Mass Transport Limited Reactions:

Now consider an electrode immersed in an electrolyte that contains two potential reactants, one described by an exchange current like that shown for i0 = 1 and the other having an exchange current like that shown for i0 = 10-4.  Also, let there be a limited amount of the first reactant.  This example might be representative of oxygen reduction  or oxide reduction.  When a cathodic current pulse is applied to the electrode, charge is added to the electrode, the potential across the electrode/electrolyte interface becomes more negative as the double layer capacitance charges, the Fermi level for the electrons in the metal begins to increase, but no significant current flows until the overpotential becomes more negative than –7 units.  As current begins to flow, the first reactant is reduced and the concentration at the electrode/electrolyte interface (Co(x=0)) begins to diminish.  If the current driving the electrode large and can not be sustained by electron exchange with the reactant the concentration of the reactant at the interface, Co(x=0), decreases until
1.)  the availability of the reactant becomes mass transport limited (the case of dissolved oxygen). The equation describing the current under mass-transport limitations is in the lower portion of the panel, mo is the mass transfer coefficient for the oxidized species.  This equation uses a linear approximation for the concentration gradient at X=0, or  2.)  the reactant becomes completely exhausted (the case of surface oxide reduction).  If current continues to be applied under these conditions the potential across the electrode/electrolyte interface begins to increase until another electron transfer process can be found to support the applied current, in this example the reaction indicated by an exchange current i0 = 10-4.  If this reaction is not mass transport limited, as is the case for water reduction, the interface overpotential stabilized at a constant value as long as the current is applied, - 20 units in this example.  
Now consider what happens when the current pulse is terminated, assuming we continue with the previous example where the overpotential is at – 20 units.  The double layer capacitance of the interface is charged to –20 units at termination of the current pulse.  The overpotential will discharge rather quickly to –15 units because the second reactant (water) is still available.  If the first reactant, which was mass transport limited, is available the overpotential can drift to – 6 units rather quickly, but more slowly than the decay observed with water reduction, because the potential on the double layer capacitance will drive that reaction until there is not sufficient overpotential to drive oxygen reduction in this example.  If the first reactant had been depleted, as in the case of reduction of oxide, then the overpotential would slowly decay to zero, the equilibrium overpotential for the redox couple.  
In the case of two current pulses as might be seen when a train of monophasic pulses is applied to an electrode of a neural prosthesis.  If a second negative current pulse were to be applied to the electrode before there was sufficient time for the overpotential to decay to zero the electrode overpotential would begin at a value negative to zero rather than zero.  It may be even possible that the second pulse comes when the overpotential is in the range that is not sufficiently negative to drive the second reaction (“B”) but while the overpotential is still capable of driving the first reaction (“A”).  If a train of current pulses were to be applied at this rate, reaction A would continue to run throughout the duration of the train application but at the mass transport limited rate.  
Now consider a case of two current pulses, the first is cathodic, negative, and the second is anodic, positive, and the second pulse begins on the falling edge of the first pulse.  The action of the second pulse will be to remove the charge on the double layer capacitance and there by bring the overpotential to a value that would not allow either reaction “A” or “B” to take place.  Now a question is, how much charge would be required to return the overpotential to h = 0?  If you said less than the charge in the first pulse, you are correct.  If you injected more you would push h to positive values because all of the charge injected in the first pulse either went into charging the double layer or in to reactions “A” and “B” which are not reversible.