Dr.A.DAYALAN, Former Prof & Head, B.V Eq (03-Different Over-potential,η)
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M.Sc (Chemistry)
ELECTRODE KINETICS – ELECTRODICS-I
03-DISCUSSION OF BUTLER-VOLMER EQUATION
For different values of 
Case-(i) : = 0 ; BV equation reduces to Nernst equation.
kc
A+e
ka
D
(kc = Cathodic & ka = Anodic rate constants]
At equilibrium (the system is not disturbed by external force i.e., no external emf),
the rates forward and reverse reactions are equal i.e., no net reaction .Therefore there is no
net current. i.e., i = ia-ic = 0
Therefore, (ia / ic ) = 1 ……………………………………………………..(1)
But, ic= nF(kbT/h) e-∆G*chem./RT e-∆e F/RT e -F/RT [A] = nF kc e -∆e F/RT[A]
ia= nF(kbT/h) e∆G*chem./RT e(1-)∆e F/RT [D] = nF ka e (1-)∆e F/RT[D]
Therefore, (ic / ia) = (kc/ka) e -∆e F/RT {[A] / [D]} = Ke -∆e F/RT{[A] / [D]}= 1
Hence, -∆e F / RT = -ln K + ln {[D] / [A]} ………………………………….(2)
But, ∆Go= -RT ln K = - nFE = -∆oe F
Hence, -RT ln K = -∆oe F
Therefore, substituting for ln K = -∆oe F/ RT in equation- (2) & rearranging we get
i.e., ∆e = ∆oe – (RT/F)ln {[D] / [A]} Nernst Equation
Case-(ii):   0.01 V, Low field approximation – LFA (Concept of polarisabilty)
i = 2io Sinh(F/2RT) (BV equation for  = 0.5)
= 2io (F/2RT); [Sinhx = x if x1 ; NB: Use a calculator, calculate & see]
 = (RT/ ioF) i similar to Ohm’s Law(V=IR). Hence, resistance  = RT/ ioF
Dr.A.DAYALAN, Former Prof & Head, B.V Eq (03-Different Over-potential,η)
∂
RT
---- = ----- = ρ
∂i
ioF
0 as io
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∞ (Non polarisalble)
Despite the passage of current across the electrode, the polarisabilty, (∂/∂i) remains
constant, the over potential tends to be zero & the interface remains at equilibrium.
On the other hand,
ρ
0 as io
∞ (Polarisalble)
The potential departs from equilibrium even for a small passage of current across the
electrode, the polarisabilty, (∂/∂i) tends to infinity
& the interface remains away from
equilibrium.
UPPER LIMIT OF LFA
F/2RT << 1 ; i.e.,  << 2RT/F
or  << 0.05 V (Best < 0.01)
@ For an electrode to be non-polarisable,  = 0.
This will be possible only if io = infinity which is not.
@ Hence, all electrodes show some degree of polarization.
NB: Greater the value of io greater is the non-polarisability.
Exchange current densities, io and symmetry factor,  for some electrodes at 25°C
System
Medium
Mercury
H+/H2
H2SO4
7.9 x10-13
0.50
Lead
H+/H2
H2SO4
5.01 x 10-12
-
Nickel
H+/H2
H2SO4
6.3 x 10-6
0.58
Tungsten
H+/H2
H2SO4
1.25 x 10-6
-
Platinum
H+/H2
H2SO4
7.9 x 10-4
-
Gold
H+/H2
H2SO4
2.51 x 10-4
-
Mercury
Cr3+/Cr2+
KCl
1 x 10-6
-
Platinum
Ce4+/Ce3+
H2SO4
3.98 x 10-5
0.75
Iridium
Fe3+/Fe2+
H2SO4
1.58 x 10-3
-
Rhodium
Fe3+/Fe2+
H2SO4
1.73 x 10-3
-
H2SO4
H2SO4
KCl
2.51 x 10-3
6.3 x 10-3
0.58
Platinum
Fe3+/Fe2+
Palladinum Fe3+/Fe2+
Calomel
Hg, Hg2Cl2

io

Metal
Highly polarisable. Does not allow charge (e-) to pass through
-
Dr.A.DAYALAN, Former Prof & Head, B.V Eq (03-Different Over-potential,η)
Discharge of H+ is difficult on Hg surface
 = (RT/ ioF) i Ohm’s Law. Resistance  = RT/ ioF
Non-polarisable.  = 0 : i.e., io = infinity, which is not-All electrode are polarizable . Lesser
the value of io greater is the polarisability
.:. Hg/H2/H+ electrode-Highest degree of polarisability (it does allow charge to flow through).
Calomel electrode is an example of a typical non-polarisable electrode.
Case(iii): 
= 0.01- 0.1 V: B.V Equation to be used as it is
TAFEL EQUATIONS
Case-v):

> 0.1 V High field approximations. HFA (Tafel Equations)-
Experimental determination of io and 
v(a)  is highly positive, i = ia = io e(1-)F/RT [Anodic]
Plot i vs +ve 
ln i = (1-)F/RT + ln io
Plot ln i vs +ve  values. Evaluate, io & (1-)
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Dr.A.DAYALAN, Former Prof & Head, B.V Eq (03-Different Over-potential,η)
v(b)  is highly negative, i = ic = io e -F/RT [Cathodic]
ln i = -F/RT + ln io
Plot ln i vs -ve  values. Evaluate, io & 
UPPER LIMIT OF HFA
e F/2RT > e -F/2RT
Let e -F/2RT < 1 % e F/2RT < 10-2 e F/2RT
10-2 e F/2RT > e -F/2RT
e F/2RT > 100 e -F/2RT
F/2RT > ln100 -F/2RT
F/RT > ln100
 > (RT/F)ln100
 > 0.25V ( ok even for  > 0.1)
HW: Work out for
(i) e -F/2RT < 5 % e F/2RT < 10-2 e F/2RT
(ii) e -F/2RT < 0.1 % e F/2RT < 10-2 e F/2RT
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Dr.A.DAYALAN, Former Prof & Head, B.V Eq (03-Different Over-potential,η)
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ALL THE ABOVE CASES for OVER POTENTIALS CAN BE SUMMARISED AS FOLLOW
DIFFERENT CASES
A comparative study of the plot of i vs  , for all possible , indicating ia.
ic & i = ia – ic (repeated)
NB:
(i) ic or ia will be zero only at infinite magnitude of 
(ii) ic = ia = io at equilibrium.  = 0
(iii) Deviations occur only after equilibrium depending on the value of 
(iv) ia will be zero (no anodic tendency) only at -ve infinite potential
(v) Similarly ic, win be zero (no cahodic tendency) only at +ve infinite potential
Dr.A.DAYALAN, Former Prof & Head, B.V Eq (03-Different Over-potential,η)
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CONDITION FOR ELECTRODE DEPOSITION
∆e = Equilibrium potential for the process considered
 = ∆ - ∆e = - ve cathodic process
∆e = Cathodic equilibrium potential;
 = ∆ - ∆e = + ve anodic process
∆e = Anodic equilibrium potential;
(a) Consider the discharge (reduction) of aq. 1.0 M Cu2+
 = ∆ - ∆e = ∆- 0.34 < 0 for cathodic; ∆ < 0.34V
i.e., the discharge of Cu2+ can occur even at no applied or +ve applied potential but less
than 0.34 V w.r.t SHE
H.W: Calculate the applied potential for the discharge of 0.01M Cu2+.
(Hint: Use Nemst eq for getting ∆e).
(b) Consider the discharge (reduction) of aq. 1.0 M Na+
 = ∆ - ∆e = ∆- (-2.71) = ∆ +2.71 < 0 for cathodic;
∆ < - 2.71 V
i.e., the discharge of Na+ can occur only at -ve applied potential and that too less than
- 2.71 V.
H.W: Calculate the applied potential for the discharge of 0.01M Na+.
(Hint:Use Nernst eq for getting ∆)
(c) Consider the following process at 1.0 M aq solutions (reduction)
½ Cl2 + e
Cl - ; SRP = +1.36 V ; Cathodic  must be –ve
 = ∆ - ∆e = ∆ - (+1.36) = ∆-1.36 < 0 for cathodic;
∆ < 1.36 V i.e., the reduction of Cl2 w.r.t SHE can occur even without any
applied potential or at any -ve applied potential.
Cl-
½ Cl2 + e; SOP = -1.36 V ; Anodic  must be + ve
 = ∆ - ∆e = ∆ - (-1.36) = ∆ + 1.36 > 0 for anodic; ∆ > - 1.36 i.e., the
discharge of Cl- w.r.t SHE can occur only at an applied potential of -1.36 V or
more.