How does this reaction occur? #2KO_2 + 2H_2O -> 2KOH + H_2O_2 + O_2­#

From a Russian tutorial on alkaline metal properties :

#2KO_2 + 2H_2O -> 2KOH + H_2O_2 + O_2­#

Should I just memorize this reaction? I can't seem to understand how it occurs.

And on Wikipedia's page for potassium peroxide the reaction is given with a different set of resulting compounds:

#4KO_2 + 2H_2O -> 4KOH + 3O_2#

1 Answer
Feb 20, 2016

Answer:

Here's what's going on here.

Explanation:

Let me start by saying that #"KO"_2# is actually potassium superoxide, not potassium peroxide, which is #"K"_2"O"_2#.

This said, the reaction between potassium superoxide and water will indeed produce hydrogen peroxide, #"H"_2"O"_2#.

However, hydrogen peroxide will then undergo decomposition to form water and oxygen gas.

#color(blue)("H"_2"O"_text(2(aq]) -> "H"_2"O"_text((l]) + 1/2"O"_text(2(g]))#

Now, the stability of hydrogen peroxide decreases with increasing pH. At high pH levels, the compound undergoes decomposition more rapidly than at low pH levels.

Notice that this reaction also produces hydroxide anions, #"OH"^(-)#, which is an indicator that the solution will be basic in nature #-># the decomposition of hydrogen peroxide is favored.

You can thus say that

#2"KO"_text(2(aq]) + 2"H"_2"O"_text((l]) -> 2"KOH"_text((aq]) + "H"_2"O"_text(2(aq]) + "O"_text(2(g])#

Taking into account the decomposition of hydrogen peroxide, you will have

#2"KO"_text(2(aq]) + color(red)(cancel(color(black)(2)))"H"_2"O"_text((l]) -> 2"KOH"_text((aq]) + color(blue)(color(red)(cancel(color(black)("H"_2"O"_text((l])))) + 1/2"O"_text(2(g])) + "O"_text(2(g])#

This will be equivalent to

#2"KO"_text(2(aq]) + "H"_2"O"_text((l]) -> 2"KOH"_text((aq]) + 3/2"O"_text(2(g])#

which will of course give you

#4"KO"_text(2(aq]) + 2"H"_2"O"_text((l]) -> 4"KOH"_text((aq]) + 3"O"_text(2(g])#

As an interesting fact, this reaction is based on the disproportionation of the superoxide anion, #"O"_2^(-)#.

A disproportionation reaction is a reaction in which the same chemical species undergoes both oxidation and reduction to form two separate products.

Oxygen has a #-1/2# oxidation state in the superoxide anion, and is being reduced to a #-2# oxidation state in the hydroxide anions, i.e. in potassium hydroxide, and oxidized to a #0# oxidation state in oxygen gas.

The net ionic equation for this reaction, with oxidation states added for oxygen, will look like this

#stackrel(color(blue)(_1/2))("O"_text(2(aq])^(-)) + 2"H"_2"O"_text((l]) -> 4stackrel(color(blue)(-2))("O")"H"_text((aq])^(-) + 3stackrel(color(blue)(0))("O")_text(2(g])#