# Nitric acid oxidation

Because, unlike other metal dissolution reactions, the $\ce{H+}$ of $\ce{HNO3}$ isn't reduced- the $\ce{NO3-}$ is.

(data/balanced reactions from Wikipedia) \begin{align} \ce{NO3- + 4 H+ + 3 e- -> NO + 2 H2O}\quad &...& E^o_{red} = 0.96 V \\ \ce{NO3- + 2 H+ + e- -> NO2 + H2O} \quad&...& E^o_{red} = 0.79 V \\ \ce{Ag+(aq) + e- -> Ag(s)} \quad&...& E^o_{red} = 0.799 V \end{align}

Since the SRP of the $\ce{NO2}$ reaction looks smaller than that of $\ce{Ag+}$ (I may be wrong, but by significant digits it can't be greater than $0.799$), one can conclude that it's the $\ce{NO}$ reaction that's occurring here. And the $\ce{NO}$ reaction has a large enough SRP to oxidise $\ce{Ag+}$.

Usually nitrogen compounds are pretty versatile when it comes to redox reactions, since Nitrogen shows many oxidation states. So the simple reason for why $\ce{HNO3}$ is so strong an oxidising agent (with respect to other acids) is that it has a different, better path availible to it to get reduced.

Note that the exact reduction path (ie final reduction products/oxidation state) depends upon the comcentration of nitric acid-so much that copper can be oxidised in three different ways. I had a table of this, I'll post it later.