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In electrochemistry, we learn of the sulphate ion getting oxidised to peroxydisulphate ion. If this was to occur in the case of the nitrate ion which has a larger discharge potential, what would be formed?

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First, let’s take a look at both sulfate and nitrate. They both show similar characteristics when you compare the oxidation states: remember that nitrogen is in group 15 so it has five valence electrons while sulfur is in group 16 so it has six — therefore, the highest reasonable oxidation states for these elements should be $\mathrm{+V}$ and $\mathrm{+VI}$, respectively.

$$\begin{align}&\ce{\overset{$\mathrm{+V}$}{N}\overset{$\mathrm{-II}$}{O3}^-}&&\ce{\overset{$\mathrm{+VI}$}{S}\overset{$\mathrm{-II}$}{O4}^2-}\end{align}$$

When oxidising sulfate $\ce{SO4^2-}$ to peroxidisulfate $\ce{S2O8^2-}$, sulfur cannot be oxidised any further; the only atom that can still be oxidised is oxygen. Hence, in an initial step, we generate $\ce{SO4^{.-}}$ with three oxygens of $\mathrm{-II}$ and a fourth of $\mathrm{-I}$. Two of these then dimerise forming a peroxo-bond.

$$\ce{^-\overset{$\mathrm{-II}$}{O3}\overset{$\mathrm{+VI}$}{S}-\overset{$\mathrm{-I}$}{O}-\overset{$\mathrm{-I}$}{O}-\overset{$\mathrm{+VI}$}{S}\overset{$\mathrm{-II}$}{O3}^-}$$

Looking at nitrate, the only obvious theoretical would be to do the same, i.e. remove one oxygen electron to intermediately generate $\ce{NO3^.}$ and then have this dimerise to $\ce{O2N-O-O-NO2}$ or $\ce{N2O6}$. Interestingly, this would be a neutral compound unlike $\ce{S2O8^2-}$ which retains its charge. I don’t know whether it has been observed or not, but I would predict its formation from what I know.

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  • $\begingroup$ $\ce{N2O5}$ is reasonably common in nitrogen oxide chemistry; I never came across $\ce{N2O6}$ in my grad work, though, so if it can occur it's a very minor player. $\endgroup$ – hBy2Py Feb 23 '17 at 15:22
  • $\begingroup$ Scifinder has a few hits for the actual molecule. Didn't really investigate further $\endgroup$ – orthocresol Feb 23 '17 at 16:02
  • $\begingroup$ That was quite helpful and I am interested in knowing the presence of N2O6. $\endgroup$ – Aaron John Sabu Feb 23 '17 at 18:59

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