If you stored $\ce{Pb(NO3)2(aq)}$ in a $\ce{Cu(s)}$ container, would any reaction occur? Lead(II) ions will not react spontaneously with the copper solid as the $E^0_\mathrm{net}$ of the reaction would be negative. However, the $E^0_\mathrm{net}$ of the reaction between nitrate ions and copper solid is positive. The reduction reaction for nitrate ions is $$\ce{2 NO3- (aq) + 4 H+(aq) + 2 e- -> N2O4(g) + 2 H2O(l)}\qquad{E^0=\mathrm{+0.80\ V}}$$

As is obvious, the reaction requires not only nitrate but hydrogen ions. However, water (the solvent) does have hydrogen/hydronium ions. Would this be enough to furnish the reaction? If not, why?

  • $\begingroup$ "Hydronium ion" and "$\ce{H^+}(\mathrm{aq})$" refer to pretty much exactly the same thing. $\endgroup$ – hBy2Py Jun 21 '16 at 2:45
  • $\begingroup$ @hBy2Py I know. That’s the underlying assumption of my question. Perhaps I should rephrase: "would the hydrogen ions in water be enough to furnish the reaction?" $\endgroup$ – lightweaver Jun 21 '16 at 2:49
  • $\begingroup$ Gotcha. My guess is "Yes," but I'm not sure enough to put an answer down on it. $\endgroup$ – hBy2Py Jun 21 '16 at 2:54
  • $\begingroup$ You are overgeneralizing. Yes, water contains enough $\ce{H+}$ for some redox reactions; throw a little sodium in it and you will see. $\endgroup$ – Ivan Neretin Jun 21 '16 at 4:42

Yes the reaction would be spontaneous at standard conditions. However standard conditions for electrochemistry are at pH 0, i.e. an $[\ce{H+}]$ concentration of 1 molar.

Solutions of lead nitrate by themselves are not going to give you pH 0. Apparently very very concentrated lead nitrate solutions will reach pH 3 or 4, due to the weakly acidic nature of hydrated lead ions.

The electrochemical potential of the reaction you are interested in at pH 3 can be calculated from the standard (i.e. pH 0) potential via the Nernst equation.

$$E_\mathrm{cathode} = E_0\;– \frac{RT}{nF}\ln{Q}$$

Here, $Q$ is the "half"-reaction quotient, given by $$ Q = \frac{\mathrm{p}_{\ce{N2O4}}}{[\ce{NO3-}]^2[\ce{H+}]^4}$$ Now assume that nitrate concentration and the partial pressure of dinitrogen tetroxide are in their standard states.

If you go through the math (see examples here or here), then you will see that there is a 118 millivolt-per-decade decrease in the potential, i.e. every increase in the pH by 1 unit decreases the potential by 118 mV. So already at pH 3 the reaction will no longer be favorable.

Copper is unstable to concentrated nitric acid, but the acid needs to be concentrated. Dilute acids cannot oxidize copper so easily.

  • $\begingroup$ Thanks! I’ve done some calculations and I understand it fully now. However, did you make a mistake with the expression for Q? Doesn’t the Nernst equation apply for the entire cell? Wouldn’t Q need to have [$\ce{Cu^2+}$]? $\endgroup$ – lightweaver Jun 21 '16 at 5:39
  • $\begingroup$ Yeah apologies my answer was confusing. You can write Nernst equations for an entire cell, or for each half-cell separately. The $Q$ I wrote was for the half-reaction you put in the OP, but the Nernst equation is mislabeled as "$E_{cell}$". It should really say $E_{cathode}$. I'll edit to correct. $\endgroup$ – Curt F. Jun 21 '16 at 21:39

It depends on your redox reaction. For example, if you add permanganate to a neutral solution containing e.g. sulphite, you will observe a redox reaction in which manganese(IV) oxide is created. If you do your redox caluculations, you will observe the following:

$$\begin{align}\tag{red}\ce{MnO4- + 3 e- + 4 H+ &-> MnO2 + 2 H2O}& | \cdot 2\\ \tag{ox}\ce{SO3^2- + H2O & -> SO4^2- + 2 e- + 2 H+} & | \cdot 3\\ \tag{redox}\ce{2 MnO4- + 2 H+ + 3 SO3^2- &-> 2 MnO2 + H2O + 3 SO4^2-}&\end{align}$$

This reaction still consumes protons but proceeds under neutral conditions. (Under acidic conditions, permanganate would reduce to $\ce{Mn^2+}$, und strongly basic conditions it would end up as $\ce{MnO4^2-}$.)


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