7
$\begingroup$

I need a way to prove/show that hydrogen peroxide was decomposed through use of catalyst.

I want to ensure that my catalyst: $\ce{Fe(NO3)3}$ or iron(III) nitrate is a catalyst, not a reactant/ consumed during the reaction.

$$\ce{2 H2O2 (aq) ->[Fe(NO3)3 (s)] 2 H2O (l) + O2 (g)}$$


  • When the reaction is happening, I will introduce a wooden glowing splint over the bubbling reaction and the wooden splint glowing brighter or reigniting will show that oxygen is being produced.

enter image description here

Question:

  • Now, how do I show that water is produced? Would I just boil the product I get after reaction (image above) and put cobalt chloride paper at the water vapour?

  • Also, if I were to use the orange solution above again as a catalyst (since it still contains iron(III) nitrate), would hydrogen peroxide decompose again? If so, is there a way to put iron(III) nitrate back to its solid state? or any way to reuse as catalyst?


Edit:

If it is easier to answer, it doesn't have to be with iron(III) nitrate. I have an option to use manganese dioxide, which is another catalyst that I can substitute for iron(III) nitrate. I think it should be OK since it does the exact same reaction.

$\endgroup$
5
$\begingroup$

Detection of oxygen: Detection of $\ce{O2}$ by a glowing splint is a good way to detect the oxygen. Also you could capture the gas by a simple fixture e.g. enter image description here and demonstrate the volume change in the receiver. This way you can actually measure moles $\ce{O2}$ produced (by $PV=nRT$) then moles $\ce{H2O2}$ decomposed stoichiometrically by the formula you've written.

Choice of catalyst: Using $\ce{MnO2}$ would be better if you want to make sure you have a catalyst. $\ce{MnO2}$ will not be consumed during the decomposition; I'm not sure about iron(III) nitrate.

Or, detecting change in $\ce{Fe(NO3)3}$ concentration: Addition of a very small concentration of potassium thiocyanate, $\ce{KSCN}$ (say 1/100 of your $\ce{Fe^{+3}}$ concentration) will yield a deep red product, iron thiocyanate ($\ce{Fe(SCN)^{+2}}$). If you have access to a spectrometer you can measure absorbance of the initial solution's product and the product after decomposing $\ce{H2O2}$.

Prove water was produced: The best way I can think of is to measure the (subtle) change of density of your solution before and after decomposition. Using $\ce{MnO2}$ would make this easy because you can remove/filter it as a solid after decomposition and thus measure mass/volume of your solution before and after. $\ce{H2O2}$ and $\ce{H2O}$ have small albeit detectably different densities at RT. $\ce{H2O2}$ is more dense than water so your density should decrease.

$\endgroup$
  • $\begingroup$ How would i remove/filter manganese dioxide as solid from water? Is there any specific equipment? Or wcan i just boil it to remove water? $\endgroup$ – didgocks Jan 25 '17 at 12:22
  • 1
    $\begingroup$ Just filter paper through a funnel, maybe a vacuum if you want to be thorough. $\endgroup$ – khaverim Jan 25 '17 at 16:02
  • $\begingroup$ Would there be a way to prove water for iron(III) nitrate? $\endgroup$ – didgocks Jan 25 '17 at 20:50
  • $\begingroup$ You can probably still detect a small decrease in density even with iron(III) nitrate in solution $\endgroup$ – khaverim Jan 25 '17 at 20:54
2
$\begingroup$

Yes, apart from potassium iodide which is commonly used as a catalyst in the decomposition of hydrogen peroxide, $\ce{Fe^3+}$ salts, manganese dioxide and nickel hydroxide can be used as a catalysts as alternatives. Since, iron nitrate contains $\ce{Fe^3+}$, it can be used as catalyst.

There are some papers that discuss the use of $\ce{Fe^3+}$ salts as catalyst. Following is the relevant information from the papers:

  1. Paper 1

We should consider the role of the Ferric Chloride ($\ce{FeCl3}$) as catalyst in the decomposition reaction of hydrogen peroxide.(...) The fact is that Iron can exist in two different oxidation states, $\ce{Fe^2+}$ (Ferrous) and $\ce{Fe^3+}$ (Ferric), allows the catalyst to break the reaction into two different redox steps, each of which has a lower energy barrier to completion than the uncatalyzed reaction:

$$\ce{H2O2(aq) + 2Fe^3+(aq) -> O2(g) + 2 Fe^2+(aq) + 2H+(aq)}$$ $$\ce{H2O2(aq) + 2 Fe^2+(aq) + 2 H+ (aq) -> 2H2(l) + 2Fe^3+(aq)}$$

Note the first step in the catalyzed reaction involves reduction of the Ferric Ion ($\ce{Fe^3+}$) to the Ferrous Ion ($\ce{Fe^2+}$), which is then re-oxidized to Ferric Ion ($\ce{Fe^3+}$) in the second step. Hence, on net, the catalyst is not consumed during the course of the decomposition.

  1. Paper 2

$\ce{Fe^3+}$ ions is actually a homogeneous catalyst. The catalytic decomposition of hydrogen peroxide can be essentially explained by two different mechanisms based on the mutual redox transition Fe(III)/Fe(V) (KREMER-STEIN mechanism) and Fe(III)/Fe(II) (HABER-WEISS mechanism), respectively.

According to the mechanism proposed by KREMER and STEIN an intermediate oxygen complex of iron with oxidation number +V is primarily formed by the reaction of $\ce{Fe^3+}$ with $\ce{H2O2}$. This complex reacts with another $\ce{H2O2}$ molecule to water and oxygen thereby reforming $\ce{Fe^3+}$.

$$\ce{Fe^3+ + H2O2 <=> [Fe^{III}OOH]^2+ + 2H+ <=> [Fe^{V}O]^3+ + H2O ->[H2O2] Fe^3+ + 2H2O + O2}$$

According to the mechanism proposed by HABER and WEISS the $\ce{Fe^3+}$ ions initiate a radical reaction, after which the chain reaction consumes the hydrogen peroxide. This mechanism can explain the high reaction rate very well.

Chain initiation: $\ce{Fe^3+ + H2O2 <=> [Fe^{III}OOH]^2+ + 2H+ <=> Fe^2+ + HOO. + H+}$

Chain propagation: $\ce{Fe^2+ + H2O2 -> Fe^3+ + 2OH.}$ $\ce{Fe^3+ + H2O2 + OH. -> Fe^3+ + HOO. + H2O -> Fe^2+ + H+ + O2 + H2O }$


Answering the update

Yes, manganese dioxide can also be used as a catalyst. From paper 2:

Manganese dioxide is an example for a heterogeneous catalyst. The surface of solid manganese dioxide provides a particularly favorable environment to catalyze the decomposition, though the mechanism is not understood very well. (...)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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