# Water on Actinide Dioxide Surfaces: A Review of Recent Progress
## 3 Water on AnO\({}_{2}\) Surfaces
### Theoretical Calculations
#### 3.2.2 H\({}_{2}\) production from H\({}_{2}\)O splitting  
The formation of H\({}_{2}\) owing to the interaction of H\({}_{2}\)O with AnO\({}_{2}\) through radiolytic process or chemical reactions can lead to potential pressurization of the containers of nuclear fuels. The formation of H\({}_{2}\) on AnO\({}_{2}\) surfaces owing to chemical reactions with H\({}_{2}\)O has rarely been studied in theoretical calculations [54; 55]. [PERSON] et al. first systematically investigated the formation and desorption mechanism of H\({}_{2}\) on reduced UO\({}_{2}\) (111) surface [54]. They found that H\({}_{2}\) formation can occur on UO\({}_{2}\) (111) surface with the oxygen vacancies [54]. Figure 10 shows the three possible reaction pathways [54]. The first formation pathway in Figure 10a involves the direct dissociation of H\({}_{2}\)O at a vacancy site, forming O-H-H structure on the surface, followed by the desorption of H\({}_{2}\) from the surface and oxygen atom healing the vacancy. This pathway has an energy barrier of 0.56 eV, and the overall reaction is exothermic by 2.62 eV [54]. The second pathway in Figure 10b involves the dissociation of H\({}_{2}\)O on the surface, forming hydroxylated surface, and then the combination of the hydrogen atoms from the hydroxyl radicals to form H\({}_{2}\) molecules. This pathway presents a large energy barrier of 1.39 eV and an overall exothermic energy of 2.56 eV. For the third pathway in Figure 10c, there are excess hydroxyl radicals near the vacancy site. The dissociation of H\({}_{2}\)O will result in a hydroxyl radical that occupies the vacancy site, and the formation of O-H-H structure. This reaction pathway is also exothermic by 1.98 eV and has an energy barrier of 0.53 eV [54]. A comparative study of H\({}_{2}\) formation on reduced ThO\({}_{2}\), UO\({}_{2}\), and PuO\({}_{2}\) (111) surfaces has been performed to show the difference in the H\({}_{2}\) formation across the AnO\({}_{2}\) series [55]. It is found that the formation of H\({}_{2}\) from catalytic splitting of H\({}_{2}\)O is endothermic for all three AnO\({}_{2}\) (111) surfaces evaluated (Table 7). In contrast, H\({}_{2}\)O can readily dissociate on the reduced ThO\({}_{2}\), UO\({}_{2}\), and PuO\({}_{2}\) (111) surfaces, while the thermodynamics of H\({}_{2}\) production is significantly different [55]. On reduced ThO\({}_{2}\) (111) and UO\({}_{2}\) (111) surfaces, the H\({}_{2}\)O adsorption and H\({}_{2}\) production is exothermic and releases 3.55 eV and 3.11 eV, respectively (Table 7). Although this process releases 0.31 eV on reduced PuO\({}_{2}\) (111) surface, there is a larger energy barrier for the H\({}_{2}\) production. This suggests that the protons of H\({}_{2}\)O will stay in the hydroxylated form on PuO\({}_{2}\) (111) surface, instead of forming H\({}_{2}\) molecule [55]. The high reactivity of H\({}_{2}\)O with reduced ThO\({}_{2}\) surface is attributed to the lone electron pair at the vacancy site on ThO\({}_{2}\) that can readily participate in the chemical reactions with H\({}_{2}\)O. On the contrary, the excess electrons are more tightly bound to the Pu 5\(f\) orbitals on reduced PuO\({}_{2}\) surfaces, thus making it harder to participate in the reaction with H\({}_{2}\)O molecule. \begin{table}
\begin{tabular}{c c c c c c c} \hline \hline \multirow{2}{*}{**In Units of eV**} & \multicolumn{2}{c}{**Perfect**} & \multicolumn{3}{c}{**Reduced**} \\ \cline{2-7}  & **ThO\({}_{2}\)** & **UO\({}_{2}\)** & **PuO\({}_{2}\)** & **ThO\({}_{2}\)** & **UO\({}_{2}\)** & **PuO\({}_{2}\)** \\ \hline [PERSON] et al. [67] & - & 0.53 & 0.40 & - & - & - \\ \hline [PERSON] et al. [53] & - & - & - & - & 2.23 & 2.10 \\ \hline [PERSON] et al. [54] & - & 0.61 & - & - & 2.20 & - \\ \hline [PERSON] et al. [38] & 0.57 & - & - & - & - & - \\ \hline [PERSON] et al. [64] & - & 1.08 & - & - & 1.88 & - \\ \hline [PERSON] et al. [42] & - & 0.85 & - & - & - & - \\ \hline [PERSON] et al. [55] & 0.56 & 0.62 & 0.58 & 2.63 & 1.90 & 2.44 \\ \hline \hline \end{tabular}
\end{table}