingroup$ This is true but to make it a better answer I'd do the actual calculation and show the real reason why this doesn't happen. $\endgroup$ – matt_black Dec 24 '14 at 14:07

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