Answer to Q 1: The equilibrium oxygen evolution reaction (OER) potential in aqueous solution is dependent on the pH not on the ionic strength of the solution. The relation based on the Nernst equation is: 1.23 - 0.059pH V(SHE). Therefore, if your solution pH is 7, then the OER redox potential will be: 1.23 - (0.059*7) = 0.817 V Vs. SHE.

Two points need to be noted here. 1) In real life OER will not occur at this equilibrium potential. You need an over-potential to kick off the reaction which is dependent on the catalytic activity of your electrode (this case stainless steel), and the IR drop due to the resistance of electrolyte, and other interfaces. 2) The pH of 3.5 wt% NaCl may not be exactly 7 even after prolonged deaeration. The pH will be typically about 5.5 due to dissolution of CO2. Deaeration by nitrogen or an inert gas may drive out the CO2 and increase the pH if carbonic acid has not formed already.

Answer to Q2: There is no relation between OER potential and 5 mA/cm2 apex current density.

The purpose of conducting cyclic polarization is to evaluate the localized corrosion behavior of stainless alloys. In oxidizing environments, these alloys show good passivity. Presence of chlorides in critical concentrations causes passivity breakdown, and the current increases during the anodic polarization signifying transpassive dissolution due to pitting which is a form of localized corrosion. The pits are grown until a current density of 5 mA/cm2 and the potential is reversed to evaluate the repassivation behavior of the material. If the pits are grown very deep, it is difficult to repassivate. Shallow pits can be repassivated. You can get this info from the cross over potential during the reverse scan.

If the material is not susceptible to pitting, then the passivation will continue as you continue the anodic polarization. If the polarization potential exceeds the OER potential, then you may see increase in the current due to oxidation of water. This is not transpassive dissolution due to pitting but just water oxidation. In order to avoid the confusion of seeing current increase due to OER with that of pitting, we limit the apex potential to redox potential of OER.

Answer to Q3: We remove dissolved oxygen to mitigate the oxygen reduction reaction as one of the cathodic reactions during the corrosion tests. It does not prevent oxygen evolution reaction at high anodic potentials, because you are oxidizing the water (see below the reactions) not removing dissolved oxygen.

2H2O = O2 +4H+ + 4e (neutral or acidic)

4OH- = O2 + 2H2O + 4e (alkaline)

The evolution of oxygen at high voltages has nothing to do with oxygen dissolved in solution. At those high voltages, you are oxidizing water into hydrogen and oxygen. You can estimate the voltage at which this will occur with a Pourbaix diagram and the pH of the solution. The oxidation of water occurs at the upper dashed line.

For example, see Figure 2 here:

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