Dr.A.DAYALAN, Former Prof & Head, B.V Eq (03-Different Over-potential,η)

M.Sc (Chemistry)

ELECTRODE KINETICS – ELECTRODICS-I

03-DISCUSSION OF BUTLER-VOLMER EQUATIONFor different values of

Case-(i) : = 0 ; BV equation reduces to Nernst equation.

(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 x1 ; NB: Use a calculator, calculate & see]

= (RT/ ioF) i similar to Ohm’s Law(V=IR). Hence, resistance = RT/ ioF

∂ RT

A + e kc

kaD

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Dr.A.DAYALAN, Former Prof & Head, B.V Eq (03-Different Over-potential,η)---- = ----- = ρ 0 as io ∞ (Non polarisalble) ∂i ioF

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

Metal System Medium io

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 -

Platinum Fe3+/Fe2+ H2SO4 2.51 x 10-3 0.58

Palladinum Fe3+/Fe2+ H2SO4 6.3 x 10-3 -Calomel Hg, Hg2Cl2 KCl

Highly polarisable. Does not allow charge (e-) to pass through

Discharge of H+ is difficult on Hg surface

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Dr.A.DAYALAN, Former Prof & Head, B.V Eq (03-Different Over-potential,η) = (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-)

v(b) is highly negative, i = ic = io e -F/RT [Cathodic]

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Dr.A.DAYALAN, Former Prof & Head, B.V Eq (03-Different Over-potential,η)

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

ALL THE ABOVE CASES for OVER POTENTIALS CAN BE SUMMARISED AS FOLLOW

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Dr.A.DAYALAN, Former Prof & Head, B.V Eq (03-Different Over-potential,η)

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

CONDITION FOR ELECTRODE DEPOSITION

∆e = Equilibrium potential for the process considered

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Dr.A.DAYALAN, Former Prof & Head, B.V Eq (03-Different Over-potential,η) = ∆ - ∆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.

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