What is the voltage dependence of the thickness of oxide coating through anodising?

In titanium anodising, solid titanium reacts with the oxygen formed at it's surface (to form titanium(IV) oxide or $$\ce{TiO2}$$) through the electrolysis of water. Experimentation and research has shown that the thickness of the coating is dependent on voltage and not current. The potential difference obviously plays some part in the reaction then (besides the water electrolysis) due to the fact that merely placing titanium in an environment with more oxygen gas does not affect the thickness of the naturally occurring oxide, but I am struggling to find out how (through both thought and research) the voltage affects the depth of the coating formed.

I understand that the actual oxidation reaction can be written as:

\begin{align} \ce{Ti &-> Ti^{4+} + 4e-}\\ \ce{O2 +4e- &-> 2O^{2-}} \end{align}

Which are added together to get:

$$\ce{Ti + O2 -> TiO2}$$

I also understand that redox reactions usually include a reduction potential or $$E^0$$ value. Now this may or may not be relevant but it is the only related value I can find with a voltage associated. I have also thought that as the coating of $$\ce{TiO2}$$ builds up, the resistance of the circuit would increase. I just cannot understand how or why the voltage affects the depth (Is this the right word, as it is converting not depositing?) of the formed oxide layer.

So with that my question is more like a request for any insight/resources that could help with my understanding of the voltage-dependency of the depth of the formed oxide layer (specifically titanium, but I'm assuming it would be relatively easy to generalise from other metals such as aluminium or niobium (which I have also searched for)).