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Electric power conversion in electrochemistry
Chemical Reactions
Electric Power
Electrolysis / Power consumption
Electrochemical battery / Power generation
battery+-
inertelectrodes
powersource
vessel
e-
e-
conductivemedium
CellConstruction
Sign or polarity of electrodes
(-) (+)
Na+ Cl-
Let’s examine the electrolytic cell for molten NaCl.
+-battery
Na (l)
electrode half-cell
electrode half-cell
Molten NaCl
Na+
Cl-
Cl-
Na+
Na+
Na+ + e- Na 2Cl- Cl2 + 2e-
Cl2 (g) escapes
Observe the reactions at the electrodes
NaCl (l)
(-)
Cl-
(+)
+-battery
e-
e-
NaCl (l)
(-) (+)
cathode anode
Molten NaCl
Na+
Cl-
Cl-
Cl-
Na+
Na+
Na+ + e- Na 2Cl- Cl2 + 2e-
cationsmigrate toward
(-) electrode
anionsmigrate toward
(+) electrode
At the microscopic level
Molten NaCl Electrolytic Cellcathode half-cell (-)
REDUCTION Na+ + e- Na
anode half-cell (+)OXIDATION 2Cl- Cl2 + 2e-
overall cell reaction2Na+ + 2Cl- 2Na + Cl2
X 2
Non-spontaneous reaction!
Definitions:
CATHODE
REDUCTION occurs at this electrode
ANODE
OXIDATION occurs at this electrode
Na+ Cl-
H2O
Will the half-cell reactions be the same or different?
Water Complications in ElectrolysisIn an electrolysis, the most easily oxidized and most easily reduced reaction occurs.
When water is present in an electrolysis reaction, then water (H2O) can be oxidized or reduced according to the reaction shown.
Electrode Ions ... Anode Rxn Cathode Rxn E°Pt (inert) H2O H2O(l)+ 2e- H2(g)+ 2OH-
(aq) -0.83 VH2O 2 H2O(l) 4e- + 4H+
(g) + O2(g) -1.23 V
Net Rxn Occurring:
2 H2 H22O O 2 H 2 H2(g)2(g)+ O+ O2 (g)2 (g) E° E° = - 2.06 V= - 2.06 V
battery+- power
source
e-
e-
NaCl (aq)
(-) (+)cathodedifferent half-cell
Aqueous NaCl
anode2Cl- Cl2 + 2e-
Na+
Cl-
H2O
What could be reduced at the
cathode?
Aqueous NaCl Electrolytic Cellpossible cathode half-cells (-)
REDUCTION Na+ + e- Na2H2O + 2e- H2 + 2OH-
possible anode half-cells (+)OXIDATION 2Cl- Cl2 + 2e-
2H2O O2 + 4H+ + 4e-
overall cell reaction2Cl- + 2H2O H2 + Cl2 + 2OH-
e-
Ag+
Ag
For every electron, an atom of silver is plated on the
electrode.Ag+ + e- Ag
Electrical current is expressed in terms of the
ampere, which is defined as that strength of current
which, when passed thru a solution of AgNO3 (aq) under
standard conditions, will deposit silver at the rate of
0.001118 g Ag/sec
1 amp = 0.001118 g Ag/sec
The mass deposited or eroded from an electrode depends on the quantity of
electricity.Quantity of electricity – coulomb (Q)
Q is the product of current in amps times time in
secondsQ = It
coulomb
current in amperes (amp)
time in seconds
1 coulomb = 1 amp-sec = 0.001118 g Ag
Ag+ + e- Ag
1.00 mole e- = 1.00 mole Ag = 107.87 g Ag
107.87 g Ag/mole e-
0.001118 g Ag/coul= 96,485 coul/mole e-
1 Faraday (F )mole e- = Q/F
mass = molemetal x MM
molemetal depends on the half-cell reaction
Examples using Faraday’s LawHow many grams of Cu will be deposited in
3.00 hours by a current of 4.00 amps? Cu+2 + 2e- Cu
The charge on a single electron is 1.6021 x 10-
19 coulomb. Calculate Avogadro’s number from the fact that 1 F = 96,487 coulombs/mole e-.
Fig 18-26 Pg 901
The electroplating of silver.
Pg 901
Courtesy International SilverPlating, Inc.
21-8 Industrial Electrolysis Processes
Prentice-Hall © 2002 General Chemistry: Chapter 21Slide 18
of 52
A series of solutions have 50,000 coulombs passed thru them, if the solutions were Au+3, Zn+2, and Ag+, and Au, Zn, and Ag were plated out respectively, calculate the amount of metal deposited at each anode.
battery- +
+ + +- - -
1.0 M Au+3 1.0 M Zn+2 1.0 M Ag+
Au+3 + 3e- Au Zn+2 + 2e- Zn Ag+ + e- Ag
e-
e- e- e-
Volta’s battery (1800)
Alessandro Volta 1745 - 1827
Paper moisturized with NaCl solution
Cu
Zn
Lesson 9NEEP 42321
Galvanic cells and electrodes
To sustain the cell reaction, the charge carried by the electrons through the external circuit must be accompanied by a compensating transport of ions between the two cells.
Relative amounts of charge can be carried by negative or positive ions (depends on their relative mobilities) through the solution.
Salt bridge, consists of an intermediate compartment filled with saturated salt solution and fitted with porous barriers at each end, is used for precise measurements. The purpose of salt bridge is to minimize the natural potential difference (junction potential).
with Galvanic Cells
19.2
spontaneousredox reaction
anodeoxidation
cathodereduction
Fig 18-9Pg 872
Diagram of a galvanic cell containing passive electrodes. the two platinum electrodes do not take part in the redoxchemistry of this cell. Theyonly conduct electrons to andfrom the interfaces.
Electrodes are passive (not involved in the reaction)
Olmsted Williams
Cu
1.0 M CuSO4
Zn
1.0 M ZnSO4
Salt bridge – KCl in agar
Provides conduction between half-cells
CellConstruction
Observe the electrodes to see what is occurring.
Cu
1.0 M CuSO4
Zn
1.0 M ZnSO4
Cu plates out or
deposits on
electrode
Zn electrode erodes
or dissolves
cathode half-cellCu+2 + 2e- Cu
anode half-cellZn Zn+2 + 2e-
-+
What about half-cell reactions?
What about the sign of the electrodes?
What happened
at each electrode
?
Why?
Cu
1.0 M CuSO4
Zn
1.0 M ZnSO4
cathode half-cellCu+2 + 2e- Cu
anode half-cellZn Zn+2 + 2e-
-+
Now replace the light bulb with a volt meter.
1.1 volts
H2 input1.00 atm
inert metal
We need a standard electrode to make
measurements against!The Standard Hydrogen Electrode (SHE)
Pt
1.00 M H+
25oC1.00 M H+
1.00 atm H2
Half-cell2H+ + 2e- H2
EoSHE = 0.0 volts
How do we calculate Standard Redox Potentials?
We must compare the half reactions to a standard
What is that standard?
2 H3O+(aq) + 2e- H2(g) + 2 H2O(l) E°= 0.00 V
This is called the standard hydrogen electrode or SHE
Now that we have a standard, we can calculate standard redox potential by using the table of standard redox potentials
19.3
• 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
Copyright 1999, PRENTICE HALL Chapter 20 30
Cell EMFCell EMFOxidizing and Reducing AgentsOxidizing and Reducing Agents
H2 1.00 atm
Pt
1.0 M H+
Cu
1.0 M CuSO4
0.34 v
cathode half-cellCu+2 + 2e- Cu
anode half-cellH2 2H+ + 2e-
KCl in agar
+
Now let’s combine the copper half-cell with the SHE
Eo = + 0.34 v
H2 1.00 atm
Pt
1.0 M H+1.0 M ZnSO4
0.76 vcathode half-cell2H+ + 2e- H2
anode half-cellZn Zn+2 +
2e-
KCl in agar
Zn
-
Now let’s combine the zinc half-cell with the SHE
Eo = - 0.76 v
Assigning the Eo
Al+3 + 3e- Al Eo = - 1.66 v
Zn+2 + 2e- Zn Eo = - 0.76 v
2H+ + 2e- H2 Eo = 0.00 v
Cu+2 + 2e- Cu Eo = + 0.34
Ag+ + e- Ag Eo = + 0.80 v
Write a reduction half-cell, assign the voltage measured, and the sign of the electrode to the
voltage.
Incr
easi
ng a
ctiv
ity
Measuring Standard Reduction Potential
Prentice-Hall © 2002 General Chemistry: Chapter 21Slide 34 of
52
cathode cathode anodeanode
What is the standard emf of an electrochemical cell made of a Cd electrode in a 1.0 M Cd(NO3)2 solution and a Cr
electrode in a 1.0 M Cr(NO3)3 solution?
Cd2+ (aq) + 2e- Cd (s) E0 = -0.40 V
Cr3+ (aq) + 3e- Cr (s) E0 = -0.74 VCd is the stronger oxidizer
Cd will oxidize Cr
2e- + Cd2+ (1 M) Cd (s)
Cr (s) Cr3+ (1 M) + 3e-Anode (oxidation):
Cathode (reduction):
2Cr (s) + 3Cd2+ (1 M) 3Cd (s) + 2Cr3+ (1 M)
x 2
x 3
E0 = Ecathode - Eanodecell0 0
E0 = -0.40 – (-0.74) cell
E0 = 0.34 V cell
19.3
Calculating the cell potential, Eocell, at
standard conditions
Fe+2 + 2e- Fe Eo = -0.44 v
O2 (g) + 2H2O + 4e- 4 OH- Eo = +0.40 v
This is corrosion or the oxidation of a metal.
Consider a drop of oxygenated water on an iron objectFe
H2O with O2
Fe Fe+2 + 2e- -Eo = +0.44 v2x
2Fe + O2 (g) + 2H2O 2Fe(OH)2 (s) Eocell= +0.84 v
reverse
• Whereas potential and free energy are related, and free energy and equilibrium are related, equilibrium and potential must be related to one another.
G° = –nFE°cell
andG° = –RT ln Keq
therefore –RT ln Keq = –nFEocell
Equilibrium Constants in Redox Reactions
RT ln Keq RTE°cell = ––––––––– = –––– ln Keq nF nF
0.025693 VE°cell = –––––––– ln Keq n
R and F are constant, therefore at 298 K:
41
Effect of Concentration on Cell Effect of Concentration on Cell EMFEMF
• A voltaic cell is functional until E = 0 at which point equilibrium has been reached.
• The point at which E = 0 is determined by the concentrations of the species involved in the
redox reaction.
The Nernst EquationThe Nernst Equation• The Nernst equation relates emf to
concentration using
and noting that
QRTGG ln
QRTnFEnFE ln
and the previous relationship:Go = -nFEo
cell
from thermodynamics:Go = -2.303RT log K
-nFEocell = -2.303RT log K
at 25oC: Eocell = 0.0591 log K
n
where n is the number of electrons for the balanced reaction
What happens to the electrode potential if conditions are not at standard conditions?
The Nernst equation adjusts for non-standard conditions
For a reduction potential: ox + ne red
at 25oC: E = Eo - 0.0591 log (red)
n (ox)Calculate the E for the hydrogen
electrode where 0.50 M H+ and 0.95 atm H2.
in general: E = Eo – RT ln (red)
nF (ox)
1) An example:
Ni(s) | Ni2+(0.600M)|| Sn2+
(0.300M) | Sn(s)
According to the reduction potentials:
2 e- + Ni2+ Ni(s) -0.230 V
2 e- + Sn2+ Sn(s) -0.140V
One of these needs to be reversed to get a positive voltage, and a spontaneous reaction.
Ni(s) 2 e- + Ni2+ +0.230 V
2 e- + Sn2+ Sn(s) -0.140V
Ni(s) + Sn2+ Ni2+ + Sn(s) 0.090 V = Eo
E) Calculation of the equilibrium constant
1) at equilibrium E = 0 ; Q = ____
From the Nernst Equation:
For the cell:
Ni(s) | Ni2+(0.600M)|| Sn2+
(0.300M) | Sn(s)
2) An example:
Sn(s) | Sn2+(1.0M)|| Pb2+
(0.0010M) | Pb(s)
According to the reduction potentials:
2 e- + Pb2+ Pb(s) E0=-0.126 V
2 e- + Sn2+ Sn(s) E0=-0.136V
One of these needs to be reversed to get a positive voltage, and a spontaneous reaction.!!!!!!!!!!!!!!!!!!!!!!!!!
Go = -nFEocell
Free Energy and the Cell Potential
Cu Cu+2 + 2e- -Eo = - 0.34
Ag+ + e- Ag Eo = + 0.80 v2x
Cu + 2Ag+ Cu+2 + 2AgEocell= +0.46 v
where n is the number of electrons for the balanced reaction
What is the free energy for the cell?
1F = 96,500 J/v
Electrolysis of Copper
A net reaction of zero, yet a process does take place.
A concentration cell based on the A concentration cell based on the Cu/Cu2+ half-reaction.Cu/Cu2+ half-reaction. AA, Even , Even
though the half-reactions involve the though the half-reactions involve the same components, the cell operates same components, the cell operates
because the half-cell concentrations are because the half-cell concentrations are different. different. BB, The cell operates , The cell operates
spontaneously until the half-cell spontaneously until the half-cell concentrations are equal. Note the concentrations are equal. Note the
change in electrodes (exaggerated here change in electrodes (exaggerated here for clarity) and the equal color of for clarity) and the equal color of
solutions.solutions.
Concentration Cells
Prentice-Hall © 2002 General Chemistry: Chapter 21Slide 53
of 52
Two half cells with identical electrodes but different ion concentrations.
2 H+(1 M) → 2 H+(x M)
Pt|H2 (1 atm)|H+(x M)||H+(1.0 M)|H2(1 atm)|Pt(s)
2 H+(1 M) + 2 e- → H2(g, 1 atm)
H2(g, 1 atm) → 2 H+(x M) + 2 e-
Concentration Cells
Prentice-Hall © 2002 General Chemistry: Chapter 21Slide 54
of 52
Ecell = Ecell° - logn
0.0592 V x2
12
Ecell = 0 - log2
0.0592 V x2
1
Ecell = - 0.0592 V log x
Ecell = (0.0592 V) pH
2 H+(1 M) → 2 H+(x M)
Ecell = Ecell° - log Qn
0.0592 V
F)The pH meter is a special case of the Nernst Equation
1/2 H2 H+ + 1 e-
Q = [H+]
pH = - log [H+]
E = Eo + 0.0592 pH
The scale on the pH meter is marked off so that a change of 1 pH unit equals 0.0592 volts or 59.2
millivolts.
The pH MeterIn practice, a special pH electrode is much more convenient than using platinum electrodes and a tank of hydrogen gas!
A stable reference electrode and a glass-membrane electrode are contained within a combination
pH electrode.
The electrode is merely dipped into a solution, and
the potential difference between the electrodes is
displayed as pH.
galvanic electrolytic
needpowersource
twoelectrodes
produces electrical current
anode (-)cathode (+)
anode (+)cathode (-)
salt bridge vessel
conductive medium
Comparison of Electrochemical Cells
G < 0G > 0
Copyright 1999, PRENTICE HALL Chapter 20 60
CorrosionCorrosionCorrosion of IronCorrosion of Iron
• Since Ered(Fe2+) < Ered(O2) iron can be oxidized by oxygen.
• Cathode: O2(g) + 4H+(aq) + 4e- 2H2O(l).
• Anode: Fe(s) Fe2+(aq) + 2e-.• Dissolved oxygen in water usually causes the
oxidation of iron.• Fe2+ initially formed can be further oxidized to
Fe3+ which forms rust, Fe2O3.xH2O(s).
• Oxidation occurs at the site with the greatest concentration of O2.
21-6 Corrosion: Unwanted Voltaic Cells
Prentice-Hall © 2002 General Chemistry: Chapter 21Slide 61
of 52
O2(g) + 2 H2O(l) + 4 e- → 4 OH-(aq)
2 Fe(s) → 2 Fe2+(aq) + 4 e-
2 Fe(s) + O2(g) + 2 H2O(l) → 2 Fe2+(aq) + 4 OH-(aq)
Ecell = 0.841 V
EO2/OH- = 0.401 V
EFe/Fe2+ = -0.440 V
In neutral solution:
In acidic solution:
O2(g) + 4 H+(aq) + 4 e- → 4 H2O (aq) EO2/OH- = 1.229 V
Copyright 1999, PRENTICE HALL Chapter 20 62
CorrosionCorrosionCorrosion of IronCorrosion of Iron
Corrosion of Iron
Corrosion of IronRust formation:
4Fe2+(aq) + O2(g) + 4H+(aq) 4Fe3+(aq) + 2H2O(l)
2Fe3+(aq) + 4H2O(l) Fe2O3H2O(s) + 6H+
(aq)
Prevention of CorrosionCover the Fe surface with a protective coatingPaint
Tin Zn
Galvanized iron
Copyright 1999, PRENTICE HALL Chapter 20 66
CorrosionCorrosionPreventing the Corrosion of IronPreventing the Corrosion of Iron
• Corrosion can be prevented by coating the iron with paint or another metal.
• Galvanized iron is coated with a thin layer of zinc.
• Zinc protects the iron since Zn is the anode and Fe the cathode:
Zn2+(aq) +2e- Zn(s), Ered = -0.76 V
Fe2+(aq) + 2e- Fe(s), Ered = -0.44 V
• With the above standard reduction potentials, Zn is easier to oxidize than Fe.
Corrosion Protection
Prentice-Hall © 2002 General Chemistry: Chapter 21Slide 67
of 52
galvanized steel (Fe)
(cathode)
(electrolyte)
(anode)
Cathodic ProtectionIn cathodic protection, an iron object to be protected is connected to a chunk of an active metal.
• The iron serves as the reduction electrode and remains metallic. The active metal is oxidized.
• Water heaters often employ a magnesium anode for cathodic protection.