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Electrochemical Equilibrium
CHAPTER 7
Electrochemical reactions involve the transfer of electrons
We recognize the importance of electrochemical equilibria in
galvanic cells for example; which can produce electriccurrent.
Battery technology relies on this knowledge where it ispossible to convert Gibbs energy change into useful form of
work
On the other hand, we also have electrolytic cells whichconsume electric power to induce a chemical reaction
This is used in the industries such as in the processelectro latin and electro-refinin Aluminum roduction
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The measurement ofelectromotive force
of an electrochemical cell at varioustemperature makes it possible to obtain thethermodynamic properties
i.e. E = f(temp)
The chemical potential (Q) of theelectrolytes, for instance, can be calculated
based on these measurements
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Equilibria Involving Potential Differences
When a small charge dQ moves through an electric potentialJ, the work done on the charge is given by
dw (joule) = J (volt or Joule/coulomb).dQ (coulomb)
If ions (that carry electric charge) with charge number ofziare transferred, the differential charge dQ can be related to thedifferential amount of mol dni by :
dQ(coul.) = zi .F(coul./mol).dni (mol)
Where F is the Faraday Constant which is equal to theavogadros number x charge of proton (or electron) :
F = Na.e = (6.0221367x1023 / mol) x (1.60217733x10-19 C )
F = 96485.309 C/mol
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So when we consider a multiphase system with phases atdifferent potential the electrical work term is of the form:
w = 7 zi F Ji ni where Ji is the potential of the phase containing species i
When we want to relate this to the thermodynamic
equation, however, it should be noted that Ji is not anatural variable of Gibbs Energy, Hence it is necessary tomake the necessary transformation :
G = G - 7 zi F Ji ni
Substituting G = 7 Qi ni gives G = 7 (Qi - ziFJi) nior G = 7 Qi.ni where Qi = Qi - zi F J i
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Based on this, the differential of the transformed Gibbs
Energy :
dG = dG - 7 ziFJi dni - 7 zi F ni dJiSubstituting dG = - SdT + VdP + 7Qidniwe then have :
dG = - SdT + VdP + 7Qidni - 7 ziFJi dni - 7 zi F ni dJiOR
dG = - SdT + VdP + (7Qi - 7 ziFJi ) dni - 7 zi F ni dJi
and as before we have : Qi = Qi - zi F Ji thendG = - SdT + VdP + Qidni - 7 zi F ni dJI
and consistently for chemical potential :i
injPTidn
Gd
QJ d!
d
,,,
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Equation for an electrochemical cell
Electrochemical cell can be classified as a galvanic cell in
which chemical reaction occur spontaneously. The Gibbs(free) energy of this reaction is harnessed to generate usefulwork (recall: it produces potential differences in Volt)
On the other hand we can also have electrochemical cell that
work in reverse, in which, chemical reaction is caused/inducedby an externally applied potential differences
(i) Cathode is the electrode in which reduction occurs; in agalvanic cell this electrode is positive since electrons are
flowing into it and reacting to form a reduced species
(ii) Anode is the electrode in which an oxidation reactionoccurs; (it is negative in a galvanic cell)
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Galvanic Cell :
At cathode : A + e- p A- (reduction, electron is supplied)At Anode : A p A+ + e- (oxidation, electron is released)
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8
Swiss German
University As the potential difference is
increased, a point will be
reached that no current is
flowing
This is the equilibrium
potential that is of primaryconcerned in this chapter
This is a thermodynamic
measurement as the direction of
the current through the cell can
be changed by an infinitesimal
change in the applied voltage
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A wide variety of galvanic cell can be constructed :
Cells without liquidjunctions :
Pt(s)H2(g)HCl(m)AgCl(s)Ag(s)
The vertical lines represent the phaseboundary
In the figure on the side:H2(g) a
non-metal is passed over a platinum
electrode (oxidized to H+), on theother side Ag+ is reduced to Ag(s)
while the Cl- is released to the solution
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Cells with liquidjunctions :
Zn(s)Zn2+(m1)1 Cu2+(m2)Cu(s)
Zn(s)Zn2+(m1)1 Zn2+(m2)Zn(s)In the above example Zn is oxidizedinto Zn2+ while Cu2+ is reduced toCu(s)
1 represent a junction betweentwo liquids
1 1 represent a salt bridgetypically made up ofpotassiumchloride orammonium chloride inwhich the anion and cations havenearly equal mobility
salt bridge prevents mixing/reaction
Note: for a galvanic
cell there is aconvention to draw
oxidation on the left
while reduction on
right
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To obtain the relationship between electromotive force (E)
and the chemical potential or activities of reactants &
products consider the following cell without liquid junction :
PtLH2(g)HCl(m)AgCl(s)Ag(s)PtRNote: m indicates molality of the species and we assume
international convention in which the left side is oxidationand the right side is reduction
So the reactions are :
2AgCl(s) + 2e-(PtR) p 2Ag(s) + 2Cl-(aq) (reduction)H2(g) p 2H+(aq) + 2e-(PtL) (oxidation)
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The sum of these two electrode reactions is :
H2(g) + 2AgCl(s) + 2e-(PtR) p 2H+(aq) + 2Ag(s) + 2Cl-(aq) + 2e-(PtL)
alternatively this can also be written (shorter version) :
H2(g) + 2AgCl(s) + 2e-(PtR) p 2HCl(aq) + 2Ag(s) + 2e-(PtL)
Consider the cell in the figurebeside; in which the electrodes are
not connected electrically (open
circuit)
Under this condition, when thesystem is at equilibrium the
relationship
7Ri Qi = 0 must be satisfied
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Such that for :
H2(g) + 2AgCl(s) + 2e-(PtR) p 2HCl(aq) + 2Ag(s) + 2e-(PtL)
We have ( for7RiQ
i= 0) :
2Q(HCl,aq) + 2Q(Ag,s) + 2Q(e-,PtL)
- Q(H2,g) - 2Q(AgCl,s) - 2Q(e-,PtR) = 0From before we have : Qi = Q+ ziFJiand for electrically neutral species zi = 0 such that Qi = Qonly for electrons: Q(e-,PtL) = Q(e-,PtL) FJL and
Q(e-,PtR) = Q(e-,PtR) FJR
But for electrons at equilibrium we have : Q(e-,PtL) = Q(e-,PtR)So the Q term can cancel out ; Therefore we have in summary :
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2Q(HCl,aq) + 2Q(Ag,s) 2FJL- Q(H2,g) - 2Q(AgCl,s) + 2FJR= 0
But we also know that change in chemical potential :
2Q(HCl,aq) + 2Q(Ag,s) - Q(H2,g) - 2Q(AgCl,s) = (rG
Therefore : (rG = 2F(JL-JR) = -2F (JR-JL) =-2FE
Hence : (rG = =-2FE note that the 2 is coming from Ri
where E = JR- JL which is the electromotive force which canbe measured using a potentiometer
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The relationship (rG = -2FE can actually be generalizedto any electrochemical reaction in which it can be written as
FEG er Y!(
Where Re is the absolute value of the stoichiometric
number of the electron in the electrochemical reaction (orsimply the number of electron being transferred during theredox reaction)
Since we are dealing with Gibbs energy change, then we
know ifthe right hand electrode has more potential thanthe left hand electrode, the value of the electromotive force(E) of the cell is positive
This will give a negative value of(rG which means that the
electrochemical reaction is spontaneous
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If the schematic diagram of the cell is reversed(i.e. the electrochemical reaction is reversed),
the sign of E and therefore the sign of(rG
would also be reversedThe reversed reaction then should have (rGwhich is positive and this means that the reverse
reaction is not a spontaneous reaction
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Relation between E and the equilibrium constant (K)
Since the value ofE can be obtained under equilibrium
condition, then we can somehow relate it to the value ofK
Recall from the previous chapter we have (rGo = -RT ln K
but before we also have (rG= -|Re|FE or equivalently forstandard state we have (rGo = -|Re|FEo (note we will seehow Eo is determined in later discussion)
Hence we have : |Re|FEo = RT ln K such that :
!!
RT
FEKorK
F
RTE
o
e
e
oY
Y
expln F = 96485.309 C/mol
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Example : Three different galvanic cells have standard
electromotive forces (Eo) of0.01 V, 0.1 V and 1.0 V at
25C. Calculate Kfor each reaction assuming |Ye| = 1 forthese reactions
Ans. :
!
RT
FEK
o
eY
exp So by entering the values, we get :
476.1)15.298)(3145.8(
)01.0)(96485(exp
.
!
!
K
VK
KmolJ
molCoul
(Note thatequivalentunitforVoltisJoule/coulomb)similarly we have K= 49.0 for Eo = 0.1 V,
K= 8.02x1016 for Eo = 1 V
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Calculate enthalpy, Entropy and Gibbs Energy. Also, if
spontaneous calculate how much energy is available for work.
Zn + Cu2+ p Zn2+ + Cu
(fHo (kJ) 0 64.77 -153.89 0 (Ho = -218.66
So (J/K) 41.63 -99.6 -112.1 33.15 (So = -21.0Therefore: (G = (H - T(S = -218.66x103 - 298x(-21.0)
= - 212.402 kJ/mol < 0 ; spontaneous
While -T(S is equal to qsurr: heat transferred to surrounding(G if negative signifies the amount of energy that is availablefor work(our purposes) hence also called Gibbs Free Energy
RECALL FROM CHAPTER FOR (USE TABLE C2 or C3
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Furthermore, knowing that (rGo = -|Re|FEo if we have dataon the variation ofGibbs energy as a function of Temp
than we can also obtain the expressions for(S and (H
recall that (S = -(HG/HT) then we have (So = |Re|F(H Eo/ HT)
moreover, (G =(H - T(S or(H =(G + T(S, thus
(Ho = -|Re|FEo + |Re|FT(H Eo/ HT)
So, if we know the variation ofEo as a function of
temperature then we can calculate the above entities.
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Example : the electrochemical reaction
Pt|H2(g)|HCl(aq)|AgCl(s)|Ag(s) has Eo (Volt) as a
function of temperature (in celcius) as follow (0 to90Celcius) :
Eo= 0.23659 - 4.8564x10-4 T - 3.4205x10-6 T2 + 5.869x10-9 T3
calculate (Go
, (So
, (Ho
for this reaction at 25C(i) for(Go we simply need to calculate Eoby substitutingthe value of T, and then calculate (rGo = -|Re|FEo
so Eo =
0.22240 V (at 298K), hence :(rGo = - (1) x (96485 Coul/mol) x (0.22240 Volt)
= - 21.458 kJ/mol
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(ii) (So = |Re|F(H Eo/ HT) =
1 x (96485 Coul/mol) x [-4.8564x10-4 - 2(3.4205x10-6)T +
3(5.869 x 10-9) T2] Volt = -62.297 J/mol.K(at 298K)
(iii) (H = (G + T(S
= -21.458 kJ/mol+ (298.15K)(-62.297x10-3 kJ/mol.K)
= -40.032 kj/mol (at 298K)
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StandardElectrode Potential(standardreduction
potential, Eo)
For both research and industrial purposes, the value of
standard electrode potential (Eo) for many substances
are tabulated
For this purpose, we need to define the referenceelectrode against which all of the other
substances/electrodes are measured
In this case, the standard
measurement is measuredagainst a hydrogen electrode in
whichH2(g) is oxidized to
H+(aq).
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Hence, the measurement is done against the oxidation of
H2(gas) at the anode i.e. H2(g) p H+(aq) + e-
To facilitate the measurement, a convention was adopted
such that the standard electrode potential of the hydrogen
(H2(g) at 1 atm on Pt electrode) is arbitrarily assigned
the value zero:
H+(aq) + e- p H2(g) Eo = 0.0000 V
(Note that it is written in terms ofreduction reaction)
for example: for the measurement of Eo
for the reductionof Ag+(aq) + e- p Ag(s) we have :
AgCl(s) + e- p Cl-(aq) + Ag(s) Eo = 0.2224 V
Hence Eo
is called the standard reduction potential
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Hence, we can also see the Eo orstandard electrode
potential is the measured tendency of an electrode reaction to
occur in the direction of reduction
A brief listing ofstandard electrode potentials (25C) can be
seen in table 7.2 (p.240)
Note that fluorine, for instance, electrode has the most positiveelectrode potential F2(g) + e
- p F- - Eo = +2.87 V
if we connect this to the expression : (rGo = -|Re|FEo andsince we expect, if G < 0, then we know this reaction is
spontaneous
Furthermore, as the value is also the highest among other
listed electrode potential for other substances
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The fluorine electrode has the highest tendency to pull
electrons from the other electrode such that it will undergo
reduction
On the other hand, Li is at the bottom of the list i.e. it has the
lowest Eo, in fact it has negative value of Eo
For such electrodes, the highest tendency is not to undergo
reduction (or to pull electrons from the other electrode), butrather they have a higher tendency to give up/donate electron
i.e. to undergo oxidation reaction
Similarly, we can see it from the point of view (Go
= -|Re|FEo
such that Gibbs energy change is positive for the reduction of
Li, therefore, we know that it is the reverse
reaction/oxidation that should occur spontaneously (Note:
against H2(g) on Pt electrode)
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Earlier we discussed the electromotive force of a cell is given
by E = JR- JL
Based on this we can express the standard electromotive forcefor a cell as : Eo = ERo - EL
o
Hence basedon this we can :
(i) Calculate the standard electrode potential of any cellinvolving any substances that are available on the list (table
7.2)
(ii) Moreover, we can also determine which electrode will act
as anode and cathode when the cell delivers current
(iii) We can also evaluate equilibrium constant K based on the
relation ln K = |Ye| FEo/(RT) ; [also (rG]
Eo = Ereduction Eoxidation
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Example: Calculate Eo for : Pt|H2(g)|HCl(aq)|Cl2(g)|Pt
Ans.: Write the relevant half reaction (reduction) i.e.
(1) Cl2 + 2e-
m 2Cl-
(aq) Eo =
1.3604 V (Eo
R)(2) H2(g) + 2e
- m 2H+(aq) Eo = 0.0000 V (EoL)
Note: The two reactions must have the same number ofelectron (donated = accepted), if not multiplied one of them
to have the same electron. The value of Eo is not affectedbythis multiplication
Then Eocell = EoR - E
oL = 1.3604 - 0.0000 = 1.3604 Volt
Hence the full reaction can be written as :
H2(g) + Cl2(aq) m 2H+(aq) + 2Cl-(aq) (i.e. HCl(ai)) Eo = 1.3604 V
NOTE: Eo = Ereduction Eoxidation
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If necessary we can calculate (Go from (Go = -
|Re|FEo
(Go= -(2)x(96458 coul)(1.3604 J/coul) = -262.44kJ/mol
we can also calculate equilibrium constant K :
451078.9
15.298.3145.8(
)3604.1()96485(2expexp x
KKmolJ
coulJxcoulx
RT
FEK
o
e !
-
!
!
i.e. volt
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KF
RTEEfromhence
HClaK
e
o
oCl
oH
ln)(
22
2
Y
!
!
!
oCl
oHe
o
P
P
P
P
HCla
F
RTEE
22
2)(
lnY
If we need to evaluate E not at standard state (eg. not at
P=1 bar)
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Activity of Electrolytes
When we deal with activity coefficient, it is customary to useconcentration in molal unit (rather than mole fraction ormolarity)
molality (m) is the amount in mole of the electrolyte in 1 kgof solvent such that the unit would be mol/kg
One of the advantage with working with molality is that theunit is not affected by temperature (unlike molarities forexample)
the activity of a solute on the m scale is defined by :
o
ii
im
ma
K!
Where mo is the standard value of
the molality ( 1 mol/kg solvent)
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For this relationship, for very low concentration (m p 0)
1lim0
!p imi
K
Such that for very dilute solution the activity aio
ii
im
ma
K!
Would simplify to ai = mi/mo (note that mo is simply 1 molal)
At very low concentration the distribution of ions in the
solution can be considered to be completely random and the
ions are too far apart to exert any interaction among
themselves, thus the Ki is unity
At higher concentration, the attraction and repulsion become
more important. Hence the K is reduced from unity (
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Fuel Cell
A fuel cell is a type of battery which is continuously supplied
with oxidant and reductant so that the redox reactions that
generate electricity can occur indefinitely
Some of the currently designed fuel cell can work at the lowerrange temperature (25-100C), mid range temperature (100-
500C), high temperature (500 to 1000C) and very high
temperature (above 1000C)
at very low temperature, the use of generally expensive catalyst
is necessary. While at high temperature, the challenge is in the
choice of material that can withstand the temperature as well as
the energy required to maintain the high temperature
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Determination of pH
In many cases in practice, chemists needs to deal with
the concentration of hydrogen ions (H+) which could range
from 10-1 M to 10-14 M
Back in 1909, a chemist named Sorenson adopted
exponential notation for [H+]. He defined pH as thenegative exponent of 10 that gives the H+ concentration
i.e. [H+] = 10-pH
Today, pH is defined to be as close as possible to the
negative base 10 logarithm ofhydrogen ion (H+
) activity :pH = -log (aH+)_
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University The pH may be measured with a hydrogen electrode
conn
ected with a calomel electrode through a salt bridge :Pt | H2(PH2) | H+(aH+) 1 1 Cl- | Hg2Cl2 | Hg
The electromotive force of this cell may be considered tobe made up of three contributions :
junctionliquidP
P
HEaE
oH
-!
21
2
log0591.02802.0
The contribution of the calomel electrode is 0.2802 V,
while the Eliquid junction can be assumed to be 0.00 V
the number 0.0591 is coming from RT/nF but still
multiplied by conversion factor from ln to log (base 10),
recall originally the middle term is :
-
21
2
lno
H
P
P
Ha
nF
RT
**
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Furthermore, to make the situation more simple we can
use PH2 = 1 bar such that the partial pressure contributio
nis reduced to 1
So after the simplification the equation (**) is :
E 0.2802 = -0.0591 log aH+
and since pH is defined as log (aH+) then
E 0.2802 = 0.0591 log aH+ or0591.0
2802.0!E
pH
Currently, the use of H2 in the measurement of pH isavoided. The use of glass electrode is typically used inpractice in most laboratories
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The basic process is shown
in the figure beside which isa hydrogen-oxygen fuel cell
with solid electrolyte
The membrane is
impermeable to reactant
gases but not to H+ ions
The (ion exchange)
membrane is also acting as
the electrolyte
For low temp operation, the
membrane is covered with
finely divided Pt catalyst
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The half cell reactions are :
O2(g) + 2H+ + 2e- p H2O(l) Eo = 1.2288 V
2H+ + 2e- p H2(g) Eo = 0 V (reversed)
H2(g) + O2(g) p H2O(l) Eo
= 1.2288 V
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