Lecture 2

Electrode potential - Nernst equation

Walther NERNST (1864-1941)

Electrochemical thermodynamics and electrode potential

Electrode potential

Fe2+ Cu

Fe2+solution 1M

Cu2+solution 1M

volmeter

Porous partition

0.78V

An electrochemical cell consisting

of iron and copper electrodes, each

of which is immersed in a 1M

solution of its ion. Iron corrodes

while copper electrodeposits.

Zn2+ Fe

Zn2+solution 1M

Fe2+solution 1M

volmeter

0.323V

An electrochemical cell consisting

of zinc and iron electrodes, each of

which is immersed in a 1M

solution of its ion. Zinc corrodes

while iron electrodeposits.

Porous partition

Standard Hydrogen Reference Electrode

2H+ + 2e = H2

Eo = 0.0 V

Standard Hydrogen Electrode = SHE

Acid Solution 1M

Reference Electrode Equation Condition E(V) vs SHE

SHE2H++2e=H2

pH=0 0,000Eo-0,059pH

Silver / silver chloride

(Ag/AgCl)

AgCl+e=Ag+Cl- aCl-=1 SHE = (Ag/AgCl) + 0,222

Eo-0,059lg(aCl-)

0,1M KCl SHE = (Ag/AgCl) + 0,288

1,0M KCl SHE = (Ag/AgCl) + 0,235

Saturated KCl SHE = (Ag/AgCl) + 0,199

Sea water SHE = (Ag/AgCl) - 0.25

Saturated Calomel Electrode

(HgCl) = SCE

Hg2Cl2+2e=2Hg+2Cl- aCl-=1 SHE = SCE+0,268

Eo-0,059lg(aCl-)

0,1M KCl SHE = SCE + 0,336

1,0M KCl SHE = SCE + 0,280

Saturated KCl SHE = SCE + 0,244

Copper/ Copper Sulphat

(Cu/CuSO4)

Cu2++2e=CuSO4 aCu2+=1 SHE = (Cu/CuSO4) + 0,340

Eo+0,0295lg(aCu2+) Saturated CuSO4 SHE = (Cu/CuSO4) + 0,318

Nernst equation

E : Half cell potential

E0: Standard half cell potential,

n: number of electrons transferred,

F Faraday constant (~96500)

R : Gas constant

T : Temperature

aOx and aRed: Activity of oxidized or reduced species COX and Cred

Red

XOo

a

aln

nF

RTEE

ne

Me M

o.eq ClnnF

RTEE

Table 1.1.

Eo

Measurement of electrode potential with SHE

This is called a galvanic coupletwo metals electrically connected in a liquidelectrolyte wherein one metal becomes an anode and being corroded, while

the other acts as a cathode polarization

Electrochemical cell consisting of standardzinc and hydrogen electrodes that has beenshort-circuited.

volmeter

M M++ e

M+H++ e H

anodecathode

2. Polarization

The displacement of each electrode potential from its equilibrium value is

termed polarization.

The magnitude of this displacement is the overvoltage

Overvoltage is normally represented by the symbol . Overvoltage is expressed interms of plus or minus volts (or millivolts) relative to the equilibrium potential.

When:

> 0 anodic polarization

< 0 cathodic polarization

polarization curve

, V

lg i

anodic polarization

cathodic polarization

= E-Ee >0

= E-Ee

Activation polarization overvoltage versus

logarithm of current density for both oxidat

ion and reduction reactions.

Distribution in the vicinity of the cathode for two types

(a) Activation polarization(b) Concentration polarization.

In solution:

Ion distribute homogeneous activation polarization

Ion distribute non homogeneous concentration polarization

(a)

iLActivation

polarization

(b)

iL

E, V

Ee

Concentration polarization

For reduction reactions, schematic plots of potential

versus logarithm of current density for (a) concentration

polarization, and (b) combined activation-concentration

polarization.

E, V

Ee

E, V

EeH2H+ + 2e H2

lgi

Activation polarization potential

versus logarithm of current density for

both oxidation and reduction

reactions.

lgi lgi

Activation polarization refers to the condition wherein the reaction rate is controlled

by the activation energy barrier

(1.12)

where and io are constants for the particular half-cell. The parameter io is termed the

exchange current density

Concentration Polarization only pour cathodic polarization

exists when the reaction rate is limited by diffusion in the solution

The mathematical expression relating concentration polarization overvoltage c and

current density i is:

(1.13)

where R and T are the gas constant and absolute temperature, respectively, n and F

have the same meanings as above, and iL is the limiting diffusion current density.

o

ai

ilog

L

Ci

iLn

nF

RT1

o

o

i

ilogEE

neMM n11

oE1

(1.8) n11 MneM

(1.9)

2121 MMMMnn

nMoe C

nF

RTEE

1

ln11

22 MneMn

2E

nMoe C

nF

RTEE

2

ln22

e

1

e

2

M

Mo

1

o

2

e

1

e

2 EEC

Cln

nF

RTEEEEE

n1

n2

If E > 0 so

1E

< then M1 is corroded.eE1

eE2

M1 M2

M1n+

conc. C1M2

n+

conc. C2

volmeter

?

An electrochemical cell consisting

of M1 and M2 electrodes, immersed

in C1 & C2 solution of its ion. M1 or

M2 corrodes .

What electrode can be oxidized or reduced ?

Be corroded ?

Metal: E stable in Ecorr= EM determinating

by experience.

E cathode = EC =?

2

2

C H2

H

Hoe Plg03,0pH059,0C

Plg

nF

RT3,2EE

2H+ + 2e H2

O2 + 4H+ + 4e 2H2O

2

2

O4

HO

oe

C Plg015,0pH059,023,1CP

1lg

nF

RT3,2EE

O2 + 2H2O + 4e 4OH-

2

2

O

O

4

OHoe

C Plg015,0pH059,023,1P

Clg

nF

RT3,2EE

( n = 4, Eo = 0,4 and lg COH- = pH-14)

Aqueous solution

EM< EC M corroded

Example problem 1.1One half of an electrochemical cell consists of a pure nickel electrode in a solution of

Ni2+ ions; the other is a cadmium electrode immersed in a Cd2+ solution.

(a) If the cell is a standard one, write the spontaneous overall reaction and calculate

the voltage that is generated.

(b) Compute the cell potential at 25oC if the CCd2+ and CNi2+ concentrations are 0.5

and 10-6M, respectively. Is the spontaneous reaction direction still the same as for the

standard cell?

2NiC

2CdC

SOLUTION

(a) From Table 1.1, the half-cell potentials for cadmium and nickel are, respectively,

-0.403 and -0.250 V. The cadmium electrode will be oxidized and nickel reduced

because cadmium is lower in the standard Eemf series (tab. 1.1); thus, the spontaneous

reactions will be:

Cd Cd2+ + 2e- (a)

Ni2+ + 2e- Ni (b)

Ni2+ + Cd Ni + Cd2+ (c)

VEEE oCdo

Ni 153.0)403.0(250.0 VEEEo

Cd

o

Ni 153.0)403.0(250.0

V153.0)403.0(250.0EEEEE oCdo

NiAC

b) Comparing electrode potential of Cd and Ni with CNi2+ =0,5M and CNi2+=10-6M

assume that in contrast to part a, nickel is oxidized and cadmium reduced according to

reaction (c) : Ni2+ + Cd Ni + Cd2+ (c)

2

2

68.314 298 10ln ( 0.25 ( 0.403) ln 0.015

2 96500 0.5

o o CdNi Cd

Ni

CRT xE E E V

nF C x

Since E is negative, the spontaneous reaction direction is the opposite to that of Equation

(c), or opposite to that of the standard cell.

Ni + Cd2+ Ni2+ + Cd Ni corrode

Other way

68.314 2980.25 (10 ) 0.4272 96500

Ni

xE ln V

x

8.314 2980.403 (0.5) 0.412

2 96500Cd

xE ln V

x

Example problem 1.2Possibility to corrosion in seawater:

(Fe, Eo = -0.44V), (Cu, Eo = 0.34V), (Pt, Eo =1.118V)

CFe2+= CCu2+= CPt2+= 10-6 M, T=25oC (298K)

SOLUTIONPossibility to corrosion in seawater: Fe, Eo=-0.44V, Cu, Eo=0.34V, Pt, Eo=1.118V

CFe2+= CCu2+= CTi2+= 10-6 M

0.9411.00E-061.118Pt

0.1631.00E-060.34Cu

-0.6171.00E-06-0.44Fe

EM,

VCMn+

Eo, VMetal

21.23 0.059* 0.015*lg( )

1.23 0.059*7 0.015*lg(0.21) 0.807

C OE pH P

V

Fe and Cu can be corrode, Pt non