[thetada]

I've done a lot of research into the equation for the "cannonfire" reaction but I've suddenly realised something doesn't add up about the accepted equation.
The below linked video quotes the following equation for the oxidation of hydrogen peroxide by potassium manganate:

https://www.youtube.com/watch?v=vvSjyPs4stc

2MnO₄^– + 3H₂O₂ → 2MnO₂ + 2OH^– + 3O₂ + 2H₂O

The half equations that would combine to give that overall equation are:

MnO₄⁻ + 2 H₂O + 3 e⁻   ⇌   MnO₂ + 4 OH⁻ +0.59

O₂ + 2H⁺ + 2 e⁻   ⇌   H₂O₂   +0.70

But that doesn't make sense because the half cell with the more negative potential should be the one that supplies the electrons, which in this case would be the Manganate cell. But it's clear that hydrogen peroxide is the oxidised species, because the demo evolves oxygen gas.

So what seems more likely is that this half equation should be used:

MnO₄⁻ + 8H⁺ + 5e⁻ ⇌ Mn²⁺ + 4H₂O   +1.49

Presumably there would be a combination of the 2+ and 4+ manganese products, but the thermodynamics seem to suggest that the 2+ species should dominate. Does anyone else have any ideas on the issue?

[Borek]

Not that I know a definitive answer, but you are for sue ignoring two things:

1. You are comparing standard potentials, when you should compare formal potentials.

2. The latter reaction requires huge amount of strong acid which is not present (which also means its formal potential will be far from the standard one).

[thetada]

That's interesting. I'm not familiar with formal potentials, but does the associated word "activity" signify something to do with the fact that manganate and peroxide are in different states?

Also, I appreciate what you're saying about the strong acid, but the Cannonfire demo doesn't need strong acid. You mix ethanol and hydrogen peroxide, ignite it and then add solid potassium manganate.

But on the flipside, since the reaction mixture is set on fire, it will be above the temperature of standard conditions and with concentrations that doubtless vary with standard conditions.

[Borek]

Yes, activity is part of the picture. Standard potential is the one observed when all substances are in the standard state - as a first approximation it means all concentrations are equal to 1 M. Thats definitely not a case here (the most important thing being, pH of water used is not 0, but most likely somewhere between 5-6). You should use Nernst equation to calculate the formal potential and then recheck the situation.

Also, I appreciate what you're saying about the strong acid, but the Cannonfire demo doesn't need strong acid. You mix ethanol and hydrogen peroxide, ignite it and then add solid potassium manganate.

And that suggests the latter reaction equation is not the one describing what is going on, doesn't it?

[thetada]

It does suggest that. I'm going to flip this whole thing on its head and get my students to suggest why the demo doesn't seem to obey the feasibility prediction from standard potentials. Thanks for your *delete me*