We have the following data:

\begin{align} \ce{NO3- + 3 e- &-> NO} &\quad E^\circ_1 &= \pu{0.96 V}\\ \ce{Hg^2+ + 2 e- &-> Hg} &\quad E^\circ_2 &= \pu{0.86 V} \end{align}

By the reasoning given in my book, nitrate oxidises mercury:

... of two substances, the one whose reduction $E^\circ$ value is greater, will be able to oxidise the other substance the other.

Then I tried calculating $E^\circ$ for overall reaction also taking into account number of electrons in each reaction. Or that we calculate $\Delta G$ for

$$\ce{NO3^- + Hg -> NO + Hg^2+}$$

$$\Delta G^\circ = -0.96 \cdot 2F + 0.86 \cdot 3 = 0.66F$$

So, $\Delta G^\circ$ is positive and so nitrate does not oxidise mercury. Which is correct?


1 Answer 1


It would really help if you copied the entire half cell reaction. The atoms must balance in a valid half cell reaction! So there are:

\begin{align} \ce{NO3− (aq) + 2 H+ + e− &<=> NO2 (g) + H2O} &\quad &E_0 = \pu{+0.80 V} \\ \ce{Hg2^2+ + 2 e− &<=> 2Hg (l)} &\quad &E_0 = \pu{+0.80 V} \\ \ce{Hg^2+ + 2 e− &<=> Hg (l)} &\quad &E_0 = \pu{+0.85 V} \\ \ce{2 Hg^2+ + 2 e− &<=> Hg2^2+} &\quad &E_0 = \pu{+0.91 V} \\ \ce{NO3−(aq) + 4 H+ + 3 e− &<=> NO (g) + 2 H2O (l)} &\quad &E_0 = \pu{+0.958 V} \\ \end{align}

The reduction of nitrate is interesting. In dissolving copper with nitric acid if the solution is highly acidic you get $\ce{NO}$, if the solution is mildly acid you get $\ce{NO2}$, and if the acidity is somewhere between you get both $\ce{NO}$ and $\ce{NO2}$.

Now if you also consider the Nernst equation, it is obvious that the concentrations of the species and the concentration of acid are important for calculating the half cell potentials. So it is impossible to say yes or no without additional information.

  • $\begingroup$ The question given in the book does not mention anything at all. But I think the last equation of reduction of nitrate applies. The third equation for Hg. And since it is standard reduction, so the concentrations must all be 1 molar right? So why can't we conclude from there? Thanks a lot. $\endgroup$
    – jonsno
    Dec 4, 2017 at 15:32
  • $\begingroup$ @samjoe - This is sort of odd case. It really depends heavily on $\ce{[H^+]}$ as well as the concentrations of the Hg and nitrate. You can assume anything you want. I'd be sure to explicitly add my assumptions to the answer. $\endgroup$
    – MaxW
    Dec 4, 2017 at 16:08
  • 1
    $\begingroup$ @andselisk - Thanks for the edit! Your edit really formats better and is much more readable. $\endgroup$
    – MaxW
    Dec 5, 2017 at 2:07
  • $\begingroup$ @MaxW You must be a huge Quake fan, I haven't seen that many \quad's anywhere else:D You are very welcomed, no big deal:) $\endgroup$
    – andselisk
    Dec 5, 2017 at 2:07
  • 1
    $\begingroup$ @andselisk - Just didn't know how else to do it, so I brute forced it. I was last in school 40 years ago. Never used Latex/MathJax until I started on this site. Done a zillion markup languages though. $\endgroup$
    – MaxW
    Dec 5, 2017 at 2:10

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