# Why is the reaction between potassium permanganate and hydrogen peroxide spontaneous?

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?

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:

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

• 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! – Corundum Jan 4 '16 at 18:30
• The Gibbs energy calculated from the cell potential is –429 kJ, which conforms more or less to the thermochemical result. – Corundum Jan 4 '16 at 18:38

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.

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

• 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 – orthocresol Jan 4 '16 at 14:08
• 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? – Toulousain Jan 4 '16 at 18:03
• Well, that is really up to you. – orthocresol Jan 4 '16 at 18:04
• Why not correct the answer? After you edit, some of those downvotes could turn to upvotes, and you might attract new upvotes too. – Curt F. Jan 4 '16 at 20:47