Dependence of electrochemical reaction rate on overpotential.

The electrode/solution current may be cathodic or anodic according to the flow of electrons which can measure the sign of overpotential i.e deviation from the equilibrium condition the two partial current densities are equal in magnitude and opposite in direction.i.e 

    η(overpotential) = ∆φ-∆φe , ∆φ = potential difference observed by adjusting the outside potential source for the change in current flow.

    ∆φe = potential difference at equilibrium

There will be polarization i.e potential variation from equilibrium

    icath = ianod , at equilibrium.

If icath = ianod = i 0

 r1 = rate of M n+ formation (i cath)

 r2=rate of M formation ( i anod)

  i 0 = exchange current density

                 icath > ianod  

       r2>r1, if E0 for (M n+ /M) at equilibrium

E'-E0(M n+ /M ) = +ve

Polarization leads to deviation of potential from equilibrium because of current flow i.e net charge flow

              icath≠ ianod  

                 i = icath - ianod  

If E is the applied potential

                              i = F. Ka . C R -F. K c. Co

= F.Ba . C R e (-∆G a * /RT) – F. Bc. Co e (-∆G c * /RT)

                             icath >ianod , i >0

We consider this as non equilibrium condition

∆Gc*=-∆G*+F∆ φ, If α = asymmetric factor

∆Gc*=-∆G*+ α F∆ φ

∆G a * =-∆G*+ (1-α )F∆ φ, there will be reduction of e∆ φ

  i = F. Ba. C R . e (-∆G*/RT). e ((1-α )F∆ φ)/RT – F. Bc. Co. e (-∆G*/RT). e ((1-α )F∆ φ)/RT

     η= ∆φ-∆φe

     ∆φ= η+∆φe


Since from Ering activation energy theory

Rate , K= B. e (-∆G*/RT), ∆G* = Gibbs free energy of activation

For a single electron transfer , F= e. NA , for 1 mole , NA =1, F =e

Anodic current, ianod = F. Ka. C R .  

Cathodic current, icath =F. Kc. Co

    C R =Reduction

   Co =Oxidation

Net current density

     i = ianod - icath

Here ∆φ<E or ∆φ>E. at equilibrium ∆φ e = E(Applied potential), K a and CR are anodic rate and reduction concentration.

KC and CO are cathodic rate and oxidation concentration

 ianod ==F. B a . CR .e(F∆φ*/RT). e((1-α )F∆ φe)/RT. e((1-α )F η ))/RT

= i 0. e((1-α )F η ))/RT.

 ianod = icath= i 0

      icath = i 0. e(-α )F η ))/RT,

We know that, i = ianod - icath

         i = i 0 . e((1-α )F η ))/RT- e((-α )F η ))/RT

This is known as B-V Equation.

Tafel equation

To derive Tafel equation we express the exponential factor

 , If η is very small η F/(RT) <<<1, So we neglect the higher power of x

       i = i 0 . (1-((1-α ) η F))/RT)- (1-((α ) η F ))/RT)

              = i 0 . (1+ η F /RT- α η F ))/RT-1+ α η F ))/RT

             = i 0 . η F /RT

If η is positive we neglect the second term of B-V equation.

                i = i 0 . ( e-((1-α ) η F ))/RT) 

i.e ln i = ln i 0 +(1-α η F ))/RT…..A

If i is negative , we can neglect the first term of B-V equation, 

                               i = i 0 . ( e(-α ) η F ))/RT)

             ln i = ln i 0 -(α η F ))/RT……………………..B

Equation A and B are known as Tafel equations

Plot of η versus log i 

This is known as Tafel plot.