What happens when oxidizing Ni(OH)2 with a strong base?

In an aqueous solution, Nickel oxyhydroxide is oxidised with a 1:1 ratio of aqueous Sodium hydroxide. My goal is to produce Nickel oxyhydroxide ($\ce{NiOOH}$).

Strong Base: $\ce{NaOH + H2O \leftrightharpoons Na+ + OH-}$
Add that to some stirred $\ce{Ni(OH)2}$:

$$\ce{ Ni(OH)2 + OH- \leftrightharpoons NiOOH + H2O + e-}$$

This is where things get difficult; wouldn't the spare electron be used by the sodium ions to make $\ce{Na (s)}$?:

$$\ce{ Na+ + e- -> Na (s) }$$

But then, $$\ce{Na(s) + H2O \leftrightharpoons NaOH }$$

That just brings everything back to the beginning.

I guess the real question is: will the initial oxidation of the $\ce{Ni(OH)2}$ happen? If not, how can I force it? I initially thought of just grounding the solution, as odd as that sounds.

Thanks for your help!

P.S. I'm a 12th grade high school student.

• @JohnSnow In $\ce{...}$ you can simply substitute \leftrightharpoons with <=> and it will render correctly. – Martin - マーチン Dec 15 '14 at 4:27

$\ce{Na (s)}$ ?: $\ce{Na+ + e- -> Na (s)}$
This reaction is, you might be knowing, the reverse of one of "the best" known favourable reaction, as the data suggests: $$\ce{Na->Na+ +e-}\tag{E^\circ=2.71\;V}\\ 2.71\;V\equiv-261.475\;\text{kJ/mol}\equiv1.0938\times10^{50}(\text{rate})$$ Just imagine the reverse reaction.
But then, $\ce{Na(s) + H2O -> NaOH }$