This question already has an answer here:

I had a lab practice in which I had to test the effectiveness of different catalysts for the decomposition of $\ce{H2O2}$. Among them, there was: $\ce{KI(aq)}$, $\ce{FeCl3(aq)}$, $\ce{MnO(s)}$, $\ce{NaOH(aq)}$, $\ce{HCl(aq)}$ and $\ce{I2(s)}$.

In different test tubes, I added some $\ce{H2O2}$ and some catalyst. After a while, I added $\ce{NaOH}$ as well. I observed an increase in the effervescence for $\ce{KI}$, $\ce{FeCl3}$, $\ce{I2}$ and $\ce{MnO}$. However, when the mixture was only $\ce{H2O2}$ + $\ce{NaOH}$, nothing happened. I've made some research, but can't find an explanation for this...

I believe $\ce{NaOH}$ increases the rate of the reaction of the other catalysts, but why?


marked as duplicate by DavePhD, Jon Custer, ringo, Todd Minehardt, ron Apr 26 '16 at 0:36

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


According to THE HOMOGENEOUS BASE-CATALYZED DECOMPOSITION OF HYDROGEN PEROXIDE J. Phys. Chem., 1961, vol. 65 , pages 304–306:

$\ce{H2O2 + OH- <=> HO2- + H2O}$

$\ce{H2O2 + HO2- ->}$ [6-membered ring with 2 "H"s bridging 2 "O-O"s, with another H bonded to one of the "O"s] $\ce{-> H2O + O2 + OH-}$

It is clear that the rate will go through a maximum when the peroxide is 50% ionized, that is $[\ce{H2O2}] = [\ce{HO2-}]$.

So OH- catalyzes decomposition of peroxide, but a maximum is reacted when the amount of hydroxide added is half the amount of peroxide. If more hydroxide than peroxide is added, the catalytic effect is lost.


Not the answer you're looking for? Browse other questions tagged or ask your own question.