The n-factor of hydrogen peroxide undergoing disproportionation to release oxygen is two.
the verdict
INSUFFICIENT LEANING
refutedsupported
the weight of evidence
6 sources for · 0 against
The retrieved sources acknowledge the disproportionation reaction of hydrogen peroxide into water and oxygen, but academic and educational discussions highlight ongoing confusion regarding whether the n-factor is one or two.
Abstract
The preparation of [Mn(saltnOCOPh)Cl]·DMF (H2saltnOCOPh: N,N′-(2-benzoyloxypropane-1,3-diyl)bis(salicylideneamine)) and kinetics and mechanisms of H2O2 disproportionation catalyzed by mononuclear Schiff base manganese(III) complexes, such as [Mn(salen)Cl] (H2salen: N,N′-ethylenebis(salicylideneamine)), [Mn(saltn)Cl] (H2saltn: N,N′-propane-1,3-diylbis(salicylideneamine)), [Mn(saltnOH)Cl] (H2saltnOH: N,N′-(2-hydroxypropane-1,3-diyl)bis(salicylideneamine)), and [Mn(saltnOCOPh)Cl] in N,N-dimethylformamide (DMF), have been investigated. The disproportionation of H2O2 to O2 and H2O proceeds coupled with the redox cycle between the Mn(III) complex and the Mn(IV) intermediate: the first step is the fast equilibrium (Km) of the Mn(III) complex and the Mn(IV) intermediate formed by the reaction of the Mn(III) complex with H2O2, followed by a slow reaction (k1) of the Mn(IV) intermediate with H2O2 to produce O2 and H2O recovering the original Mn(III) complex. The Km values decrease in the following order: [Mn(salen)Cl] (728 mol−1 dm3) >> [Mn(saltnOH)Cl] (28.0 mol−1 dm3) > [Mn(saltn)Cl] (6.28 mol−1 dm3) > [Mn(saltnOCOPh)Cl] (1.83 mol−1 dm3), reflecting an increased distortion along the axis containing the coordination of H2O2 to the Mn(III) complex. On the other hand, the rate constants (k1) fall into the following sequences: [Mn(saltnOH)Cl] (4.29 × 105 mol−2 dm6 s−1) > [Mn(salen)Cl] (1.67 × 105 mol−2 dm6 s−1) > [Mn(saltnOCOPh)Cl] (3.34 × 104mol−2 dm6 s−1) > [Mn(salen)Cl] (6.15 × 103 mol−2 dm6 s−1). In spite of the small Km values for saltnOH, saltnOCOPh, and saltn complexes with the 1,3-diamine ligand, compared to that for the salen complex, the large k1 value for the saltnOH complex strongly suggests stabilization of the transition state for the formation of hydrogen-bondings among the Mn(IV) intermediates and H2O2. Futhermore, the effect of the OH− ion on the H2O2 disproportionation catalyzed by [Mn(salen)Cl] has been reported. On account of the formation of [Mn(salen)OH] coordinated by the OH− ion, the appearance of the reaction path involving not only the Mn(III)–Mn(IV) cycle, but also the Mn(II)–Mn(III) cycle, is shown based on the ESR and visible spectral studies. The activity for the Mn(II)–Mn(III) cycle is 20 times larger than that for the Mn(III)–Mn(IV) cycle.
Heme and nonheme dimanganese catalases are widely distributed in living organisms to participate in antioxidant defenses that protect biological systems from oxidative stress. The key step in these processes is the disproportionation of H2O2 to O2 and water, which can be interpreted via two different mechanisms, namely via the formation of high-valent oxoiron(IV) and peroxodimanganese(III) or diiron(III) intermediates. In order to better understand the mechanism of this important process, we have chosen such synthetic model compounds that can be used to map the nature of the catalytically active species and the factors influencing their activities. Our previously reported μ-1,2-peroxo-diiron(III)-containing biomimics are good candidates, as both proposed reactive intermediates (FeIVO and FeIII2(μ-O2)) can be derived from them. Based on this, we have investigated and compared five heterobidentate-ligand-containing model systems including the previously reported and fully characterized [FeII(L1−4)3]2+ (L1 = 2-(2′-pyridyl)-1H-benzimidazole, L2 = 2-(2′-pyridyl)-N-methyl-benzimidazole, L3 = 2-(4-thiazolyl)-1H-benzimidazole and L4 = 2-(4′-methyl-2′-pyridyl)-1H-benzimidazole) and the novel [FeII(L5)3]2+ (L5 = 2-(1H-1,2,4-triazol-3-yl)-pyridine) precursor complexes with their spectroscopically characterized μ-1,2-peroxo-diiron(III) intermediates. Based on the reaction kinetic measurements and previous computational studies, it can be said that the disproportionation reaction of H2O2 can be interpreted through the formation of an electrophilic oxoiron(IV) intermediate that can be derived from the homolysis of the O–O bond of the forming μ-1,2-peroxo-diiron(III) complexes. We also found that the disproportionation rate of the H2O2 shows a linear correlation with the FeIII/FeII redox potential (in the range of 804 mV-1039 mV vs. SCE) of the catalysts controlled by the modification of the ligand environment. Furthermore, it is important to note that the two most active catalysts with L3 and L5 ligands have a high-spin electronic configuration.
Disproportionation reactions do not need begin with neutral molecules, and can involve more than two species with differing oxidation states (but rarely). Disproportionation reactions have some practical significance in everyday life, including the reaction of hydrogen peroxide, \(\ce{H2O2}\) poured over a cut. This a decomposition reaction of hydrogen peroxide, which produces oxygen and water. Oxygen is present in all parts of the chemical equation and as a result it is both oxidized and reduced. The reaction is as follows:
\[\ce{2H2O2(aq) -> 2H2O(l) + O2(g)} \nonumber \]
Dicussion
On the reactant side, \(\ce{H}\) has an oxidation state of +1 and \(\ce{O}\) has an oxidation state of -1, which changes to -2 for the product \(\ce{H2O}\) (oxygen is reduced), and 0 in the product \(\ce{O2}\) (oxygen is oxidized). Which element undergoes a bifurcation of oxidation states in this disproportionation reaction:
\[\ce{HNO2 -> HNO3 + NO + H2O} \nonumber \]
- Answer
-
The \(\ce{N}\) atom undergoes disproportionation. You can confirm that by identifying the oxidation states of each atom in each species. References
- Petrucci, et al. General Chemistry: Principles & Modern Applications.
On the mechanism of the pseudocatalatic degradation of hydrogen peroxide by lactoperoxidase/iodide.
Hydrogen peroxide is catalytically disproportionated by lactoperoxidase in the presence of iodide ions, Km = 55 microM in 100 mM sodium phosphate, pH 7.00, 25 degrees C. Products formed are water and molecular oxygen. The reaction is competitively inhibited by hydrogen sulfite, Ki = 0.24 mM in 100 mM sodium phosphate, pH 7.00, 25 degrees C. The stoichiometry of the reaction is identical with the corresponding catalase reaction but the mechanism differs. A mechanistic model for lactoperoxidase-iodide dismutation of hydrogen peroxide is discussed.
Published in Acta chemica Scandinavica. Series B: Organic chemistry and biochemistry (1986)
We study how time-dependent optical measurements of spectra, scattering, and imaging can be used to add to the understanding of heterogeneous reactions, compared to work performed using tools developed for homogeneous reactions. Using hydrogen peroxide disproportionation by potassium permanganate as a model reaction, we measure the entire spectrum over reaction time, enabling a clear and useful correlation analysis and assignment of chemical species in heterogeneous conditions. We measure time-dependent dynamic light scattering to study oxygen nanobubble product formation kinetics and equilibrium. We perform macroscopic video-rate reaction imaging and information-theoretic analysis to characterize reaction and transport contributions to the observed signal. To illustrate the differences arising from measuring sample subsets vs the entire system, we integrate stochastic and macroscopic numerical simulations of reaction-diffusion to study homogeneous and heterogeneous reaction conditions. We hope the tools presented here may help understanding other chemical reactions in heterogeneous conditions.
# What is the n factor of H2O2 undergoing disproportionation reaction liberating oxygen
Tags: inorganic-chemistry, redox, oxidation-state
- Score: 9
- Views: 19734
- Answers: 2
- Answered: yes
- Asked by: gurdeep singh (111 rep)
- Asked: 2021-02-26
- Edited: 2022-06-08
- Site: chemistry
- Closed: closed
## Question
There is a lot of confusion about the n-factor of hydrogen peroxide in its disproportionation into water and oxygen.
$$\ce{2H2O2 -> 2H2O + O2}$$
Some sources(1,2) say n-factor of $\ce{H2O2}$ is 2
and other say n-factor of $\ce{H2O2}$ is 1 for this reaction.
Actually my main problem is "the relationship between normality and volume strength of $\ce{H2O2}$". According to my respected teacher:
$$\pu{Normality = \frac{Volume~strength}{5.6}}$$
and
$$\pu{Molarity = \frac{Volume~strength}{11.2}}$$
implying that n-factor of $\ce{H2O2}$ is 2 but I am calculating n-factor of $\ce{H2O2}$ as 1 because the definition of n-factor is "the no of electron gained or lost per mole".
As we can see 2 electrons are participating in this reaction so electron exchanged per mole is 1 which is equal to n-factor. Please clarify.
Sources
https://socratic.org/questions/what-is-equivalent-weight
Everything we examined (6)
This check searched the claim as stated. It did not run a separate search for evidence against it.