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2

We can split the reaction $$\ce{H2O2 -> H2O + O2}$$ into the respective reduction and oxidation half-reactions. $$\ce{H2O2 -> O2 + 2H+ + 2e-}$$ $$\ce{H2O2 + 2H+ + 2e- -> 2H2O}$$ Since the n-factor of $\ce{H2O2}$ for both these half-reactions is 2, the n-factor is: $$\frac{1}{n_f} = \frac{1}{2} + \frac{1}{2} = 1$$ $$n_f = 1$$ Based on the comments ...

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X-ray diffraction experiments, regardless if about small molecules or proteins / RNA / DNA, are usually one element characterizing a sample. On occasion, you may infer the oxidation state from the transition metals environment: how many closest neighbors, their distance, their number, their spatial arrangement. You are free to complement findings by any ...

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If we try to calculate the oxidation state of nitrogen in $\ce{N2O}$ using the familiar algebraic method, we get oxidation state $+1$ for both nitrogen atoms and that's what I found when I looked it up on the internet. Well … you get an average oxidation state. This calculation arguably implicitly assumes that all nitrogen atoms be equivalent. In some cases ...

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