1
$\begingroup$

I've done a Google search regarding this question, and have found that there is no place online which answers this question to the degree that I want. Unfortunately, every site online stops at the place where they say that this reaction occurs via two steps:

  1. $\ce{H2O2 + I- -> OI- + H2O}$
  2. $\ce{H2O2 + OI- -> H2O + O2 + I-}$

My question is, is the iodide ion abstracting the O in step 1, or is it simply bonding after the O has already been given off? Since H2O2 decomposition happens incredibly slowly, the latter seems very unlikely. On the other hand, I can't seem to visualize why an anion would abstract the uncharged oxygen either. Yes, O has a lone pair, but that should give it a slight negative dipole moment and make it even harder to abstract, in my mind.

Secondly, in step two, the OI- appears to catalyze the decomposition of H2O2 further. Why? Again, wouldn't the negative charge on the anion cause it to have no interaction with the neutral hydrogen peroxide?

$\endgroup$
2
  • 1
    $\begingroup$ Don't know the mechanism, but the oxygen in hydrogen peroxide is in an oxidized state (-1) relative to its common oxidation state (-2), so electrophilic. Maybe the iodide stabilizes the oxygen dianion and the latter is what causes the peroxide to fall apart? $\endgroup$
    – Buck Thorn
    Commented Feb 14, 2023 at 7:18
  • 2
    $\begingroup$ For some reason you seem to think these are elementary reaction, while they obviously aren't. This is just the two reactions happening, whatever their mechanisms are is another thing. $\endgroup$
    – Mithoron
    Commented Feb 14, 2023 at 12:59

1 Answer 1

2
$\begingroup$

OP stated in the question that:

I've done a Google search regarding this question (How does $\ce{KI}$ catalyze the reaction with $\ce{H2O2}$?), and have found that there is no place online which answers this question to the degree that I want.

I think OP hasn't put enough time to find out that this experiment (Decomposition of hydrogen peroxide catalyzed by iodide) is a very popular experiment amongst first year laboratory programs around most of universities in United States and around the world. For example, see Ref.1, in which it states that $\ce{H2O2}$ can act either as a oxidizing or reducing reagent: $$\ce{H2O2 + 2H+ + 2e- <=> 2H2O} \qquad \mathrm{E^\circ} = \pu{1.776 V} \tag1$$ $$\ce{O2 + 2H+ + 2e- <=> H2O2} \qquad \mathrm{E^\circ} = \pu{0.815 V} \tag2$$

The equations $(1)$ and $(21)$ shows that $\ce{H2O2}$ can undergo spontaneous self-oxidation-reduction, and is therefore, thermodynamically unstable. Accordingly, $\ce{H2O2}$ decomposes spontaneously and exothermically, but slowly to oxygen and water at room temperature. The slow reaction rate is due to its relatively large activation energy of $\pu{76 kJ mol-1}$ (Ref.2). However, it would decompose at a much faster rate when a catalyst is introduced. For example, the large activation energy is lowered to ~$\pu{41 kJ mol-1}$ when $\ce{MnO2}$ is used as the catalyst and proceeds with a heat of reaction of $\pu{−99 kJ mol-1}$ (Ref.2). The iodide-catalyzed decomposition reaction proceeds with a heat of reaction of $\pu{−98.3 kJ mol-1}$ (Ref.3).

According to Ref.1, $\ce{I-/IO-}$ redox couple with $\mathrm{E^\circ} = \pu{0.55 V}$ and $\ce{I2/I-}$ redox couple with $\mathrm{E^\circ} = \pu{0.535 V}$ are ideal systems to catalyze $\ce{H2O2}$ decomposition. If first couple is used, the rate determining step would be:

$$\ce{H2O2 + I- -> IO- + H2O} \tag3$$

The same the rate determining step is given in an older reference (Ref.4), so that the mechanism might be correct. The same group, Liebhafsky & Coworkers has mentioned in later publication (Ref.5) that:

The first successful quantitative investigation in chemical kinetics, completed in 1866 by Harcourt and Esson, proved that the rate of the reaction, $$\ce{H2O2 + 2H+ + 3I- -> 2H2O + I3-} \tag4$$ is proportional to the amounts of peroxide and iodide present, and that it is “accelerated" by the addition of acid. A quarter century later Magnanini in a remarkable investigation, which we shall subsequently consider in some detail, demonstrated how this “acceleration" is related to the acid added.

The authors have measured the rate of the reaction (given in the equation $(4)$) for the first time under the simplest experimental conditions, with the result that the law, previously established (by Magnanini): $$\frac{d[\ce{I3-}]}{dt} = k_1^\circ \frac{[\ce{I-}]}{[\ce{H2O2}]} + k_1[\ce{H+}][\ce{I-}][\ce{H2O2}] \tag5$$

As a result, they have concluded that the simplest kinetic interpretation of this rate law assumes following two rate determining steps: $$\ce{H2O2 + I- ->[k_1^\circ] H2O + IO-} \tag6$$ $$\ce{H2O2 + H+ + I- ->[k_1] H2O + HIO} \tag7$$ And, both steps proceed simultaneously and independently. At $\pu{25 ^\circ C}$, they have reported measured values of $k_1^\circ $ and $k_1$ as $0.69$ and and $10.5$, respectively.


References:

  1. John C. Hansen, "The Iodide-Catalyzed Decomposition of Hydrogen Peroxide: A Simple Computer-Interfaced Kinetics Experiment for General Chemistry," J. Chem. Educ. 1996, 73(8), 728-732 (ODI: https://doi.org/10.1021/ed073p728).
  2. William Sweeney, James Lee, Nauman Abid, and Stephen DeMeo, "Efficient Method for the Determination of the Activation Energy of the Iodide-Catalyzed Decomposition of Hydrogen Peroxide," J. Chem. Educ. 2014, 91(8), 1216–1219 (ODI: https://doi.org/10.1021/ed500116g).
  3. Tomoyuki Tatsuoka and Nobuyoshi Koga, "Energy Diagram for the Catalytic Decomposition of Hydrogen Peroxide," J. Chem. Educ. 2013, 90(5), 633−636 (ODI: https://doi.org/10.1021/ed400002t).
  4. Herman A. Liebhafsky, "The Catalytic Decomposition of Hydrogen Peroxide by the Iodine-Iodide Couple. II & III. The Rate of Oxidation in Neutral, and in Acid, Solution of Hydrogen Peroxide by Iodine," J. Am. Chem. Soc. 1932, 54(9), 3499–3508 (ODI: https://doi.org/10.1021/ja01348a003).
  5. Herman A. Liebhafsky and Ali Mohammad, "The Kinetics of the Reduction, in Acid Solution, of Hydrogen Peroxide by Iodide Ion," J. Am. Chem. Soc. 1933, 55(10), 3977–3986 (ODI: https://doi.org/10.1021/ja01337a010).
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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