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Electrode Potential
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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