# Explain the mechanism for the "Pt"-catalyzed formation of water from "H"_2 and "O"_2?

Jun 7, 2017

This is gone into great detail here:
http://onlinelibrary.wiley.com/doi/10.1002/fuce.200500201/pdf

I will try to distill it down to the main details, but if you find the time and are eager to know the full details, you can go read that article.

DISCLAIMER: VERY LONG ANSWER!

The mechanism has two possible pathways. After reading the abstract, we can conclude that the so-called $\boldsymbol{\text{OOH}}$ formation mechanism is more energetically favorable, so that is what we will cover here. Both happen, but this pathway will occur more easily.

We will indicate adsorbed species as "ads". The overall mechanism (pg. 19 of 23), performed on top of a ${\text{Pt}}_{14.13 .8}$ cluster, is based on the most energetic pathway (as there may be multiple structures for each intermediate):

$\boldsymbol{\left(1\right)} \text{ }$ $\text{H"_2(g) -> 2"H} \left(a \mathrm{ds}\right)$,
$\text{ "" }$ $\Delta {H}_{1} = - \text{14.38 kcal/mol}$

$\boldsymbol{\left(2\right)} \text{ }$ ${\text{O"_2(g) -> "O}}_{2} \left(a \mathrm{ds}\right)$,
$\text{ "" }$ $\Delta {H}_{2} = - \text{11.28 kcal/mol}$

$\boldsymbol{\left(3\right)} \text{ }$ $2 \text{H"(ads) + "O"_2(ads) -> "H"(ads) + "OOH} \left(a \mathrm{ds}\right)$,
$\text{ "" }$ $\Delta {H}_{3} = - \text{13.76 kcal/mol}$

$\boldsymbol{\left(4\right)} \text{ }$ $\text{OOH"(ads) -> "O"(ads) + "OH} \left(a \mathrm{ds}\right)$,
$\text{ "" }$ $\Delta {H}_{4} = - \text{28.26 kcal/mol}$

$\boldsymbol{\left(5\right)} \text{ }$ $\text{OH"(ads) + "H"(ads) -> "H"_2"O} \left(a \mathrm{ds}\right)$,
$\text{ "" }$ $\Delta {H}_{5} = - \text{24.14 kcal/mol}$

$\boldsymbol{\left(6\right)} \text{ }$ $\text{H"_2"O"(ads) -> "H"_2"O} \left(g\right)$,
$\text{ "" }$ $\Delta {H}_{6} = + \text{13.89 kcal/mol}$

$\text{--------------------------------------------------------}$

$\text{H"_2(g) + "O"_2(g) -> "H"_2"O"(g) + "O} \left(a \mathrm{ds}\right)$,

$\Delta {H}_{r x n} = - \text{77.93 kcal/mol}$

$= - \text{326.06 kJ/mol}$

Below, we focus on the particular intermediates with the lowest energy, and we'll add some extra explanation as to what's going on:

1) Suppose the plane of the adsorption sites (assumed horizontal) is the $y z$ plane, and the $x$ axis is perpendicular to it.

Under that convention, it is likely that the platinum $4 {d}_{x z}$ orbital donates electron density into the antibonding ${\sigma}^{\text{*}}$ molecular orbital of ${\text{H}}_{2}$, increasing its antibonding character and thus homolytically cleaving the bond.

Then, the two hydrogen atoms (white) bind in separate spots on the surface.

2) The ${\text{O}}_{2}$ binds to the platinum surface in a four-membered ring upon homolytically cleaving its $\pi$ bond by the same kind of interaction (with ${\text{O}}_{2}$'s ${\pi}^{\text{*}}$) discussed with ${\text{H}}_{2}$ from $\left(1\right)$.

3) One of the adsorbed $\text{H}$ atoms breaks its $\sigma$ interaction with a platinum atom's $4 {d}_{x z}$ orbital and tries to bond in a $\sigma$ fashion to the ${\text{O}}_{2}$.

Likely, the ${\text{O}}_{2}$ breaks one side (the left, let's say) of its $\pi$ interaction, and $\sigma$ bonds with the $\text{H}$ atom, forming an $\text{OOH}$ peroxyl radical.

4) Then, the $\text{OOH}$ radical dissociates into $\text{O}$ and $\text{OH}$ (breaking an $\text{O"-"O}$ peroxide bond). The dissociated $\text{OH}$ then binds back to the surface via the oxygen atom (forming a $\text{Pt"-"O}$ bond), then slightly rotates itself to stabilize.

5) The formed $\text{OH}$ radical then associates with an adsorbed $\boldsymbol{\text{H}}$ atom to form an adsorbed water molecule (bound by the oxygen atom in a $\text{Pt"-"O}$ bond).

6) The water molecule then desorbs from the platinum surface. As for the remaining adsorbed $\text{O}$ atom, it might associate with another $\text{O}$ atom to form ${\text{O}}_{2} \left(a \mathrm{ds}\right)$ (the article goes into a bit further detail about that).