Polarographic Equation:

Let's take a common electrode reaction in that any procedure of reduction at the cathode is accompanied through a simultaneous oxidation as free electrons cannot exist in solution.

Ox + ne ↔ Re d

An electrode potential for such reversible reaction is given through

E = E⁰ + (RT/nF)ln (a_ox/a_red)                                                           ... (1)

aox and ared are the activities of the oxidant and the reductant respectively as they exist at the electrode-solution interface (denoted through the subscript 'o' in Eq. 1).  By substituting all constants the equation becomes

E = E⁰ + 0.059/n log [Ox]0/[Re d]0                                              ... (2)

Before the commencement of the polarographic wave only a small residual current flows and the concentration of electroactive material is same in the whole solution.  As soon as the applied e.m.f. exceeds the decomposition potential of the electroactive material, some of the reducible substance at the electrode-solution interface is reduced and its concentration starts to decrease.  Because there is a concentration gradient among the concentration of the substance at the interface and the bulk of the solution some ions will move in from the bulk of the solution to the interface through means of diffusion.  The current observed depends upon the rate of the diffusion of oxidant and is given by

i = K₁ ([Ox] - [Ox] 0)                                                                                                                       ... (3)

so that according to Ilkovic equation K₁ is equivalent to 607 n _ox D _ox  ^1/2m^2/3t^1/6 .

While the current attains the limiting value represented through the diffusion current plateau the concentration of oxidant at the electrode-solution interface will be essentially zero, as it is reduced as soon as it reached the interface from the bulk of the solution. The current therefore is i_d and Eq. 8.7 might be written as,

i = K₁[Ox] = i_d                                                                                                               ... (4)

From Eq. (3) & (4) it follows that

[Ox]0 = (i_d - i ) / K₁                                                                                                 ... (5)