Cathodic protection of active metals

Question

Cathodic protection prevents the removal of positive form the metal surface by applying a negative charge to the metal.

Is is possible to prevent the reaction of an active metal (i.e. sodium) with water by aggressively forcing current into the metal using a DC source? We allow the "water" to be a dilute solution (~1%) of the metal hydroxide. There is also an anode that completes the circuit but its OK if the anode disintegrates.

Edit: I will let the answer stand but it begs one question: If the power supply forces the sodium metal to be at a strongly positive potential, it may end up coated in layer(s) of hydroxide (or even oxide?), which may prevent the water from reaching it. But if the metal is liquid (NaK) then it may cause the metal to have "negative surface tension" making it impossible to maintain a high potential as the capacitance runs away to "infinity". So it's not completely settled.

Answer 1

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:

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

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:

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

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.