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Let's consider the following two redox reactions;

  • $\ce{C_2H_4O_2 + 2\;O_2 -> 2\;HCO_3^- + 2\;H^+}$

  • $\ce{C_2H_4O_2 + 1.6\;NO_3^- -> 2\;HCO_3^- + 0.8\;N_2 + 0.8\;H_2O + 0.4\;H^+}$

Considering that the standard Gibbs energies of formation of the involved chemical species are the following;

  • $\ce{C_2H_4O_2}$: -393.8 kJ.mol-1

  • $\ce{HCO_3^-}$: -586.9 kJ.mol-1

  • $\ce{NO_3^-}$: -111.3 kJ.mol-1

  • $\ce{H_2O}$: -237.2 kJ.mol-1

  • $\ce{O_2$, $N_2$, $H^+}$: 0 kJ.mol-1

Then the standard Gibbs energy variation associated to those reactions are respectively -804.4 kJ.mol-1 for the oxidation of acetic acid by oxygen and -816.0 kJ.mol-1 for the oxidation of acetic acid by nitrate.

It appears that the oxidation by nitrate is more exergonic that the oxidation by oxygen in standard conditions, when computed this way.

However, the redox potential of $\ce{0.25\;O_2 + H^+ + e^- -> 0.5\;H_2O}$ is +0.811 V, while the redox potential of $\ce{1/5\;NO_3^- + 6/5\;H^+ + e^- -> 1/10\;N_2 + 3/5\;H_2O}$ is 0.746 V. I haven't found any data for the redox potential of acetic acid, however the redox potential of its electron acceptors indicates that an electron transfer from acetic acid to oxygen should imply a larger Gibbs energy variation than an electron transfer from acetic acid to nitrate (according to the $\Delta G = nFE$ formula), which contradicts what has been previously computed from Gibbs energies of formation.

I am new to electrochemistry, so what did I misunderstand here?

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  • $\begingroup$ Assuming that you have made no maths errors in calculating $\Delta G$ from the Gibbs energies of formation, I think the first thing I would check is your reduction potential for $\ce{NO3-}/\ce{N2}$. That half-equation you have written is not balanced and I couldn't find the value for the reduction potential (I did a brief search in Atkins as well as Wikipedia). $\endgroup$
    – orthocresol
    Commented Sep 27, 2016 at 15:35
  • $\begingroup$ Apart from that, I can't see anything wrong with your line of reasoning. There are a couple of typos ($\ce{NO3+}$, and $\Delta G = -nFE$ (minus sign is missing)) but they don't affect the logic. Perhaps I am being dense too, I don't know. $\endgroup$
    – orthocresol
    Commented Sep 27, 2016 at 15:43
  • $\begingroup$ Yes you are right, there were some typos in the $NO3^-/N_2$ half-reaction. However this doesn't affect the conclusion. For the record, those redox potentials come from Stumm and Morgan, 1996 $\endgroup$
    – scutigera
    Commented Sep 27, 2016 at 16:07

2 Answers 2

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Looking up the redox potentials in tables vs. Hydrogen as standard at zero volts the redox for Oxygen is

$$\ce{O2 + 4H^+ + 4e <=> 2H2O} \qquad \qquad E^\circ = +1.229~\mathrm{V}$$

so the tables you use may be vs. calomel electrode as standard.

The redox potential for the nitrate reaction has to be constructed from others as

$$\begin{align} \ce{NO3^- + 3H^+ + 4e}\, &\ce{<=> HNO2 + H2O} & E_1^\circ &= +0.934~\mathrm{V} \\ \ce{2HNO2 + 4H^+ + 4e}\, &\ce{<=> N2O + 3H2O} & E_2^\circ &= +1.297~\mathrm{V} \\ \ce{N2O + 2H^+ + 2e}\, &\ce{<=> N2 + H2O} & E_3^\circ &= +1.766~\mathrm{V} \end{align}$$

The overall half-reaction, after multiplying the nitrate reaction by 2 (to balance) and adding is

$$\ce{2NO_3^- + 12H^+ + 10 e <=> N2 + 6H2O} \tag{1}$$

The redox potential for the reaction $(1)$ is calculated from $n_\mathrm{r}E_\mathrm{r}^\circ = n_1E_1^\circ + n_2E_2^\circ +....$ where the n's are the number of electrons in each half reaction and $n_\mathrm{r}$ for the new reaction.

The $\ce{NO_3^- to N2}$ redox potential is therefore $1.246~\mathrm{V}$, calculated as

$$E_\mathrm{r}^\circ = \frac{(4 \times 0.934) + (4 \times 1.297) + (2 \times 1.766)}{10}$$

The two values, $1.229~\mathrm{V}$ and $1.246~\mathrm{V}$, would now seem to agree with your $\Delta G$ values. They may be the same within error as may your $\Delta G$ values.

(Redox Values are from the CRC Handbook of Chemistry and Physics, 75 ed.)

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  • $\begingroup$ Wonderful, I will openly admit I was too lazy to fact check :) $\endgroup$
    – orthocresol
    Commented Sep 28, 2016 at 9:58
  • $\begingroup$ "so the tables you use may be vs. calomel electrode as standard" no, it is that the OP is using redox potentials at pH 7 $\endgroup$
    – DavePhD
    Commented Sep 28, 2016 at 12:52
  • $\begingroup$ I biochemistry it is normal for reactions involving protons to use pH 7 as a standard state thus $\ce{O2 + 4H^+ + 4e <=> 2H2O}$ has a potential of $ E^{0'} = 0.816 $ V, but this does not matter if all potentials are on the same standard state. We only need the difference in values. $\endgroup$
    – porphyrin
    Commented Sep 29, 2016 at 9:49
  • $\begingroup$ @porphyrin it matters because the number of moles of H+ is different in the two equations, so changing 1M to 10^-7M affects them differently. The difference is different at pH 7 vs pH 0. I will post data later. $\endgroup$
    – DavePhD
    Commented Sep 29, 2016 at 12:17
  • $\begingroup$ @porphyrin so at pH 7, according to table 11-1 here books.google.com/… the redox potential are as OP is saying. And the reason the differences are different is as explained here www1.lsbu.ac.uk/water/water_redox.html . $\endgroup$
    – DavePhD
    Commented Sep 29, 2016 at 14:02
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The standard states of the Gibbs energies of formation (usually 1M concentration for ions) must match the standard states for the redox potentials.

The redox potentials you used are for pH=7, which means [H+]= 10^-7 M.

See this link for sample calculations of the effect of [H+] on redox potentials.

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