Transcript Chapter 20 Electrochemistry - University of the Witwatersrand

Chapter 20
Electrochemistry
CHEM 180/181
Chapter 20
Topics in Electrochemistry
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Oxidation-Reduction (Redox) Numbers
Balancing Redox Reactions
Voltaic Cells
Cell EMF
Spontaneity of Redox Reactions
Batteries
Corrosion
Electrolysis
CHEM 180/181
Chapter 20
Oxidation and Reduction
Oxidation (loss of e-)
Na
Na+ + e-
Reduction (gain of e-)
Cl + e-
Cl-
Oxidation-reduction (redox) reactions occur when electrons
are transferred from an atom that is oxidized to an atom that
is reduced.
Electron transfer can produce electrical energy spontaneously,
but sometimes electrical energy is used to make them occur
(nonspontaneous).
CHEM 180/181
Chapter 20
CHEM 180/181
Chapter 20
Oxidation-Reduction (Redox) Reactions
BOTH reduction and oxidation must occur.
A substance that gives up electrons is oxidized and is
called a reducing agent or reductant (causes another
substance to be reduced).
A substance that accepts electrons is reduced and is
therefore called an oxidizing agent or oxidant (causes
another substance to be oxidized).
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Chapter 20
Oxidation-reduction (Redox) Reactions
Determine if it’s a redox reaction by keeping track of the
oxidation states of all elements involved.
Zn(s) + 2H+(aq)
(0)
(+1)
Zn2+(aq) + H2(g)
(+2)
(0)
Quick hint: If a reaction includes an ELEMENT that turns into
an ion, it’s a redox reaction.
CHEM 180/181
Chapter 20
Oxidation Number Guidelines
1. Atoms in elemental form, oxidation number is zero.
(Cl2, H2, P4, Ne are all zero)
2. Monoatomic ion, the oxidation number is the charge on
the ion. (Na+: +1; Al3+: +3; Cl-: -1)
3. O is usually -2. But in peroxides (like H2O2 and Na2O2) it
has an oxidation number of -1.
4. H is +1 when bonded to nonmetals and -1 when bonded
to metals.
(+1 in H2O, NH3 and CH4; -1 in NaH, CaH2 and AlH3)
5. The oxidation number of F is -1.
6. The sum of the oxidation numbers for the molecule is the
charge on the molecule (zero for a neutral molecule).
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Chapter 20
Example
Determine the oxidation state of all elements in
ammonium thiosulfate (NH4)2(S2O3).
(NH4)2(S2O3)
NH4+
S2O32-
-3 +1
sum = overall
charge on ion
CHEM 180/181
+2 -2
NH4+
-3 +4
S2O32-
total
+4 -6
Chapter 20
Determining Oxidation States
What is the oxidation state of Mn in MnO4-?
Answer: +7
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Chapter 20
Balancing oxidation-reduction equations
Balancing chemical equations follows law of conservation of
mass.
AND, gains and losses of electrons must also be balanced.
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Chapter 20
Half-Reactions
Separate oxidation and reduction processes in equation,
Sn2+(aq) + 2Fe3+(aq)
Sn4+(aq) + 2Fe2+(aq)
Oxidation:
Sn2+(aq)
Reduction: 2Fe3+(aq) + 2e-
Sn4+(aq) + 2e2Fe2+(aq)
Overall, the number of electrons lost in the oxidation half
reaction must equal the number gained in the reduction
half reaction.
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Chapter 20
Balancing Equations by the Method of
Half-Reactions: Acidic
Consider :
MnO4-(aq) + C2O42-(aq)
Mn2+(aq) + CO2(g)
Unbalanced half-reactions: MnO4-(aq)
C2O42-(aq)
(acidic)
Mn2+(aq)
CO2(g)
First, balance everything EXCEPT hydrogen and oxygen.
Deal with half-reactions SEPARATELY.
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Chapter 20
Balancing Equations by the Method of
Half-Reactions: Acidic
To balance O:
Add 4H2O to products to balance oxygen in reactants.
MnO4-(aq)
Mn2+(aq) + 4H2O(l)
To balance H:
Add 8H+ to reactant side to balance the 8H in water.
8H+(aq) + MnO4-(aq)
Mn2+(aq) + 4H2O(l)
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Balancing Equations by the Method of
Half-Reactions: Acidic
Balance charge: Add up charges on both sides.
Add 5 electrons to reactant side.
5e- + 8H+(aq) + MnO4-(aq)
Mn2+(aq) + 4H2O(l)
Mass balance of C in oxalate half-reaction.
C2O42-(aq)
2CO2(g)
Balance charge by adding two electrons to the products.
C2O42-(aq)
2CO2(g) + 2eLast step: Cancel electrons and add reactions together.
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Chapter 20
Balancing Equations by the Method of
Half-Reactions: Acidic
5e- + 8H+(aq) + MnO4-(aq)
C2O42-(aq)
Mn2+(aq) + 4H2O(l)
2CO2(g) + 2e-
Top reaction times 2. Bottom reaction times 5. ALL PARTS!!
10e- + 16H+(aq) + 2MnO4-(aq)
2Mn2+(aq) + 8H2O(l)
5C2O42-(aq)
10CO2(g) + 10e.
16H+(aq) + 2MnO4-(aq) + 5C2O42-(aq)
2Mn2+(aq) + 8H2O(l) + 10CO2(g)
**Electrons have cancelled out**
CHEM 180/181
Chapter 20
Balancing Equations by the Method of
Half-Reactions: Acidic Summary
1. Divide equation into two incomplete half-reactions.
2. Balance each half-reaction
(a) balance elements other than H and O.
(b) balance O atoms by adding H2O.
(c) balance H atoms by adding H+ (basic conditions will
require further work at this step).
(d) balance charge by adding e- to the side with greater
overall positive charge.
CHEM 180/181
Chapter 20
Balancing Equations by the Method
of Half-Reactions: Summary
3.
Multiply each half-reaction by an integer so that the number
of electrons lost in one half-reaction equals the number
gained in the other half-reaction.
4.
Add the two half-reactions and cancel out all species
appearing on both sides of the equation.
5.
Check equation to make sure there are same number of
atoms of each kind and the same total charge on both sides.
Errors can be caught!!
CHEM 180/181
Chapter 20
Balancing Equations by the Method of
Half-Reactions: Basic
Balancing process is started using H+ and H2O, then adjusting
with OH- to uphold reaction conditions
(H+ does not exist in basic solutions).
Balance the following reaction:
H2O2(aq) + ClO2(aq)
ClO2-(aq) + O2(g)
CHEM 180/181
Chapter 20
(basic)
Balancing Equations by the Method of
Basic Redox Reactions
Half-Reactions: Basic
Split into two half-reactions.
H2O2(aq)
O2(g)
ClO2(aq)
ClO2-(aq)
Balance elements, then oxygen by adding H 2O. Then, add H+
to balance H, just like an acidic redox reaction.
H2O2(aq)
ClO2(aq)
CHEM 180/181
O2(g) + 2H+
ClO2-(aq)
Chapter 20
Balancing
Equations
by
the
Method
of
Basic Redox Reactions
Half-Reactions: Basic
Add OH- to both sides, enough to neutralize all H+ (basic
reactions cannot support H+).
2OH- + H2O2(aq)
O2(g) + 2H+ + 2OHClO2(aq)
ClO2-(aq)
Combine H+ and OH- to form H2O.
2OH- + H2O2(aq)
O2(g) + 2H2O
CHEM 180/181
Chapter 20
Balancing
Equations
by
the
Method
of
Basic Redox Reactions
Half-Reactions: Basic
Balance charge by adding e-.
2OH- + H2O2(aq)
O2(g) + 2H2O + 2e1e- + ClO2(aq)
ClO2-(aq)
Multiply each reaction so both have same e -. Then add them
together, cancelling where possible.
2OH- + H2O2(aq) + 2ClO2(aq)
O2(g) + 2H2O + 2ClO2-(aq)
Double-check your answer!
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Chapter 20
Balance this redox equation
Cu(s) + NO3-(aq)
Ans: Cu(s) + 2NO3-(aq) + 4H+(aq)
CHEM 180/181
Cu2+(aq) + NO2(g)
(acidic)
Cu2+(aq) + 2NO2(aq) + 2H2O(l)
Chapter 20
Balance this redox equation
NO2-(aq) + Al(s)
NH3(aq) + Al(OH)4-(aq)
2Al(s) + NO2-(aq) +OH-(aq) + 5H2O(aq)
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Chapter 20
(basic)
2Al(OH)4-(aq) + NH3(aq)
Voltaic Cells
Voltaic (aka galvanic) cells: electrochemical reactions in which
electron transfer occurs via an external circuit.
The energy released in a voltaic cell reaction can be used to
perform electrical work.
Reactions are spontaneous.
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Voltaic Cells: Components
For example
Anode: Zn(s)  Zn2+(aq) + 2eCathode: Cu2+(aq) + 2e-  Cu(s)
(oxidation half-reaction)
(reduction half reaction)
Salt bridge: cations move from anode to cathode, anions
move from cathode to anode.
External circuit (wire): electrons move from anode to
cathode (between two solid metal electrodes). Electron
transfer can naturally occur in OTHER forms besides a
wire (direct electron transport between solution and
metal).
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Voltaic Cells: Ion Flow
Anions and cations move through a porous barrier or salt
bridge.
Cations move into the cathodic compartment to neutralize the
excess negatively charged ions.
Anions move into the anodic compartment to neutralize the
excess Zn2+ ions formed by oxidation.
Anode: Zn(s)  Zn2+(aq) + 2e-
CHEM 180/181
Chapter 20
(oxidation half-reaction)
Simplified Voltaic Cell
In any voltaic cell, the
electrons flow from the
anode through the
external circuit to the
cathode.
Anode thus labeled with
negative sign (-) and
cathode with positive
sign (+).
Anions always migrate
toward the anode and
cations towards the
cathode.
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Chapter 20
Voltaic Cells: Electron Flow
As oxidation occurs, Zn(s) is converted to Zn 2+ and 2e-.
The electrons flow from anode (where they’re produced)
towards the cathode where they are consumed in the
reduction reaction.
We expect the Zn
electrode to lose mass
and the Cu electrode
to gain mass.
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Chapter 20
One-Pot Voltaic Cells
If a strip of Zn is placed in a solution of CuSO 4, Cu is deposited
on the Zn and the Zn dissolves, forming Zn 2+.
Zn is spontaneously oxidized to Zn2+ by Cu2+ (Cu is the
oxidizing agent).
The Cu2+ is spontaneously reduced to Cu0 by Zn (Zn is the
reducing agent).
The entire process is spontaneous.
Zn(s) + Cu2+(aq)
Zn2+(aq) + Cu(s)
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Chapter 20
The Atomic Level
A Cu2+(aq) ion comes into contact with a Zn( s) atom on the
surface of the electrode. Two electrons are directly transferred
from the Zn(s), forming Zn2+(aq), to the Cu2+(aq) forming Cu(s).
CHEM 180/181
Chapter 20
Cell Electromotive Force (EMF)
Electromotive force (emf): the force required to push electrons
through the external circuit.
Potential difference: difference per electrical charge between
two electrodes. Measured in volts.
One volt is the potential required to impart one joule of energy
to a charge of one coulomb:
1J
1V 
1C
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Chapter 20
Cell EMF
Cell potential (Ecell): the emf of a cell. Aka cell voltage and is
positive for spontaneous cell reactions.
Ecell strictly depends on reactions that occur at the cathode and
anode, the concentration of reactants and products, and the
temperature and is described by the equation:
Ecell = Ered(cathode) - Ered(anode)
For 1M solutions at 25˚C and 1 atm (standard conditions),
the standard emf (standard cell potential), Ecell, is written as
E ˚cell.
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Chapter 20
Standard Reduction Potentials
Emf of a voltaic cell depends on the particular cathode and
anode half-cell reactions involved.
Standard reduction potential measured for each half-cell and
then used to determine E˚cell (emf).
Convenient and extensive tabulation of electrochemical data.
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Chapter 20
Standard Reduction Potentials
Potential associated with each half-cell reaction is
chosen to be the potential for reduction to occur at
that electrode. Hence, Standard Reduction
Potentials.
Standard reduction potentials, E˚red, are measured
relative to the standard hydrogen electrode (SHE).
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Standard Hydrogen Electrode (SHE)
SHE is assumed as the cathode (for consistency). Pt
electrode in a tube containing 1M H+ solution. H2(g) is
bubbled through the tube and equilibrium is established.
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Chapter 20
Determining Standard Reduction
Potentials
For the SHE:
2H+(aq, 1M) + 2e-  H2(g, 1 atm)
E˚red = 0.00 V
Standard reduction potentials can then be calculated using the
SHE (E˚= 0.00 V) as E˚red(cathode).
Ecell  Ered cathode  Ered anode
Each calculated E˚ is rewritten as a reduction and tabulated.
CHEM 180/181
Chapter 20
Finding the Standard Reduction Potential
E˚cell is measured, 0.00 V (SHE) is used for E˚(cathode), and the
reduction potential for Zn is found.
CHEM 180/181
Chapter 20
Zn Standard Reduction Potential
We measure E˚cell relative to the SHE:
E˚cell = E˚red(cathode) - E˚red(anode)
0.76V = 0.00 V - E˚red(anode).
Therefore, E˚red(anode) = -0.76V
And we find that -0.76V can be assigned to reduction of zinc.
Zn2+(aq) + 2e-  Zn(s)
E˚red = -0.76 V
CHEM 180/181
Chapter 20
Zn Standard Reduction Potential
Since E˚red = -0.76 V (negative!) we conclude that the
reduction of Zn2+ in the presence of the SHE is not
spontaneous.
However, the oxidation of Zn with the SHE is spontaneous.
Reactions with E˚red > 0 are spontaneous reductions relative to
the SHE. E˚red < 0 are spontaneous oxidations.
Changing the stoichiometric coefficient does not affect E˚red.
2Zn2+(aq) + 4e-  2Zn(s)
E˚red = -0.76 V
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Chapter 20
CHEM 180/181
Chapter 20
Full list in Appendix E
EMF Trends
The larger the difference between E˚red values, the larger E˚cell.
Spontaneous voltaic cell: E˚red(cathode) is more positive than
E˚red(anode), resulting in a positive E˚cell.
Recall
Ecell  Ered cathode  Ered anode
CHEM 180/181
Chapter 20
Calculating E˚red from E˚cell
Zn(s) + Cu2+(aq, 1M)
Given: Zn2+ + 2e-
Zn2+(aq, 1M) + Cu(s) E˚cell = 1.10V
Zn(s)
E˚red = -0.76V
Calculate the E˚red for the reduction of Cu2+ to Cu.
Cu2+ + 2eCu(s)
E˚cell = E˚red(cathode) – E˚red(anode)
1.10V = E˚red(cathode) – (-0.76V)
E˚red(cathode) = 1.10V – 0.76V = 0.34V
Q: What about E˚red for the oxidation of Cu(s) – reverse
reaction?
CHEM 180/181
Chapter 20
Oxidizing and Reducing Agents
We can use E˚red values to understand aqueous
reaction chemistry.
e.g. Zn(s) + Cu2+(aq)
Zn2+(aq) + Cu(s)
Since Cu2+ is responsible for the oxidation of Zn(s),
Cu2+ is called the oxidizing agent.
Since Zn(s) is responsible for the reduction of Cu 2+,
Zn(s) is called the reducing agent.
E˚red for Cu2+ (0.34V) indicates that, compared to
E˚red for Zn (-0.76V), it will be reduced.
CHEM 180/181
Chapter 20
Example Problem: Cell emf
Calculate the standard emf for the following:
Ni(s) + 2Ce4+(aq)
Ni2+(aq) + 2Ce3+(aq)
Ask yourself…What are the half-reactions?
Which one is oxidized? Reduced?
Ni2+(aq) + 2eCe4+(aq) + 1e-
Ni(s)
Ce3+(aq)
E˚red = -0.28V
E˚red = 1.61V
E˚cell = E˚red(cathode) - E˚red(anode)
E˚cell = 1.61V-(-0.28) = 1.89V
is the oxidizing agent.
CHEM 180/181
Chapter 20
Example Problem: Cell emf
Which reaction will undergo reduction? What reaction occurs at the
anode?
Sn4+(aq) + 2eMnO4-(aq) + 8H+(aq) + 5e-
Sn2+(aq)
Mn2+(aq) + 4H2O(l)
E˚red = 0.154V
E˚red = 1.51V
MnO4-(aq) reaction has more positive E˚red, so it will be reduced more
favorably.
Sn2+ will be oxidized at the anode.
E˚cell = 1.51-0.154 = 1.36V
CHEM 180/181
Chapter 20
Oxidizing and Reducing Agents: EMF
The more positive E˚red, the greater the tendency for the
reactant in the half-reaction to be reduced. Therefore, the
stronger the oxidizing agent.
The more negative E˚red, the greater the tendency for the
product in the half-reaction to be oxidized. This means the
product is a reducing agent.
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Chapter 20
A species at higher
left of the table of
standard reduction
potentials will
spontaneously oxidize
a species that is lower
right in the table.
F2 will oxidize H2 or Li.
Ni2+ will oxidize Al(s).
CHEM 180/181
Chapter 20
Example Problem: Redox Agents
Which is the strongest reducing agent? Oxidizing agent?
Ce4+, Br2, H2O2, or Zn?
E˚red = 1.61V for Ce4+ (aq)
1.776 for H2O2 (aq)
1.065V for Br2 (l)
-0.763V for Zn(s)
Most easily reduced = most positive = H2O2. So, H2O2 is the
strongest oxidizing agent.
Most easily oxidized = most negative = Zn. So, Zn is the
strongest reducing agent.
CHEM 180/181
Chapter 20
Spontaneity of Redox Reactions
In a spontaneous voltaic (galvanic) cell, E˚red(cathode) is more
positive than E˚red(anode).
Ecell  Ered cathode  Ered anode
A positive E˚cell indicates a spontaneous voltaic cell process.
A negative E˚cell indicates a non-spontaneous process.
It all relates back to Gibbs Free Energy.
CHEM 180/181
Chapter 20
Spontaneity of Redox Reactions
Cell emf and free-energy change are related by
G  nFE
G is the change in free-energy, n is the number of moles of
electrons transferred, F is Faraday’s constant, and E is the
emf of the cell. G units are J (assumed per mole).
1F = 96,500 C/mol = 96,500 J/Vmol
If standard conditions: G˚ = -nFE˚cell
If Ecell > 0 then G < 0, both of which indicate spontaneous
processes.
CHEM 180/181
Chapter 20
Example Problem: Cell spontaneity
Calculate ΔG˚ for the following reaction. Is it spontaneous?
Cl2(g) + 2I-(aq)
2Cl-(aq) + I2(s)
E˚cell = 0.823V
ΔG˚ = -nFE˚cell
ΔG˚ = -(2 mol)(96,500 J/Vmol)(0.823 V) = -156 kJ
Since ΔG˚ is negative (and E˚cell is positive), it is spontaneous.
CHEM 180/181
Chapter 20
Batteries
Defn: Self-contained
electrochemical power source
with one (or more) complete
voltaic cell.
When multiple cells or multiple
batteries are connected in
series, greater emfs can be
achieved.
Cathode labeled with a plus sign;
the anode with a minus sign.
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Chapter 20
How Batteries Work
Spontaneous redox reactions, as in voltaic cells, serve as the
basis for battery operation.
Specific reactions at anode and cathode determine the voltage
of the battery, and the usable life of the battery depends on
the quantity of these substances.
Primary cells cannot be recharged. Secondary cells are
rechargeable using an external power source after its emf
has dropped below a usable level.
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Chapter 20
Lead-Acid Batteries
A 12V car battery consists of 6 cathode/anode pairs, connected
in series, each producing 2V.
Cathode: PbO2(s) on a metal grid in sulfuric acid
PbO2(s) + SO42-(aq) + 4H+(aq) + 2e-  PbSO4(s) + 2H2O(l)
Anode: Pb(s) in sulfuric acid
Pb(s) + SO42-(aq)  PbSO4(s) + 2e-
The overall electrochemical reaction is
PbO2(s) + Pb(s) + 2SO42-(aq) + 4H+(aq)  2PbSO4(s) + 2H2O(l)
CHEM 180/181
Chapter 20
Lead-Acid Batteries
Ecell = Ered(cathode) - Ered(anode)
= (+1.685 V) - (-0.356 V)
= +2.041 V.
Wood or fiberglass spacers prevent
electrode contact.
H2SO4 consumed during discharge,
(voltage may vary with use).
Recharging reverses forward reaction:
PbO2(s) + Pb(s) + 2SO42-(aq) + 4H+(aq)  2PbSO4(s) + 2H2O(l)
CHEM 180/181
Chapter 20
Alkaline Batteries
Most common nonrechargeable (primary) battery (100 billion
produced annually).
Anode: Powdered Zn in gel in contact with cKOH solution
Zn(s) + 2OH-(aq)  Zn(OH)2(s) + 2eCathode: MnO2 and C paste
2MnO2(s) + 2H2O(l) + 2e-  2MnO(OH)(s) + 2OH-(aq)
CHEM 180/181
Chapter 20
Rechargeable Batteries
Lightweight batteries for use in cell phones, notebook
computers, etc.
Nickel-cadmium (NiCad) battery still common.
Cadmium oxidized at anode, nickel oxyhydroxide [NiO(OH)(s)]
reduced at cathode.
Cathode:
2NiO(OH)(s) + 2H2O(l) + 2e-  2Ni(OH)2(s) + 2OH-(aq)
Anode:
Cd(s) + 2OH-(aq)  Cd(OH)2(s) + 2eCHEM 180/181
Chapter 20
Rechargeable Batteries
Solid reaction products adhere to the electrodes, permitting
electrode reactions to be reversed during recharging.
Drawbacks of NiCad battery:
Toxicity of Cd (disposal problems), relatively heavy.
CHEM 180/181
Chapter 20
NiMH and Li-ion Batteries
Cathode reaction in nickel metal hydride (NiMH) battery is
same as for Ni-Cd battery.
2NiO(OH)(s) + 2H2O(l) + 2e-  2Ni(OH)2(s) + 2OH-(aq)
Anode consists of a metal alloy, e.g. ZrNi 2, capable of
absorbing hydrogen atoms. During oxidation, the hydrogen
atoms lose electrons, and the resulting H+ ions react with
OH- to form water (drives cathode reaction to the right).
Another type:
Li-ion battery - lightweight with very high energy density.
CHEM 180/181
Chapter 20
Fuel Cells
Direct, efficient production of electricity.
NOT batteries as they are not self-contained (some products
must be released).
On Apollo moon flights, the H2-O2 fuel cell was the primary
source of electricity.
Cathode: 2H2O(l) + O2(g) + 4e-  4OH-(aq)
Anode: 2H2(g) + 4OH-(aq)  4H2O(l) + 4eWhat’s the overall reaction?
CHEM 180/181
Chapter 20
Corrosion
Generally the result of an undesirable redox reaction.
Spontaneous! A metal is attacked by some substance in its
environment to form an unwanted compound.
For many metals oxidation is thermodynamically favored at RT.
Corrosion can form an insulating protective coating, preventing
further oxidation. (e.g., Al forms a hydrated form of Al 2O3)
~20% of iron produced annually in USA is used to replace
rusted items!!
CHEM 180/181
Chapter 20
Corrosion of Iron
Rusting of iron requires both oxygen and water.
Other factors that can accelerate rusting of iron:
- pH of solution (acidic)
-presence of salts (ocean)
-contact with metals more difficult to oxidize than iron
-stress on the iron
Rusting of iron is electrochemical and iron itself conducts
electricity.
CHEM 180/181
Chapter 20
Corrosion of Iron
Recall: E˚red Fe2+ (-0.44V) < E˚red O2 (1.23V), so iron can be
oxidized by oxygen.
Cathode: O2(g) + 4H+(aq) + 4e-  2H2O(l) E˚red=1.23V
Anode:
Fe(s)  Fe2+(aq) + 2e- E˚red=-0.44V
Dissolved oxygen in water usually causes Fe oxidation.
Fe2+ initially formed can be further oxidized to Fe3+ which forms
rust, Fe2O3 . xH2O(s).
CHEM 180/181
Chapter 20
Corrosion of Iron
CHEM 180/181
Chapter 20
Corrosion of Iron
Oxidation occurs more readily at the site with the greatest
concentration of O2 (water-air junction).
Salts produce the necessary electrolyte required to complete
the electrical circuit.
Preventing Corrosion of Iron
Corrosion can be prevented by coating the iron with paint or
another metal.
Galvanized iron (commonly used for nails) is coated with a thin
layer of zinc.
CHEM 180/181
Chapter 20
Corrosion Prevention: Galvanization
Zinc protects iron since Zn is more easily oxidized (E˚redZn is
more negative than E˚redFe).
Anode: Zn2+(aq) +2e-  Zn(s)
Cathode: Fe2+(aq) + 2e-  Fe(s)
E˚red = -0.76 V
E˚red = -0.44 V
Zn is oxidized INSTEAD OF IRON, preventing corrosion.
Protects even if the surface coat is broken.
CHEM 180/181
Chapter 20
Corrosion
Preventing
Prevention:
Corrosion
Galvanization
of Iron
CHEM 180/181
Chapter 20
Corrosion Prevention: Sacrificial Anodes
To protect underground pipelines.
The water pipe is the cathode and a more active metal is used
as the anode (cathodic protection).
Often, Mg is used as the sacrificial anode:
Mg2+(aq) +2e-  Mg(s) E˚red = -2.37 V
Fe2+(aq) + 2e-  Fe(s)
E˚red = -0.44 V
Since E˚red of Mg is more negative, it will oxidize first.
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Corrosion Prevention: Sacrificial Anodes
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Chapter 20
Electrolysis
Defn: Use of electrical energy to drive a non-spontaneous
redox reaction.
In voltaic and electrolytic cells
– reduction occurs at the cathode, and
– oxidation occurs at the anode.
However, in electrolytic cells, electrons are forced to flow
from the anode to cathode.
Anode is now designated as POSITIVE(+).
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Electrolysis of Molten Salts
Used industrially to produce metals such as Na and Al.
Requires high temperatures.
Example: Electrolysis of molten NaCl: NaCl(s)  Na+(l) + Cl-(l)
Two reactants: Na+ and ClCathode: 2Na+(l) + 2e-  2Na(l) E˚red = -2.71V
Anode: 2Cl-(l)  Cl2(g) + 2e-
-
E˚red = 1.359V
This is clearly a non-spontaneous reaction.
E˚cell = -2.71- (1.359) = -4.069V (which is a + ΔG˚cell)
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Chapter 20
Electrolysis of Molten Solutions
Chapter 20to cathode, but are forced.
Electrons still flow from anode
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Electrolysis of Aqueous Solutions
Aqueous solution electrolysis is complicated by water.
Ask: Is water oxidized (to form O2) or reduced (to form H2) in
reference to the other components?
e.g. For an aqueous solution of NaF in an electrolytic cell,
possible reactants are Na+, F- and H2O.
Both Na+ and H2O can be reduced but not F-.
Thus possible reactions at cathode are:
Na+(aq) + eNa(s)
2H2O(l) + 2eH2(g) + 2OH-(aq)
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Chapter 20
E˚red = -2.71 V
E˚red = -0.83 V
Electrolysis of Aqueous Solutions
More positive the E˚red favors reduction reactions.
Reduction of H2O (E˚red = -0.83V) occurs at cathode with H2
gas produced.
At anode, either F- or H2O must be oxidized since Na+ cannot
lose additional electrons.
2F-(aq)
F2(g) + 2eE˚red = +2.87 V
4OH-(aq)
O2(g) + 2H2O(l) + 4eE˚red = +0.40
Thus oxidation of H2O occurs at anode (more negative E˚red
favors oxidation reactions).
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Chapter 20
Electrolysis of Aqueous Solutions
Cathode 4H2O(l) + 4e-
Anode
4OH-(aq)
2H2O(l)
2H2(g) + 4OH-(aq)
O2(g) + 2H2O(l) + 4e- E˚red = -0.40 V
2H2(g) + O2(g)
E˚cell = -1.23 V
Process is non-spontaneous (as expected).
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E˚red = -0.83 V
Chapter 20
Please note – sign on
anode half-reaction!!
Electroplating
Active electrodes - take part in electrolysis.
Example: electrolytic plating.
Defn: Electrolysis used to deposit a thin layer of one metal on
another in order to improve beauty or resistance to
corrosion. e.g. electroplating nickel on a piece of steel.
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Chapter 20
Electroplating
Consider an active Ni electrode and another metallic electrode
placed in an aqueous solution of NiSO 4.
Anode: Ni(s)  Ni2+(aq) + 2eCathode: Ni2+(aq) + 2e-  Ni(s)
E˚cell = 0V, so outside
source of voltage
needed!
With voltage supplied, Ni plates on the steel electrode.
Electroplating is important in protecting objects from corrosion
(chrome).
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Chapter 20
Quantitative Aspects of Electrolysis
How much material can we obtain with electrolysis?
Consider the reduction of Cu2+ to Cu(s).
– Cu2+(aq) + 2e-  Cu(s).
– 2 mol of electrons will plate 1 mol of Cu
– The charge of 1 mol of electrons is 96,500 C (=1F).
– Using Q = It
The amount of Cu can be calculated from the
current used (I) and time (t) allowed to plate.
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Chapter 20
Example
Calculate the number of grams of aluminum
produced in 1.00h by the electrolysis of molten
AlCl3 if the electrical current is 10.0 A.
Q = It Q = (10.0 A)(3600 s) = 3.60x104 C
Moles e- = CF = (3.60x104 C)(1 mol e- / 96,500 C) =
0.373 mol eAl3+ + 3e-
Al
Moles Al = (0.373 mol e-)(1 mol Al / 3 mol e-) = 0.124 mol Al
Grams Al = (0.124 mol Al)(27.0 g / mol) = 3.36 g Al
Ans. 3.36g
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Chapter 20
Electrical Work
Free-energy is a measure of the maximum amount of useful
work that can be obtained from a system.
We know
G  wmax .
G   nFE .
 wmax   nFE
If work is negative, then work is performed by the system and E
is positive (which means it’s spontaneous)
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Chapter 20
Electrical Work
The emf can be thought of as a measure of the driving force for
a redox reaction to proceed.
In order to drive non-spontaneous, electrolytic reactions, the
external, applied emf must be greater than Ecell.
If Ecell = -1.35 V, then external emf must be at least +1.36 V.
Note: work can be expressed in Watts (W)
1 W = 1 J/s.
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Chapter 20
Recommended Problems
DO as many problems as possible!! Not just these…
Redox reactions: 20.3, 20.7, 20.9
Voltaic Cells: 20.13, 20.21, 20.23
Batteries & Corrosion: 20.66, 20.68, 20.71, 20.72
Electrolysis: 20.75, 20.79
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Chapter 20
End of Chapter 20
Electrochemistry
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Chapter 20