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When hydrogen peroxide is mixed with potassium permanganate, oxygen gas and water vapour are formed, according to the reaction (source):

$$\ce{2MnO4- + 3H2O2 -> 2MnO2 + 2H2O + 3O2 + 2OH-}$$

This reaction is spontaneous, and exothermic. It is an example of a redox reaction, with the following half reactions occurring (data from Vanýsek):

$$ \begin{align} \ce{MnO4- + 2 H2O + 3 e- &-> MnO2 + 4 OH-} &\quad E^\circ_\mathrm{red} &= 0.595~\mathrm{V} \\ \ce{H2O2 &-> O2 + 2 H+ + 2 e-} &\quad E^\circ_\mathrm{ox} &= -0.695~\mathrm{V} \end{align} $$

$E^\circ_\mathrm{cell}$ is equal to the sum of the oxidation potential and the reduction potential of the two half reactions; in this case, it would be $-0.1~\mathrm{V}$. A redox reaction is spontaneous if $E^\circ_\mathrm{cell}$ is positive — how can it be, then, that hydrogen peroxide spontaneously reacts with permanganate ions?

Using thermodynamical data (from NIST), I have calculated that the $\Delta G^\circ_\mathrm{m}$ of the reaction is $-463.576~\mathrm{kJ}$. The reaction should indeed be spontaneous. How can it be, then, that the results of the thermodynamical approach and the electrochemical one differ drastically?

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3 Answers 3

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For both half-reactions, the actual potentials depend on $\mathrm{pH}$; however, the values given in the question apply to different $\mathrm{pH}$.

The given reaction of hydrogen peroxide and the corresponding potential apply to $\mathrm{pH=0}$ or $\left[\ce{H+}\right]=1$:

$$\ce{O2 + 2 H+ + 2e- <=> H2O2}\qquad E^\circ=0.695\ \mathrm V$$

Whereas the given reaction of permanganate and the corresponding potential apply to $\mathrm{pH=14}$ or $\left[\ce{OH-}\right]=1$:

$$\ce{MnO4- + 2 H2O + 3e- <=> MnO2 + 4 OH-}\qquad E^\circ=0.595\ \mathrm V$$

The corresponding reaction and potential for $\mathrm{pH=0}$ or $\left[\ce{H+}\right]=1$ are:

$$\ce{MnO4- + 4 H+ + 3e- <=> MnO2 + 2H2O}\qquad E^\circ=1.695\ \mathrm V$$

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  • $\begingroup$ But then, the net reaction would be $\ce{2MnO4- + 3H2O2 + 2H+ -> 2MnO2 + 4H2O + 3O2}$, which doesn’t correspond to the reaction cited above. Thus, I would use for $\ce{H2O2 + 2OH- <=> O2 + 2H2O + 2e-} (E^\circ=0.146\ \mathrm{V})$, which yields the given equation. Thank you so much for your help! $\endgroup$
    – Corundum
    Commented Jan 4, 2016 at 18:30
  • $\begingroup$ The Gibbs energy calculated from the cell potential is –429 kJ, which conforms more or less to the thermochemical result. $\endgroup$
    – Corundum
    Commented Jan 4, 2016 at 18:38
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The calculations you have done (if correct, I didn't check them) suggest that the reaction would not be spontaneous at standard conditions, which for electrochemistry means a pH of 0. If you want to model the process at pH 7, you would need to adjust the potential used for both reactions: permanganate reduction produces $\ce{OH-}$ and peroxide oxidation produces $\ce{H+}$. Thus, the equilibrium potential for either half-reaction will depend on pH.

You can use the Nernst equation to do this adjustment. Since standard conditions have a concentration of $\ce{OH-}$ of $10^{-14}$ molar, and at pH 7 this concentration will be $10^7$-fold higher, the $E_\mathrm{red}$ of permanganate should be less reducing than the standard value $E_\mathrm{red}^\circ$. The opposite is true of peroxide.

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The potential of a cell is the reduction potential minus the oxidation potential. In your case, it would be $E^\circ_\mathrm{cell} = 1.19 - (-2.08) = 3.27~\mathrm{V}$ for the whole reaction or $1.64~\mathrm{V/mol}$ of permanganate. You must remember to take into account the stoichiometric coefficients of the half reactions to obtain the overall potential

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    $\begingroup$ 1) Corundum already took the negative sign in the oxidation potential, i.e. it is a true oxidation potential, not a reduction potential for an oxidation half-reaction. The electrochemical values are correct. 2) When calculating $E^\circ_\mathrm{cell}$ you do not multiply electrode potentials by the stoichiometric coefficients $\endgroup$
    – orthocresol
    Commented Jan 4, 2016 at 14:08
  • $\begingroup$ So right. I think I am still coming into the new year. Should I correct the answer or just leave it so that people can see why it is wrong? $\endgroup$
    – Toulousain
    Commented Jan 4, 2016 at 18:03
  • $\begingroup$ Well, that is really up to you. $\endgroup$
    – orthocresol
    Commented Jan 4, 2016 at 18:04
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    $\begingroup$ Why not correct the answer? After you edit, some of those downvotes could turn to upvotes, and you might attract new upvotes too. $\endgroup$
    – Curt F.
    Commented Jan 4, 2016 at 20:47

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