# In brown ring test, how is NO^- formed?

$$\ce{NO3- + 3[Fe(H2O)6]^{2+} + 4H+ -> NO + 3[Fe(H2O)6]^{3+} + 2H2O}$$

$$\ce{NO + [Fe(H2O)6]^{2+} -> [Fe(NO)(H2O)5]^{2+} + H2O}$$

I read that in the brown ring complex $\ce{[Fe(NO)(H2O)5]^{2+}}$ , $\ce{NO^{-}}$ is the ligand.

What I don't understand is that :

How is $\ce{NO^{-}}$ formed? Nitrate ion got reduced to NO while $\ce{Fe^2+}$ was oxidized to $\ce{Fe^{3+}}$. The electrons that were released in the conversion of ferrous to ferric ion were used in converting nitrate ion to $\ce{NO}$. So where do these extra electrons come up that convert $\ce{NO}$ to $\ce{NO^{-}}$ ? I don't want to go into the mechanism; I just want to have an intuitive understanding of what's happening in the reaction.