Cathodic protection of active metals
It is unlikely that a potential applied to a sodium electrode in water would prevent the spontaneous oxidation of sodium by water.  Consider the following reduction potentials:

$$\ce{2Na+ + 2e- -&gt; 2Na(s)}\ E^0=-2.71\ \mathrm V\tag{1}$$ 
$$\ce{2 H2O + 2e- -&gt;H2 + 2 OH-}\ E^0=-0.83\ \mathrm V\tag{2}$$

The reduction of water will always occur before the reduction of sodium ions.  If we reverse equation (1) so we can combine the two and get the full redox equation for your system, we get:

$$\ce{2 Na(s) + 2 H2O &lt;=&gt; 2Na+(aq) + H2 + 2 OH-}\ E^0= 1.88\ \mathrm V\tag{3} $$

A positive potential indicates a spontaneous reaction under the given conditions so thermodynamically, it will be near impossible to prevent the reaction from being spontaneous.

How impossible?  We can do a thought experiment and ask the question at what pH would this reaction be non spontaneous? In other words, when would the Nernst Equation for this system evaluate to zero:

$$E=E^0-\frac{RT}{nF}\ln\frac{\text{Products}}{\text{Reactants}}\tag{4}$$

Entering our values, converting Ln to Log and substituting $\mathrm{pH} = -\log[\ce{H}]$ (and assuming all other concentrations are fixed at 1 M or 1 atm):

$$E=1.88+0.059(14-\mathrm{pH})\tag{5}$$

The reaction would not be spontaneous until the pH reached 45 units.  Something that can't be done in water.