Chpt 20: Electrochemistry
Chpt 20: Electrochemistry
CEM 152 – SS2012
Cell Potential
and Free Energy

When both reactants and products are in their standard states,
and under constant pressure and temperature conditions
DGo = –nFEo
where





DGo is the standard free energy change of the reaction
n is the number of mole of electrons transferred in the reaction
F is Faraday’s constant
Eo is the standard cell potential
Faraday’s constant is the quantity of electrical charge on 1 mol
of electrons
1 F = 96,500 C/mol e1 F = 96,500 J/Vmol e-
Chpt 20: Electrochemistry
CEM 152 – SS2012
DGo from Cell emf I

Example: For the reaction
I2(s) + 5Cu2+(aq) 
2IO3-(aq) + 5Cu(s) + 12H+(aq)
calculate the standard free energy change for the reaction.
The oxidation reaction is
I2(s)  2IO3-(aq) + 10e- Eored = +1.195 V
The reduction reaction is
5Cu2+(aq) +10e-  5Cu(s) Eored = +0.337 V
Eo = Eored(reduction) - Eored(oxidation)
Eo = +0.337 V - (+1.195 V)
Eo = -0.858 V
from the half-reactions we see that 10 moles of electrons are transferred in the
reaction
n = 10
Chpt 20: Electrochemistry
CEM 152 – SS2012
DGo from Cell emf II
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
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Example: For the reaction
I2(s) + 5Cu2+(aq) 
2IO3-(aq) + 5Cu(s) + 12H+(aq)
calculate the standard free energy change for the reaction.
DGo = – nFEo
DGo =-(10 mole-)(96,500 J/Vmole-)(-0.858 V)
DGo = +828 kJ
Since DGo > 0, reaction is non-spontaneous (we could also note this since Eo
< 0)
If reactants and products are not in their standard states, the more general
relation
DG = –nFE
can be applied
Chpt 20: Electrochemistry
CEM 152 – SS2012
Example

1.
2.
A – 4 B – 1, 2
C – 2 D – 2, 3
E – 1, 4
3.
4.
Chpt 20: Electrochemistry
Over time, you notice that the tin electrode
seems to be disappearing while there are
deposits forming on the silver electrode.
Which of the following is a correct
statement?
The silver electrode is the cathode and the
tin electrode is the anode.
Electrons are flowing from the tin electrode
to the silver electrode.
Nitrate ions are flowing through the salt
bridge to the silver solution.
The half-reaction occurring at the tin
electrode is: Sn4+ + 2e- → Sn2+.
CEM 152 – SS2012
Example





Chlorine dioxide is used to treat municipal water
supplies.
2 NaClO2 (aq) + Cl2 (g) → 2 ClO2 (g) + 2 NaCl (aq)
Cl2 + 2e- → 2ClEored = 1.36 V
ClO2 + e- → ClO2Eored = 0.954 V
Calculate DGo for the reaction.
A – 78 kJ
B – -39 kJ
C – -78 kJ
D – -156 kJ
E – 39 kJ
Chpt 20: Electrochemistry
CEM 152 – SS2012
The Nernst Equation I


The cell emf will depend reactant/product concentration
The dependence of cell emf on concentration can be derived
from the dependence of the free energy change on
concentration
DG = DGo + RTlnQ




R is the gas constant (8.314 J/Kmol)
T is absolute temperature
Q is the reaction quotient, which depends on reactant and
product concentrations
We can derive an expression for E by replacing DGo with –
nFEo and DG with –nFE
Chpt 20: Electrochemistry
CEM 152 – SS2012
The Nernst Equation II




Following the substitution
-nFE = -nFEo + RTlnQ
E = Eo - (RT/nF) lnQ
This relation is the Nernst Equation. The equation is typically
expressed in base 10 logarithms
2.303 log x = ln x
E = Eo - (2.303RT/nF) logQ
At standard temperature (T = 298 K) the quantity in parenthesis takes
the form
E = Eo - (0.0592/n) logQ @ 298 K
The Nernst equation can be used in two ways


to determine the cell emf under non-standard conditions
to find [product] or [reactant] from E
Chpt 20: Electrochemistry
CEM 152 – SS2012
Cell emf Under Non-Standard
Conditions

Example: Calculate the standard emf for a cell that employs the
following reaction
2Al(s) + 3I2(s)  2Al3+(aq) + 6I-(aq)
when [Al3+] = 4.0x10-3 M and [I-] = 0.010 M
from the the standard reduction potentials the cell potential is Eocell =
+2.20 V and n = 6
E = Eo - (0.0592/n) logQ
Q = [Al3+]2[I-]6
Q = (4.0x10-3)2(0.010)6
Q = 1.6 x 10-17
E = (+2.20V) - (0.0592/6) log (1.6x10-17)
E = 2.36 V
Chpt 20: Electrochemistry
CEM 152 – SS2012
Product Concentration from
Cell emf I

Example: A voltaic cell is constructed that uses the following
reaction and operates at 298 K
2Al(s) + 3Mn2+(aq)  2Al3+(aq) + 3Mn(s)
What is [Al3+] to produce a cell emf of +1.00 V?
Al(s)  Al3+(aq) + 3e- Eored = -1.66 V
Mn2+(aq) + 2e-  Mn(s) Eored = -1.18 V
Eo = Eored(reduction) - Eored(oxidation)
Eo = -1.18V - (-1.66V)
Eo = +0.48 V
Chpt 20: Electrochemistry
CEM 152 – SS2012
Example

1.
2.
3.



A galvanic cell is constructed from the following two
half reactions
Ag+ (aq) + e- → Ag (s)
Eored = 0.80 V
H2O2 (aq) + 2H+ (aq) + 2e- → 2H2O(l) Eored = 1.78 V
In which scenario is Ecell greater than Eocell?
[Ag+] = 1.0 M, [H2O2] = 2.0 M, [H+] = 2.0 M
[Ag+] = 2.0 M, [H2O2] = 1.0 M, [H+] = 1.0x10-7 M
[Ag+]= 2.0 M, [H2O2] = 0.25 M, [H+] = 4.0 M
A–1
B–2
C–3
D – 2, 3
E – 1, 3
Chpt 20: Electrochemistry
CEM 152 – SS2012
Example

1.
2.
3.
4.
A galvanic cell is constructed from the following two half
reactions
Cu2+ (aq) + 2e- → Cu (s)
Eored = 0.34
Ag+ + e- → Ag (s)
Eored = 0.80
The electrodes in this cell are Ag(s) and Cu(s). For which of
the following scenarios does the cell potential decrease?
CuSO4(s) is added to the copper half-cell compartment
(assume no volume change)
NH3 (aq) is added to the copper half-cell compartment [Cu2+
reacts with NH3 to form Cu(NH3)42+ (aq)]
NaCl (s) is added to the silver half-cell compartment [Ag+
reacts with Cl- to form AgCl(s)]
Water is added to both half-cells until the volumes are doubled
Chpt 20: Electrochemistry
CEM 152 – SS2012
A – 1, 3
B – 2, 3, 4
C – 1,4
D – 2, 4
E – 1, 3, 4
Equilibrium Constants for
Redox Reactions
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
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At equilibrium, DG = 0, Q = K and
0 = Eo - (0.0592/n) logK
logK = nEo/0.0592
The above expression allows us to calculate the equilibrium
constant for a redox reaction from the standard emf for the
reaction
Example: What is the equilibrium constant for reaction
2Al(s) + 3Mn2+(aq)  2Al3+(aq) + 3Mn(s)
We have already that determined Eo = 0.48V and n = 6
logK = (6)(0.48V)/0.0592
logK = 48.64
K = 4.45 x 10+48
Chpt 20: Electrochemistry
CEM 152 – SS2012
Corrosion
Corrosion reactions are redox reactions in which a metal is attacked
by some substance in its environment and converted to an unwanted
compound

Corrosion of iron (two-step process)
I. 2Fe(s) + O2(g) + 4H+ 
2Fe2+(aq) + 2H2O()
II. 4Fe2+(aq) + O2(g)+ 4H2O() 
2Fe2O3(s) + 8H+(aq)

The half reactions for the first step
O2(g) + 4H+ + 4e-  2H2O() Eored = 1.23 V
Fe(s)  Fe2+(aq) + 2e- Eored = -0.44 V

Salts enhance the corrosion as the ions serve as electrolytes to
complete the electrical circuit

Chpt 20: Electrochemistry
CEM 152 – SS2012
Corrosion
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Multiple ways to prevent corrosion.


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Some metals form a protective oxide layer over its
surface and prevent further oxidation


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Paint or metal coating on metal
alloying
Cathodic protection
Al
Zn (coating a metal in zinc is referred to as galvanizing)
Stainless steel, a steel alloy with Cr, also forms a
protective oxide layer over its surface.
Chpt 20: Electrochemistry
CEM 152 – SS2012
Cathodic Protection




To prevent corrosion, cathodic protection is
sometimes employed.
Protect a metal by ensuring it will act as the
cathode in an electrochemical cell.
A sacrificial anode, typically Zn, is attached to the
cathode that is being protected.
Compare the following half reactions



Fe2+ (aq) + 2e- → Fe(s)
Zn2+(aq) + 2e- → Zn(s)
Mg2+ (aq) + 2e- → Mg(s)
Chpt 20: Electrochemistry
Eored = -0.44 V
Eored = -0.76 V
Eored = -2.37 V
CEM 152 – SS2012
Electrolysis
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Electrical energy can be used to induce non-spontaneous redox
reactions to occur
The electrolysis of water
2H2O()  O2(g) + 2H2(g)
Ox: 2H2O()  O2(g) + 4H+ + 4eRed:
4H2O() + 4e-  2H2(g) + 4OHThe cell emf for this reaction is
Eo = Eored(reduction) - Eored(oxidation)
Eo = –0.83 - (+1.23 V)
Eo = –2.06 V
Since Eo is negative, the reaction in the forward direction is nonspontaneous
If work is done on the system (in the form of electrical energy) the
reaction can proceed
Chpt 20: Electrochemistry
CEM 152 – SS2012
Electrochemical Cell


A power supply can provide
electrical energy to drive the
electrochemical cell
Oxidation occurs at the anode


Reduction occurs at the cathode

Chpt 20: Electrochemistry
The anode is positive since e- are
withdrawn from this electrode
the cathode is negative since e- are
supplied to this electrode
CEM 152 – SS2012
Electrolysis and Stoichiometry

Example: What volume of H2(g) at STP is produced when 5.0 A is passed
through a water electrolysis cell for 1 hour?
5 A = 5 C/s
# Coulombs = (5 C/s)(1 hr)(3600 s/hr)
= 18,000 C
mol e- = (18,000 C)/(96,500 C/mol e-)
= 0.186 mol e2H2O()  O2(g) + 2H2(g)
2 mol H2(g) liberated for every 4 mol emol H2(g) = (0.186mol e-)(2mol H2)/(4mol e-)
= 0.093 mol H2(g)
V = nRT/P
V = (0.093 mol)(0.082 Latm/molK)(298 K)/
(1 atm) = 2.28 L
Chpt 20: Electrochemistry
CEM 152 – SS2012
Isolation of Al



Aluminum is abundant on Earth (behind O2 and Si)
but until recently was incredibly expensive
Aluminum is readily oxidized and is found in oxide
compounds
Al3+ (aq) + 3e- → Al(s) Eored = -1.66 V
Aluminum cannot be plated out of a solution of
aqueous Al3+ ions since the electrolysis of water
would occur first
2 H2O + 2e- → H2 + 2OH- Eored = -0.83 V
Chpt 20: Electrochemistry
CEM 152 – SS2012
Isolation of Al




Aluminum isolated
electrochemically from
Al2O3.
Cathode: AlF63- (aq) + 3e- →
Al(s) + 6FReduction: 2Al2OF62- + 12F+ C → 4AlF63- + CO2 + 4eProcess consumes a few
percent of the power
produced in the US
annually.
Chpt 20: Electrochemistry
CEM 152 – SS2011
Electrolysis and Electrical
Work


The change in free energy for a chemical system provides a
measure of the maximum useful work extracted from the
reaction
wmax = DG = -nFE
We should remember that work done on a system will result in
w>0



since electrolysis involves a non-spontaneous process, E < 0
and hence w > 0
Electrical work is typically expressed as a product of power
times time
energy (J) = power (J/s)  time (s)
The unit of electrical power is the watt
1 watt (W) = 1 J/s
Chpt 20: Electrochemistry
CEM 152 – SS2012
Electrical
Work Example


Calculate the number of kWh of electricity required to produce 1 kg of
Mg from electrolysis of MgCl2 if applied emf = 5.0V
mol Mg = (1x103 g Mg)/(24.3 g/mol)
= 41.1 mol
mol e- = (41.1 mol Mg)(2 mol e-)/(1 mol Mg)
= 82.3 mol ew = -nFE
w = -(82.3 mol e-)(96,500 J/Vmol e-)(-5.0 V)
w = 3.97 x 107 J
1 kWh = 3.6 x 106 J
w = (3.97 x 107 J)/(3.6 x 106 J/kWh)
w = 11.0 kWh
This is assuming the electrolytic cell is 100 percent efficient
Chpt 20: Electrochemistry
CEM 152 – SS2012
Electroplating
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Electrolytic processes which involve a metal electrode which
participates in the cell reaction can be used to deposit the
active metal onto another
Oxidation at M (active metal) anode
M(s)  M2+ + 2e- ; Eored < 1.23 V
Reduction of M at metallic cathode
M2+ + 2e-  M(s) ; Eored > -0.83
The ions from the active metal which go into solution are
“plated” on the cathode electrode (where the reduction occurs)
For reference


O2(g) + 4H+(aq) + 4e- → H2O (l)
2 H2O(l) 2e- → H2(g) + 2OH-
Chpt 20: Electrochemistry
CEM 152 – SS2012
Eored = +1.23 V
Eored = -0.83 V