Prediction of ionic reduction?

Question

Here are some standard electrode potentials:-

$$\ce{SO_4^{2-} + 4H^+ + 2e^- -> SO_2 + 2H_2O}\\E^o=0.17\mathrm{V}\\ \ce{NO_3^- + 4H^+ + 3e^- -> NO + 2H_2O}\\E^o=0.97\mathrm{V}\\ \ce{NO_3^- + 2H^+ + e^- -> NO_2 + H_2O}\\E^o=0.78\mathrm{V}$$ When nitric acid reacts with a metal, either $\ce{NO_2}$ or $\ce{NO}$ is formed? How do you predict which is thermodynamically favourable and which reaction will actually take place? For example, when dilute nitric acid reacts with copper, Nitric oxide is liberated and when it is concentrated, Nitrogen dioxide is evolved? As Nitric acid formation is more favourable, shouldn't it proceed in all cases as increasing the concentration will decrease/increase the potential of both the reaction simultaneousy(Nernst equation)?

In the case of sulfate ion, although the reduction of sulfate is more favourable than that of hydrogen, it is hydrogen ions which are getting reduced when a metal reacts at dilute concentrations? why is this so? Moreover, at high concentrations, once again sulfur dioxide is evolved? How can we understand such occurrences? Is overpotential involved in these observances?

In the case of permanganate ions, there are so many possible redox pathways, but a different one is preferred in a different case based on the conditions. For example, in acidic medium, permanganate is reduced to $\ce{Mn^{2+}}$ while in neutral/slightly alkaline solutions, it reduces to $\ce{MnO_2}$. Why is this so? Here are the redox potential values for permanganate conversions (latimer diagrams)

I want to know the general way to explain all these phenomena, as to what drives the reaction to give one product under certain condition and another in different conditions. Explaining any one example would be enough to set the general method which I can extrapolate to understand the remaining Examples.I am comfortable with basic electrochemical equations and those of thermodynamics.

Answer 1

Fortunately, the explanation is not as complex as it might seem. It is an oft-made analysis, but one must be very careful when determining reaction spontaneity by looking at the standard free energies $\Delta G_r^o$ alone, as most reactions are not performed in standard conditions and therefore the actual reaction free energy change $\Delta G_r$ may be quite different.

You are correct that, in standard conditions, $\Delta G_r = \Delta G^o_r$ and therefore that $\ce{NO}$ is the most stable reduction product of $\ce{NO_3-}$. However, standard conditions entail, among other things, that $\ce{[H+]\ = \ 1\ M}$ (at least approximately). This corresponds to quite an acidic aqueous solution, with pH zero. If one were to examine the reaction in a more dilute acidic medium, then the value of $\Delta G_r$ for each reduction reaction would change. The reaction that produces $\ce{NO_2}$ has a $\Delta G_r$ that depends on $\ce{[H+]^2}$, while for $\ce{NO}$ the $\Delta G_r$ varies with $\ce{[H+]^4}$. It is clear then that, as one goes to a more and more dilute acidic medium, $\Delta G_r$ decreases more strongly for the reaction that produces $\ce{NO}$, and at sufficiently weak acidic solutions the relative stabilities of the products actually switches, so then $\ce{NO_2}$ becomes the most stable product. A very similar argument can be made for the switch between $\ce{H+}$ and $\ce{SO^{2-}_4}$ reduction in media with sulfuric acid. For $\ce{MnO^{-}_4}$, note that the conversion to $\ce{Mn^{2+}}$ implies the consumption of eight protons, and hence if it can happen, it will likely do so in acidic media. As the acidity of the solution decreases, the conversion to $\ce{Mn^{2+}}$ is quickly suppressed, favouring reactions for which the $\Delta G_r$ depends less strongly on $\ce{[H+]}$, and so we find that $\ce{MnO^_2}$ becomes more favourable.