Electrochemistry MAE-212
Dr. Marc Madou, UCI, Winter 2016
Class II Thermodynamics of Electromotive Force
(II)
Table of Content
 Standard Redox Potentials Table
 Thermodynamics Ecell, ΔG, and Keq
 Derivation of the Nernst Equation
 Some example problems
 Structure of the solid/electrolyte interface
Standard Redox Potentials Table
•
E0 is for the reaction as written
•
The more positive E0 the greater the tendency
for the substance to be reduced
•
The half-cell reactions are reversible
•
The sign of E0 changes when the reaction is
reversed
•
Changing the stoichiometric coefficients of a
half-cell reaction does not change the value of
E0 (Intrinsic vs extrinsic properties)
Relation between Equilibrium constant, Gibbs free
energy and EMF of a cell
 The free energy function is the key to assessing the way in which a chemical system will
spontaneously evolve. In the Gibbs-Duhem formalism of the Gibbs free energy we can
write:
dG = -SdT + V dP + å mi dni + g dA + f dl
constant T
constant P
don’t change
shape
dG = å mi dni
don’t stretch it
Relation between Equilibrium constant, Gibbs free energy
and EMF of a cell
 Every substance has a unique propensity to contribute to a system’s energy. We call this
property Chemical Potential:
m
 When the substance is a charged particle (such as an electron or an ion) we must include
the response of the particle to an electrical field in addition to its Chemical Potential.
We call this the Electrochemical Potential (F is the Faraday constant, z the charge
on the particle and f the potential):
m= m + z F f
Relation between Equilibrium constant, Gibbs free energy
and EMF of a cell
 The chemical potential or electrochemical potential
(if we are dealing with a charged particle) is the
measure of how all the thermodynamic properties
vary when we change the amount of the material
present in the system. Formally we can write:
Relation between Equilibrium constant,
Gibbs free energy and EMF of a cell
• Start with the First Law of Thermodynamics and some standard thermodynamic relations and we
find:
dU = dq + dw
dq = T dS
dw = -P dV + dwelectrical
dU = T dS - PdV + dwelectrical
dGT = dHT - T dS
dH P = dU P + PdV
= dUT, P + P dV - T dS
= T dS - P dV + dwelectrical + P dV - T dS
dGT,P = dwelectrical
•And therefore, the Gibbs function is at the heart of electrochemistry, for it identifies the
amount of work we can extract electrically from a system.
Relation between Equilibrium constant, Gibbs free
energy and EMF of a cell
• Now we can easily see how this Gibbs function relates to a potential.
welectrical = V Q
since Q = n F
= nF E
• By convention, we identify work which is negative with work which is being done by the
system on the surroundings. And negative free energy change is identified as defining a
spontaneous process.
DGT,P = -welectrical = -n F E
• Note how a measurement of a cell potential directly calculates the Gibbs free energy
change for the process.
Relation between Equilibrium constant, Gibbs free
energy and EMF of a cell
 The propensity for a given material to contribute to a reaction is measured by
its activity, a.
 How “active” is this substance in this reaction compared to how it would behave
if it were present in its standard state?
• Activity scales with concentration or partial pressure.
a  C/C˚ (solution) and a  P/P˚ (gas)
BUT…
• intermolecular interactions
• deviations from a direct correspondence with pressure or concentration
Relation between Equilibrium constant, Gibbs free
energy and EMF of a cell
 Definition of activity is then:
•
Activity coefficients close to 1 for dilute solutions and low partial pressures.
• Activity changes with concentration, temperature, other species, etc. Can be very
complex.
• Generally, we ignore activity coefficients for educational simplicity, but careful work
always requires its consideration.
Relation between Equilibrium constant, Gibbs free
energy and EMF of a cell
 How does chemical potential change with activity?
 Integration of the expressions for the dependence of amount of material on the Gibbs
function, leads to the following relationship :
Derivation of the chemical potential
from the Gibbs free energy
𝑑𝐺 = −𝑆𝑑𝑇 + 𝑉𝑑𝑃 + ∑𝜇𝑖 𝑑𝑛𝑖 + 𝛾𝑑𝐴 + 𝑓𝑑𝑙
The total temperature and pressure 𝑃 remain constant, but the partial pressure 𝑃𝑖 of each
specie changes. Therefore, if we don’t change the geometry of our cell, we have
𝛿𝐺
𝛿𝐺
𝛿𝐺 𝛿𝜇𝑖
= 𝜇𝑖 → 𝐺 = 𝜇𝑖 𝑛𝑖 ,
=𝑉 →
=
𝑛 =𝑉
𝛿𝑛𝑖 𝑇,𝑃,𝑛
𝛿𝑃 𝑇,𝑛
𝛿𝑃 𝛿𝑃 𝑖
𝑗
𝑖
𝑉
Ideal gas law 𝑃𝑉 = 𝑛𝑖 𝑅𝑇 → 𝑛 =
𝑖
𝑅𝑇 𝛿𝜇
=
→
𝑃
𝛿𝑃
𝜇
𝑅𝑇
. So:
𝑃
𝑃
1
𝑃
0
𝑑𝜇 = 𝑅𝑇
𝑑𝑃 → 𝜇 = 𝜇 + 𝑅𝑇 ln 0
𝑃
𝑃
0
0
𝜇
𝑃
𝑃
𝑎 = 𝛾 0 → 𝜇 = 𝜇0 + 𝑅𝑇 ln 𝑎
𝑃
Relation between Equilibrium constant,
Gibbs free energy and EMF of a cell
•
How does Gibbs free energy change with concentration/activity? Same dependence as
for the chemical potential:
•
When we apply this to a reaction, the reaction quotient
comes into to play, giving us:
•
Say we have the reaction :
wA + xB ® yC + zD
Relation between Equilibrium constant, Gibbs free energy and EMF of
a cell
 The above reaction is a generic reaction and in order to analyze this chemical
process mathematically, we formulate the reaction quotient Q:
aCy a zD
Q= w x
a A aB
• It always has products in the numerator and reactants in the denominator
• It explicitly requires the activity of each reaction participant.
• Each term is raised to the power of its stoichiometric coefficient.
Relation between Equilibrium constant, Gibbs free
energy and EMF of a cell- Nernst Equation
 Take the expression for the Gibbs dependence on activity and rewrite this in terms of
cell potential:
•
The relation between cell potential E and free energy gives:
•
Rearrange and obtain the Nernst Equation:
Relation between Equilibrium constant,
Gibbs free energy and EMF of a cell- Nernst
Equation
 The equation is often streamlined by restricting discussion to T = 25 °C and inserting
the values for the constants, R and F.
• Note the difference between using natural logarithms and base10 logarithms.
• Be aware of the significance of “n” – the number of moles of electrons transferred in the
process according to the stoichiometry chosen.
Relation between Equilibrium constant,
Gibbs free energy and EMF of a cell Chemical Equilibrium
• The reaction proceeds, Q changes, until finally DG=0. At that moment the overall
reaction stops. This is equilibrium.
• This special Q* (the only one for which we achieve this balance) is renamed Keq, the
equilibrium constant.
Relation between Equilibrium constant, Gibbs free
energy and EMF of a cell -Chemical Equilibrium
 When the system is at equilibrium, the proper quotient of equilibrium concentrations is
equal to the equilibrium constant:
 The system is at equilibrium when the concentrations are in their equilibrium values, so
[A] = [A]e, [B] = [B]e, etc. and thus:
Q* = Keq
Relation between Equilibrium constant,
Gibbs free energy and EMF of a cell Chemical Equilibrium
Example problems:
 Cu is cathode (it is reduced). Zn is anode (it is oxidized).
2+
–
2+
Cu ( aq ) + 2 e ® Cu( s)
Zn ( s ) ® Zn ( aq ) + 2 e
Cu 2+( aq ) + Zn( s) ® Zn 2+( aq ) + Cu (s)
• Note that n=2 for this reaction.
=1
=1
• Activity for solid materials is 1; replace activities with concentrations.
–
Example problems:
 What is the potential in the cell if [Cu2+] = 0.01 M and [Zn2+] = 1.00 M?
1.00
E = 1.10 - 0.01285 ln
= 1.10 - 0.01285 ln (100)
0.01
= 1.10 - 0.01285 ( 4.6052 ) = 1.041V
• Note that the cell potential decreased by about 60mV. This was a change in
concentration of TWO orders of magnitude, but since it was also a TWO electron
process, we saw the same 60 mV change in potential.
Example Problems:
 Nernst equation demonstrates that potential depends upon concentration.
 A cell made of the same materials, but with different concentrations, will also produce a
potential difference.
Cu | Cu2+ (0.001 M) || Cu2+ (1.00 M) | Cu
 What is standard cell potential E˚ for this cell?
 What is the cell potential E? What is “n”, the number of electrons transferred? Which
electrode, anode or cathode, will be in numerator?
Example Problems:
 The equations we have derived allow us to relate measured cell potentials to Standard
Gibbs Free Energies of reaction. These in turn are related to a reaction’s equilibrium
constant.
 Consider the cell
 Pt | I– (1.00 M), I2 (1.00 M) || Fe2+ (1.00 M), Fe3+ (1.00 M) | Pt
 Standard Cell Potential is (from tables) = 0.771 V - 0.536 V = +0.235 V
This is the free energy change. It leads to the equilibrium constant for the reaction.
Example Problems:
Fe2+ + 2e–  Fe
-0.44
Sn2+ + 2e–  Sn
V2+ + 2e– V
-1.19
Ag+ + e–  Ag
To get a final positive cell potential, the
more negative half-reaction (V) must
act as the anode.
Fe2+ + V  Fe + V2+
Ecell = -0.44 - (-1.19) = +0.75 V
-0.14
+0.80
More negative potential reaction is the
anode.
Multiply the Ag reaction by 2, but don’t
modify the cell potential.
2 Ag+ + Sn  2 Ag + Sn2+
Ecell = +0.80 - (-0.14) = +0.94 V
Example Problems:
A fuel cell is an electrochemical
cell that requires a continuous
supply of reactants to keep
functioning
Anode:
Cathode:
2H2 (g) + 4OH- (aq)
4H2O (l) + 4e-
O2 (g) + 2H2O (l) + 4e-
4OH- (aq)
2H2 (g) + O2 (g)
2H2O (l)
Example Problems:
Corrosion
Example Problems:
Cathodic Protection of an Iron Storage Tank
Example Problems:
Electrolysis is the process in which electrical energy is used to cause a non-spontaneous chemical
reaction to occur.
Example Problems:
Electrolysis of Water
Example Problems:
Electrolysis and Mass Changes
charge (Coulombs) = current (Amperes) x time (sec)
1 mole e- = 96,500 C = 1 Faraday
How much Ca will be produced in an electrolytic cell of molten CaCl2 if a current of 0.452 A is
passed through the cell for 1.5 hours?
2Cl- (l)
Anode:
Cathode:
Ca2+ (l) + 2e-
Cl2 (g) + 2eCa (s)
Ca2+ (l) + 2Cl- (l)
Ca (s) + Cl2 (g)
2 mole e- = 1 mole Ca
mol Ca = 0.452
C
s
= 0.0126 mol Ca
= 0.50 g Ca
x 1.5 hr x 3600
s
hr
x
1 mol e96,500 C
x
1 mol Ca
2 mol e-
Structure of the solid/electrolyte
interface
 Every solid/liquid has a interface that one
would like to control --the better one can
control that interface the better one can say
something about the environment
 Interfaces are very complex often involving
fractals (beach, trees, snow flakes, etc.)
rather than smooth transitions, this implies ,
for example, that perfect selectivity will be
hard to achieve (too many different binding
sites)
Structure of the solid/electrolyte interface
 Charge carriers in electrode materials:
 Metals (e.g. Pt) : electrons
 Semiconductors (e.g. n-Si) :




electrons and holes
Solid electrolytes (e.g. LaF3 ) : ions
Insulators (e.g. SiO2):no charge
carriers
Mixed conductors (e.g. IrOx) : ions
and electrons
Solution (e.g. 1 M NaCl in H2O):
solvated ions
Double layer-(in case of a metal 10-40 µF cm-2)
Inner Helmholtz plane (IHP)
Outer Helmholtz plane (OHP)
Gouy-Chapman layer (GCL)