Is a metal hydroxide fully or partly oxidised in insufficient base? Firstly, my question only applies to transition metal compounds with more than one hydroxide group, otherwise it would be more obvious. 
The reason I'm asking is in an effort to oxidise some nickel dihydroxide into nickel oxyhydroxide for practical use in an electrode. I've only come across somewhat laborious methods (any help in this would be appreciated) so I've considered the oxidation like in a NiMH cell (the second equation).
With nickel in a basic solution:
My question could stem from the following situation; If I stir a 2:1 ratio of NaOH:Ni(OH)2 together I would assume the following reaction would be favoured:
$$\ce {4 Ni(OH)2 + 8OH- -&gt; 1O2 + 2 Ni2O3 + 8H2O}$$
But, if I stir a 1:1 ratio of NaOH:Ni(OH)2 and ground the solution:
$$\ce {Ni(OH)2 + OH- -&gt; NiOOH + H2O + e-}$$
*I may or may not be misssing some electrons...I'm just starting out in electro-chemistry. 
Is this assumption true? Why or why not? Am I missing the bigger picture? Cue Gibbs?  I did try it, but with the setup I had, I couldn't check for gaseous oxygen, and unfortunately both Ni2O3 and NiOOH are black and powdery.
Similarly with iron in an acidic solution:
In insufficient acid, which one is what would happen and why?:
$$\ce {Fe(OH)3 + H+ -&gt; [Fe(OH)2]+ + H2O}$$ vs.
$$\ce {Fe(OH)3 + H+ -&gt; Fe^{3+} + 3 H2O + excess \,\,\, Fe(OH)3}$$

I guess the question boils down to; How do reduction / oxidation potentials change with multiple individual reduction / oxidations? 
I hope everything makes sense. If not I will gladly try to explain my question further. 
Thank you for your help!
What you are worried seems to be in your comment, which I'm explaining[maybe not]

Oxidations:


  $$\ce{NiOOH + H2O + e- -&gt; Ni(OH)2 + OH-} \quad E^{\circ}=+0.49\\
\ce{Ni2O3(s) + 6H+ + 2e- -&gt; 2Ni^2+ + 3H2O} \quad E^{\circ}=+1.753\\$$


I suppose if you know about electrode potential you might be able to understand now,If you wish I might explain it, if you can't on your own.