Oxidation Numbers Ambiguous

[horizon]

The solution to a chem question says that in that the following reaction...
H-(aq) + H₂O(l) → OH-(aq) + H₂(g)
...is a redox reaction because the oxidation number of H increases (from -1 in H^- to 0 in H₂) and also decreases (from +1 in H₂O to 0 in H₂).

Why is it that the  -1 in H^- increases to 0 in H₂ and not to +1 in OH-?

[Starlight]

H- itself is -1 oxidation state

For OH-

(-2) + (H) = -1

therefore H = 1

i.e. oxidation state for H- increases from -1 to 1 (oxidation)

for H2

In H20:

2(H) - 2 = 0

therefore H= 1

In H2 = 0

therefore the oxidation number for H2 decreases from 1 to 0 (reduction)

You have to realise that there are 2 seperate processes occuring, reduction + oxidation= redox

[charmanderp]

The OH- group is present because H2O donates a proton (H+) to H-. H- (which originates from a metal hydride which has been dissolved in water) easily attracts protons from the water it is dissolved in because of its negative charge.

Therefore, H2O has been reduced to form H2 It can't have been reduced to OH-, otherwise the oxidation number of H would not change (it would be +1 in both cases). H- has been oxidised (essentially augmented, it has lost an electron but gained a proton) to form H2.

[horizon]

Sorry, I don't think I made it quite clear.  I was wondering why the solution says that the oxidation number in H- increases from -1 to 0 in H2 and not +1 in OH-, how do we know it is between H- and H2 and not H- and OH-?

[Somye]

Okay, so like charmanderp said, let's think about how the OH group originated from the reactants of H2O and H-. Now, ovbiously, the H2O group has had had to donate a proton (H+) in order to form the OH group, and hence the OH corresponds to the H20. This H+ bonds with the H- in forming H2, and hence your redox process is evident.

For what you're saying to be true, the H- would have had to accept an oxygen atom to form the OH group, which clearly did not happen