What happens when oxidizing Ni(OH)2 with a strong base?
This is where things get difficult; wouldn't the spare electron be used by the sodium ions to make 
  $\ce{Na (s)}$ ?: $\ce{Na+ + e- -&gt; Na (s)}$


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


  But then, 
  $\ce{Na(s) + H2O -&gt; NaOH }$


This then would never actually then happen. Intersetingly see this