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):

$$\ce{2MnO4- + 3H2O2 -&gt; 2MnO2 + 2H2O + 3O2 + 2OH-}$$

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

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

$E^\circ_\mathrm{cell}$ 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 $\left[\ce{H+}\right]=1$:

$$\ce{O2 + 2 H+ + 2e- &lt;=&gt; H2O2}\qquad E^\circ=0.695\ \mathrm V$$

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

$$\ce{MnO4- + 2 H2O + 3e- &lt;=&gt; MnO2 + 4 OH-}\qquad E^\circ=0.595\ \mathrm V$$

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

$$\ce{MnO4- + 4 H+ + 3e- &lt;=&gt; MnO2 + 2H2O}\qquad E^\circ=1.695\ \mathrm V$$