The first reaction (autoprotolysis of water) isn't a redox reaction; the second reaction you describe IS a
redox reaction. So they're different in a very fundamental way.
The hydrogen is being both oxidized (H2 to H+) and reduced (H2 to H-).
Such reactions are called 'disproportionation reactions' and are well known for the halogens. For example,
Cl2(g) + H2O(l) HCl(aq) + HOCl(aq)
But hydrogen has a very low electron affinity. It's much less electronegative than the halogens, and the formation of
a hydride is strongly endothermic (compared to the formation of a chloride or a bromide, which are strongly exothermic).
That makes disproportionation much less likely for hydrogen than for the halogens.
H2 would be an extremely weak acid.
Br2(l) + H2O(l) HBr(aq) + HOBr(aq)
Consequentially hydride ions react violently with anything that contains hydrogen ions. Some hydride salts will
spontaneously combust when exposed to air because they react so exothermically with the water vapor it contains!
So you must be cautious about proposing such a reaction in an H+ rich environment like water. I'd like to know a little more about your measurements. If the reaction occured to any appreciable extent, electrolysis of hydrogen-saturated water would produce hydrogen at the cathode (from reduction
of the hydrogen ions from both the H2 and the H2O) and also at the anode (from oxidation of the hydride
ions). Is this what you tried? The amount of anodic hydrogen produced would be extremely low even if the mechanism you
propose is correct. How did you ensure that the hydrogen produced at the anode was from hydride oxidation and not just hydrogen escaping from the saturated solution?
Author: Fred Senese firstname.lastname@example.org