Comment on “Radical-mediated proton transfer enables hydroxyl radical formation in charge-delocalized water” by R. Zhao, Q. Zhang, N. Yang, L. Li, Z. Li and C. Cui, Chem. Sci., 2025, 16, 11954

Abstract

Cui and coworkers claim that H₂O˙⁺ and HO˙ are formed spontaneously by lowering the pH to below 4. This is thermodynamically most unlikely. Thus, proton transfer does not involve HO˙.

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Cui and coworkers¹ reported recently that “proton transfer proceeds via a ˙OH⋯H⁺⋯H₂O intermediate, enabling a radical-mediated proton transfer pathway” and “that thermal energy at room temperature is sufficient to drive the formation of ˙OH radicals in acidic solutions”.¹ They hypothesize that HO˙ originates from H₂O˙⁺[dihydridooxygen(˙⁺) or oxydaniumyl], reaction (1):

H₂O˙⁺ → HO˙ + H⁺ (1)

The fate of the electron that was removed from H₂O is not mentioned. Possibly, Cui and coworkers assume that a hydrated electron (e⁻_aq) is formed, as has been suggested elsewhere,²

2H₂O → H₂O˙⁺ + e⁻_aq (2)

In any case, mass balance was not preserved. In support of their hypothesis, they claim to have detected a crown ether–H₂O˙⁺ complex.

Science progresses by building on what has been established. For over 60 years, species like hydrated electrons, hydroxyl radicals, and other short-lived species have been studied; their thermochemical and kinetics properties are known and have been compiled (https://kinetics.nist.gov/solution/).^3,4 When one consults this body of knowledge, one must conclude that formation of H₂O˙⁺ and HO˙ as proposed is thermodynamically most unlikely. One may, of course, question this body of knowledge, but Cui and coworkers¹ do not do so. They acknowledge that HO˙ is very oxidizing, quoting the standard electrode potential (E°) at pH 14, but not the implication that HO˙ is energetically difficult to obtain from water. Two minor criticisms: (1) that one-electron potential should be called a reduction potential, not an oxidation potential,⁵ and (2), it cannot be compared to the two-electron reduction potential of the Cu²⁺/Cu couple. Below I show that H₂O˙⁺ and HO˙ are unlikely to have been present in the solutions examined by Cui and coworkers.¹

The species H₂O˙⁺ and e⁻_aq are formed by ionizing radiation.⁶ H₂O˙⁺ has a very short lifetime τ of 46 fs, because reaction (1) is very fast, k = 2 × 10¹³ s⁻¹.⁷ Whether H₂O˙⁺ can be scavenged depends on how far it can diffuse. To provide an upper limit on that distance, I make the assumption that H₂O˙⁺ has the same diffusion coefficient (D) as H⁺, 9.31 × 10⁻⁵ cm² s⁻¹ at 25 °C.⁸ During 2τ, after which 14% of H₂O˙⁺ is left, the root mean square displacement, calculated with (6Dt)^1/2 is 72 pm, or half the radius of a water molecule. Thus, unless produced directly next to a scavenger,⁹ H₂O˙⁺ decays to HO˙ and H₃O⁺. The conclusion is clear: H₂O˙⁺ cannot be trapped or captured.

Little is known about the thermodynamic properties of H₂O˙⁺. It has never been included in reviews of O₂ and its reduced forms,^4,10,11 because of its short lifetime. E°(H₂O˙⁺/H₂O) has been estimated at >4 V,⁹ which makes H₂O˙⁺ the most oxidizing species known. Use of a thermochemical cycle that consists of the reactions (3) and (4), see Fig. 1, allows one to make a more precise estimate of E°(H₂O˙⁺/H₂O) and of Δ_fG°(H₂O˙⁺_aq):

H₂O˙⁺ + e⁻ → H₂O  ΔG° = −900 kJ mol⁻¹ (3)

½H₂ → H⁺ + e⁻  ΔG° = +423 kJ mol⁻¹ (4)

Reaction (4) is necessary because standard electrode potentials are referenced to the normal hydrogen electrode.⁵ This approach is identical to that described by Phillips and Williams.¹² To provide a lower limit on E°(H₂O˙⁺/H₂O), I assume that H₂O˙⁺ is hydrated, that is, the solvating H₂O molecules are oriented to H₂O˙⁺ in an energetically most favourable way. Whether, or how fast, H₂O˙⁺ becomes hydrated, is a topic for another time. The standard Gibbs hydration energy of H₂O˙⁺ is unknown. Because the radius of H₂O is close to that of K⁺, I use the standard Gibbs hydration energy of the latter, −326 kJ mol⁻¹, for H₂O˙⁺.¹³ The ionisation potential of H₂O is 12.621 eV,³ and the hydration energy of H₂O is −8.6 kJ mol⁻¹.¹⁴ For reaction (4) I use the Gibbs hydration energy of H⁺, −1098 kJ mol⁻¹,¹⁵ the ionisation potential of H˙, 13.60 eV,³ and 1/2 of the Gibbs dissociation energy of H₂, +203 kJ mol⁻¹.¹⁴ The sum of the energetics of reactions (3) and (4) is −477 kJ mol⁻¹, which results in E°(H₂O˙⁺/H₂O) = +4.95 V and Δ_fG°(H₂O˙⁺_aq) = +240 kJ mol⁻¹. Both ionisation potentials are enthalpies, but as used, the two errors caused by neglect of the respective entropies are expected to cancel. Given the assumption that the Gibbs hydration energy of H₂O˙⁺ is close to that of K⁺, I assign an error of 0.10 V to E°(H₂O˙⁺/H₂O) and an error of 10 kJ mol⁻¹ to Δ_fG°(H₂O˙⁺_aq). Thus, reaction (1) is indeed very favourable, −214 kJ mol⁻¹. The implication of the value of Δ_rxn1G° may not be immediately clear: if one imagined a solution that was 1 M in H⁺ and HO˙ each, 2.2 molecules of H₂O˙⁺ would be present at equilibrium in 1 × 10¹⁴ liter of water. This number is a maximum, because of the assumption that H₂O˙⁺ is solvated. If it is not, the solvation Gibbs energy of −326 kJ mol⁻¹ disappears from the left of Fig. 1 which results in E°(H₂O˙⁺/H₂O) ≅ +8.3 V, and Δ_fG°(H₂O˙⁺_aq) ≅ +570 kJ mol⁻¹.

Fig. 1 Thermochemical diagram for the calculation of E°(H₂O˙⁺/H₂O). It shows the energetics in kJ mol⁻¹ of reactions (3) and (4), which are −900 kJ mol⁻¹ and +423 kJ mol⁻¹, respectively, as calculated from data cited in the text. The sum of the energetics of these two reactions yields E°(H₂O˙⁺/H₂O) = +4.95 V via ΔG° = −nFΔE°, in which n is 1.

The absorption spectrum and properties of e⁻_aq. were described already in the early 1960s.^16,17 Δ_fG°(e⁻_aq) = +278 kJ mol;⁴ with Δ_fG°(H₂O˙⁺) = +240 kJ mol⁻¹, one calculates that the comproportionation reaction, reaction (2), is endergonic by +922 kJ mol⁻¹, again, a best case scenario based on H₂O˙⁺ being hydrated. Thus the formation of the hydrated electron does not compensate, by far, for the one-electron oxidation of water, as was suggested by Head-Gordon and coworkers.¹⁸ The statement that “pure water contains 0.3 M of fully charged water molecules”² is incorrect, as are the results of the ab initio calculations of Cui and coworkers¹ on the presence of H₂O˙⁺ in acidified water. In contrast, my thermochemical analysis is straightforward and yields accurate results.

I will not address in detail the evidence that, according to the authors,¹ supports the formation of HO˙ and H₂O˙⁺ at room temperature in acidified water, except to point out, again, that H₂O˙⁺ cannot be scavenged; that the reaction of HO˙ with benzoate¹⁹ and other aromatic compounds^20,21 leads to hydroxylated products; decarboxylation is a very minor reaction, and that the oxidation of I⁻ may be due to traces of O₂. The latter reaction takes place near and below pH 4, a common observation during iodometric titrations. Deoxygenation by passing Ar through a solution is rarely effective. To achieve full deoxygenation, use of the Schlenk technique and a glove box are essential. Other explanations for the results of the scavenging experiments of HO˙ must be found.

In conclusion, and contrary to the report of Cui and coworkers¹ and others,^2,22 the short lifetime of H₂O˙⁺ prevents it from playing a role in any reaction. The Grotthuss mechanism of proton transfer²³ does not involve HO˙.

Author contributions

WHK is the sole author.

Conflicts of interest

No competing interests.

Data availability

No experiments were carried out. All results shown were calculated from thermodynamics and kinetics data available in the literature. All sources have been referenced.

Acknowledgements

I thank Dr P. L. Bounds for discussions on hydration of short-lived species.

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