Procedure:

If both couples (sample and titrant) included within titration are reversible, we get curve of Figure. At the last point the current is zero or near zero. Eg:  Fe (II) solution in sulphuric acid medium titrated along with standard Ce(IV) solution.  Fe (II) gets oxidized to Fe (III) as:

Fe2+ ↔ Fe³⁺   + 1 e^-

and Ce (IV) reduced to Ce (III) as :

Ce⁴⁺ + 2e ↔   Ce³⁺

If the potential difference applied among two electrodes is about 0.2 V no current will be flow.

Because the potential difference is extremely small than the potential difference actually needed between Fe(II), Fe(III) couple and Ce(IV), Ce(III) couple, no reaction takes place and no current will flow before the titration started. Only while the potential difference among two electrodes is about 0.7 V, the oxidized form Fe (III) at anode will decrease back to Fe (II) at the cathode. But in titrations while Ce(IV) solution is added from the burette, equivalent amount of Fe(II) from the electrolytic cell will be generating Fe(III) and could be reduced at the cathode.  In this, Fe (III) is produced not through electrode reaction of oxidation of Fe (II) but through the reaction along with Ce(IV) that could be reduced to Fe(II) at the cathode. Because the couple Fe (II) and Fe (III) is present in solution, Fe (III) is reduced at cathode and Fe (II) is oxidized at anode. This is since the Fe (III) - Fe (II) couple is reversible and even little potential difference among two electrodes is sufficient while both are in the solution.  As more and more Ce(IV) solution is added from the burette more and more Fe(III) being produced and Fe(II) concentration in the solution is reduced. So, along with more Fe (III) ion being produced, the current will be rising in proportion until the concentration of Fe (III) ion equals in which of Fe (II) ion. If the titration is continued additional, the current will decrease since the amount of Fe (II) ion available for reaction at the anode becomes smaller.

At last, while all Fe (II) is erased through titration with Ce(IV), at the equivalence point, the current will be almost zero because no Fe(II) ion is present in the solution to oxidize at the anode. At this stage, cathode will reduce Ce(IV) ion rather than Fe(III) ion.  This is because its couple Fe (II) is not existing in solution and anode will be able to oxidize Ce(III) ion. Within more and more Ce(IV) ion added,  current further increases.