LECTURE 10: REDOX
AND
ELECTROCHEMISTRY
Figure 4.9
The redox process in compound formation.
Sample Problem 4.7
Determining the Oxidation Number of an
Element
PROBLEM: Determine the oxidation number (O.N.) of each element in these
compounds:
(a) Zinc
(b) Sulfur trioxide (c) Nitric acid
PLAN:
Thechloride
O.N.s of the ions in a polyatomic ion add up to the charge of
the ion and the O.N.s of the atoms or ions in a compound add up to
zero.
SOLUTION:
(a) ZnCl2. The O.N. for zinc is +2 and that for chloride is -1.
(b) SO3. Each oxygen is an oxide with an O.N. of -2. Therefore,
the O.N. of sulfur must be +6.
(c) HNO3. H has an O.N. of +1 and each oxygen is -2.
Therefore, the N must have an O.N. of +5.
Figure 4.10
Highest and lowest oxidation
numbers of reactive maingroup elements.
Sample Problem 4.8
Recognizing Oxidizing and Reducing Agents
PROBLEM: Identify the oxidizing agent and reducing agent in each of the
(a) following:
2Al(s) + 3H2SO4(aq)
Al2(SO4)3(aq) + 3H2(g)
(b) PbO(s) + CO(g)
Pb(s) + CO2(g)
(c) 2H2(g) + O2(g)
2H2O(g)
PLAN: Assign an O.N. for each atom and see which atom gained and which
atom lost electrons in going from reactants to products.
An increase in O.N. means the species was oxidized (and is the
reducing agent) and a decrease in O.N. means the species was
reduced (is the oxidizing agent).
SOLUTION:
0
+1 +6 -2
+3 +6 -2
0
(a) 2Al(s) + 3H2SO4(aq)
Al2(SO4)3(aq) + 3H2(g)
The O.N. of Al increases; Al is oxidized; it is the reducing
The O.N. of H decreases; H is reduced; H2SO4 is the oxidizing
agent.
agent.
Sample Problem 4.8
continued
+2 -2
Recognizing Oxidizing and Reducing Agents
+2 -2
(b) PbO(s) + CO(g)
0
+4 -2
Pb(s) + CO2(g)
The O.N. of C increases; C is oxidized; CO is the reducing
The O.N. of Pb decreases; Pb is reduced; PbO is the oxidizing
agent.
agent.
0
0
+1 -2
(c) 2H2(g) + O2(g)
2H2O(g)
The O.N. of H increases; it is oxidized; H2 is the reducing
agent.
The O.N. of O decreases; it is reduced; O2 is the oxidizing
agent.
Figure 4.11 A summary of terminology for oxidation-reduction (redox) reactions.
Figure 4.12
An active metal displacing hydrogen from water.
Figure 4.13
Displacing one metal by another.
Figure 4.14
The activity series
of the metals.
TYPE OF REDOX REACTIONS:
1)
2)
3)
4)
Combination Reactions
Decomposition Reactions
Displacement reactions
Combustion reactions
Sample Problem 4.9
Identifying the Type of Redox Reaction
PROBLEM: Classify each of the following redox reactions as a
combination, decomposition, or displacement reaction, write a
balanced molecular equation for each, as well as total and net
ionic equations for part (c), and identify the oxidizing and
reducing agents:
(a) Magnesium(s)
+ nitrogen(g)
magnesium nitride (aq)
(b) Hydrogen peroxide(l)
water(l) + oxygen gas
(c) Aluminum(s) + lead(II) nitrate(aq)
aluminum nitrate(aq) + lead(s)
PLAN: Combination reactions produce fewer products than reactants.
Decomposition reactions produce more products than
reactants.
Displacement reactions have the same number of products and
reactants.
Identifying the Type of Redox Reaction
Sample Problem 4.9
continued
0
+2 -3
(a) Combination 0
Mg3N2(aq)
3Mg(s) + N2(g)
Mg is the reducing agent; N2 is the oxidizing
agent.
+1 -1
+1 -2
0
(b) Decomposition
H2O2(l)
H2O(l) + 1
2 O2(g) or
2H2O2(l)
2H2O(l) + O2(g)
H2O2 is the oxidizing and reducing agent.
(c) Displacement
0
+2 +5 -2
+3 +5 -2
0
Al(s) + Pb(NO3)2(aq)
Al(NO3)3(aq) + Pb(s)
2Al(s) + 3Pb(NO3)2(aq)
2Al(NO3)3(aq) + 3Pb(s)
Pb(NO3)2 is the oxidizing agent; Al is the reducing
agent.
REDOX TITRATIONS –
In redox titration, known concentration of the oxidizing agent is used to se the
concentration of the reducing agent ( or vice versa).
Redox Titration.
Redox titrations are used to find out information about one reactant, using known information about the other.
Worked example 1.
Iron(II) sulphate can be oxidised using acidified potassium permanganate solution. Calculate the mass of
iron(II) sulphate which will completely react with 200 ml of 0.25 mol l-1 acidified permanganate solution.
Write the redox equation:2MnO4- + 16H+ + 5 Fe2+ 
8H2O + 5Fe3+
2Mn2+ +
Calculate the number of moles of the ‘known’ reactant
No. of moles = C x V(litres)
= 0.25 x 0.2 = 0.05 mol
Use mole ratio in equation to calculate the number of moles of the ‘unkown’ reactant.
0.05 mol of MnO4- reacts with 5/2 x 0.05 = 0.125 mol of Fe2+
Use mass = no. of mole x gfm to calculate the mass of iron(II) sulphate
Mass of FeSO4
= number of moles x gfm
Mass of FeSO4 = 0.125 x 152 = 19 g
Calculations for you to try.
1.
Iron(II) ions react with acidified dichromate solution as shown below:6Fe2+ + Cr2O72- + 14H+  6Fe3+ + 2Cr3+ + 7H2O
Calculate the number of moles of iron(II) ions which will completely
react with 25cm3 of 0.4 mol l-1 dichromate solution.
Number of moles of dichromate = C x V(litres)
= 0.4 x 25/1000
= 0.01
From the mole ratio in the balanced equation
number of moles of iron(II)
Higher Grade Chemistry
= 0.06
= 6 x 0.01
2.
Hydrogen peroxide reacts with acidified permanganate solution as shown below:5H2O2 + 2MnO4- + 6H+  2Mn2+ + 5O2 + 8H2O
100 cm3 of hydrogen peroxide solution reacts with 10 cm3 of
0.2 mol l-1 permanganate solution.
Calculate the concentration of the hydrogen peroxide solution.
Number of moles of permanganate = C x V(litres)
= 0.2 x 10/1000
= 0.002
From the mole ratio in the balanced equation
number of moles of H2O2
= 5/2 x 0.002
= 0.005
Use Concentration = number of moles/Volume (litres)
=
0.005
/ 0.1
= 0.05 mol l-1
Higher Grade Chemistry
ELECTRON TRANSFER REACTIONS
• Electron transfer reactions are OXIDATION-REDUCTION or
REDOX reactions.
• Results in the generation of an electric current (electricity) or be
caused by imposing an electric current.
• Therefore, this field of chemistry is often called
ELECTROCHEMISTRY.
ELECTROCHEMICAL CELLS
ANODE:
Zn (s)
Zn+2(aq) + 2e-
CATHODE:
Cu+2(aq) + 2e-
Cu(s)
ELECTROCHEMICAL CELL: CELL DIAGRAM
Zn(s)|ZnSO4(1.00M)||CuSO4(1.00 M)|Cu(s)
Vertical lines separates phase boundary
Double vertical lines denotes salt bridge
Anode written first (left of the salt bridge)
Concentration of solution, pressures of gases are indicated in cell
diagrams
ELECTROMOTIVE FORCE, EMF (E)
The potential difference between the anode and cathode in a cell
is called the electromotive force (emf).
It is also called the cell potential, and is designated E.
For the cell in the diagram E = 1.104 v at 25oC and 1.0 molar
solutions of Zn+2 and Cu+2
EMF of a cell is normally measured using potentiometers.
SINGLE ELECTRODE POTENTIALS
Potential of all electrodes are measured in reference to a
STANDARD HYDROGEN ELECTRODE.
By definition, the reduction potential for hydrogen is 0 V:
2 H+ (aq, 1M) + 2 e−  H2 (g, 1 atm)
STANDARD REDUCTION POTENTIALS
Reduction
potentials for
many
electrodes
have been
measured
and
tabulated.
STANDARD CELL POTENTIAL
Because cell potential is based on the potential
energy per unit of charge, it is an intensive property.
The reverse of a half-cell reaction will have the same
Eo value but of opposite sign.
The cell potential at standard conditions can be found
through this equation:
E E
o
o
cathode
E
o
anode
CELL POTENTIAL
• For the oxidation in this cell,
 Ered = −0.76 V
• For the reduction,
 Ered = +0.34 V
CELL POTENTIAL
E E
o
o
cathode
E
o
anode
= +0.34 V − (−0.76 V)
= +1.10 V
OXIDIZING AND REDUCING AGENTS
• The strongest oxidizers
have the most positive
reduction potentials.
• The strongest reducers
have the most negative
reduction potentials.
OXIDIZING AND REDUCING AGENTS
The greater the difference
between the two, the
greater the voltage of the
cell.
THERMODYNAMICS OF ELECTROCHEMICAL CELLS
G for a redox reaction can be found by using the
equation
G = −vFE
where v is the number of moles of electrons
transferred, and F is a constant, the Faraday.
1 F = 96,485 C/mol = 96,485 J/V-mol (96,500)
FREE ENERGY:
Under standard conditions :
o

G
o
r
E 
vF
RT ln K
E 
vF
vFE o
ln K 
RT
thus if we know E o we can calculate the equilibrium
constant of redox cell reaction
o
SAMPLE PROBLEM:
Predict whether the following reaction would occur
spontaneously under standard state conditions,
calculate the equilibrium constant at 25oC.
Sn(s) + 2Ag+(aq)
Sn+2(aq) + 2Ag(s)
SAMPLE PROBLEM:
Based on the following electrode potentials
Fe+2(aq) + 2eFe(s)
Eo = -0.447 v
Fe+3(aq) + eFe+2(aq)
Eo = 0.771 v
Calculate the standard reduction potential for the halfreaction:
Fe+3(aq) + 3eFe(s)
Eo = ???
THE NERNST EQUATION:
• Remember that
G = G + RT lnK
• This means
−vFE = −vFE + RT lnK
THE NERNST EQUATION
Dividing both sides by −nF, we get the Nernst equation:
RT
EE 
ln K
vF
o
at equilibrium, E  0
RT
E 
ln K , at standard conditions
vF
0.0257V
o
EE ln K
v
o
CONCENTRATION CELLS
• Notice that the Nernst equation implies that a cell could be created
that has the same substance at both electrodes.
• For such a cell,

Ecell
would be 0, but K would not.
• Therefore, as long as the concentrations are different, E
will not be 0.
SAMPLE PROBLEM:
Predict whether the following reaction would proceed
spontaneously as written at 298 K:
Cd(s) + Fe+2(aq)
Cd+2(aq) + Fe(s)
Given that [Cd+2] = 0.15M and [Fe+2] = 0.68M
TYPES OF ELECTRODES:
METAL ELECTRODES:
Piece of metal immersed
in a solution containing
the cations of the metal.
Galvanic cells employ gas
electrodes.
Electrode reaction:
M+z(aq) + ze-
M(s)
Ex. Copper, Zinc, Ag,
TYPES OF ELECTRODES:
GAS ELECTRODES: A gas (1 atm) over a solution of cation/anion of the
gas in an inert metal electrode.
Example is the Standard Hydrogen Electrode with Pt metal.
Pt|H2(g)|H+(aq)
TYPES OF ELECTRODES:
METAL INSOLUBLE SALT
ELCTRODES: Coating a
piece of metal with an
insoluble salt of the same
metal.
Example is a Ag-AgCl
electrode.
Ag(s)|AgCl(s)|Cl-(aq)
TYPES OF ELECTRODES:
Glass Electrodes: Electrode
consist of a very thin
membrane made of special
type of glass that is permeable
to H+ ions
TYPES OF ELECTRODES:
ION SELECTIVE
ELECTRODES: Specific for
cations such as Li+, Na+, K+
Ag+, and Cu+2 and for anions
such as S2- and CN-.
TYPES OF ELECTROCHEMICAL CELLS:
CONCENTRATION
CELLS: Concentration
cells contain electrodes
made of the same metal
and solutions containing
the same ions but at
different concentrations.
TYPES OF ELECTROCHEMICAL CELLS:
FUELS CELLS:
CELLS
HYDROGEN FUEL
Hydrogen and oxygen are bubbled
through an electrolyte solution (NaOH
or H2SO4) with inert electrodes also
serving as catalysts.
Anode Reaction:
H2(g) + 2OH-(aq)
2H2O(l) + 2e-
Cathode Reaction:
1/2O2(g) + ½O2(l) + 2e
2OH-
APPLICATIONS OF EMF MEASUREMENTS
DETERMINATION OF ACTIVITY COEFFICIENTS
Example:
Pt|H2(1bar)|HCl(1m)|AgCl(s)|Ag(s)
Overall Reaction:
1/2H2(g) + AgCl(s)
Ag(s) + H+(aq) + Cl-(aq)
And emf of the at 298 k is given as:
E  E  0.257 ln
o
aH  aCl  aAg
1/ 2
H2
f
aAgCl
E  Eo  0.257 ln aH  aCl 
APPLICATIONS OF EMF MEASUREMENTS
DETERMINATION OF ACTIVITY COEFFICIENTS
aH  aCl    2 m2   2 m 2
E  E o  0.257 ln(   m )2
E  E o  0.514 ln m  0.514 ln  
E  0.0514 ln m  E o  0.514 ln  
Plot of (E  0.514lnm) vs m of HCl, extrapolate m to 0, we get E o
since at m  0,
   1, the mean activity coefficient can now be
found at a particular m.
APPLICATIONS OF EMF MEASUREMENTS
pH Determination
Ag(s)|AgCl(s)|HCl(aq),NaCl(aq)||HCl(aq)||KCl(sat’d)|Hg2Cl2(s)|Hg(l)
The overall emf E for this arrangement is:
E  E ref  0.0591 log aH 
E  E ref  0.0591pH
E  Eref
pH 
0.0591
POTENTIOMETRIC REDOX TITRATIONS
v 2Ox1  v1Rd2  v 2 Rd1  v1 Ox 2
Ox1  V1e -

Rd1
Ox 2  v 2 e -  Rd2
0.0257 Rd1 
E1  E ln
Ox1 
v1
o
1
0.0257 Rd2 
E2  E ln
Ox 2 
v2
o
2
POTENTIOMETRIC REDOX TITRATIONS
at equivalence point;
Eeq
v 1E  v 2 E

(v1  v 2 )
o
1
o
2
POTENTIOMETRIC REDOX TITRATIONS
example :
Ce4   Fe2   Ce3   Fe3  , v1  v 2  1
Ce
4
 e  Ce
-
3
Fe  3  e -  Fe  2
Eeq
E1  1.72 v
E2  0.771 v
(1.72  0.771)

 1.25V
2
POTENTIOMETRIC REDOX TITRATIONS
From the standard emf’s of the Fe2+|Fe3+ and Ce3+|Ce4+ couples, calculate the
equilibrium constant for the following reaction at 298 K.
The reaction in part (a) is employed in a redox titration. Calculate the emf of the
cell after the addition of 10.0 mL of a 0.10 m Ce+4 solution to a 50 mL of a 0.10
Fe+2 solution.
Applications of OxidationReduction Reactions /
Electrochemistry
Batteries
Alkaline Batteries
HYDROGEN FUEL CELLS
Hydrogen fuel cells
CORROSION AND…
CORROSION PREVENTION
Corrosion
• Rusting - spontaneous oxidation.
• Most structural metals have reduction potentials that
are less positive than O2 .
• Fe  Fe+2 +2e-
Eº= 0.44 V
Eº= 0.40 V
• O2 + 2H2O + 4e- 4OH• Fe+2 + O2 + H2O Fe2O3 + H+
• Reactions happens in two places.
Salt speeds up process by increasing conductivity
Water
Fe2+
Iron DissolvesFe  Fe+2
Rust
eO2 + 2H2O +4e-  4OH-
Fe2+ + O2 + 2H2O  Fe2O3 + 8 H+
Preventing Corrosion
• Coating to keep out air and water.
• Galvanizing - Putting on a zinc coat
• Has a lower reduction potential, so it is more easily
oxidized.
• Alloying with metals that form oxide coats.
• Cathodic Protection - Attaching large pieces of an
active metal like magnesium that get oxidized instead.
Dry Cell Battery
Anode (-)
Zn ---> Zn2+ + 2eCathode (+)
2 NH4+ + 2e- --->
2 NH3 + H2
Alkaline Battery
Nearly same reactions as in
common dry cell, but under
basic conditions.
Anode (-): Zn + 2 OH- ---> ZnO + H2O + 2eCathode (+): 2 MnO2 + H2O + 2e- --->
Mn2O3 + 2 OH-
Mercury Battery
Anode:
Zn is reducing agent under basic conditions
Cathode:
HgO + H2O + 2e- ---> Hg + 2 OH-
Lead Storage Battery
Anode (-) Eo = +0.36 V
Pb + HSO4- ---> PbSO4 + H+ + 2eCathode (+) Eo = +1.68 V
PbO2 + HSO4- + 3 H+ + 2e---> PbSO4 + 2 H2O
Ni-Cad Battery
Anode (-)
Cd + 2 OH- ---> Cd(OH)2 + 2eCathode (+)
NiO(OH) + H2O + e- ---> Ni(OH)2 + OH-
ELEC
Harnessing the Power of Voltaic Cells
Batteries and Corrosion
COMMERCIAL VOLTAIC CELLS
• Voltaic Cells are convenient energy sources
• Batteries is a self-contained group of voltaic cells arranged in series.
•
Advantage: Portable
•
Disadvantage: Very Expensive (US$1.20 / Kwatt-h)
•
Need cells in series to provide power
The Processes occurring during the discharge
and recharge of a lead-acid battery. When the
lead-acid battery is discharging (top) it behaves
like a voltaic cell: the anode is negative
(electrode-1) and the cathode is positive
(electrode-2). When it is recharging (bottom), it
behaves like an electrolytic cell; the anode is
positive (electrode-2) and the cathode is
negative (electrode-1).
DRY CELL OR LeClanche cell
Dry Cells
Invented in the 1860’s the common dry cell or LeClanche cell, has become a familiar household item. An active zinc anode in
the form of a can house a mixture of MnO2 and an acidic electrolytic paste, consisting of NH4Cl, ZnCl2, H2O and starch
powdered graphite improves conductivity. The inactive cathode is a graphite rod.
Anode (oxidation)
Zn(s) g Zn2+(aq) = 2eCathode (reduction). The cathodic half-reaction is complex and even today, is still being studied. MnO2(s)
is reduced to Mn2O3(s) through a series of steps that may involve the presence of Mn2+ and an acid-base
reaction between NH4+ and OH- :
2MnO2 (s) + 2NH4+(aq) + 2e- g Mn2O3(s) + 2NH3(aq) + H2O (l)
The ammonia, some of which may be gaseous, forms a complex ion with Zn2+, which crystallize in contact
Cl- ion:
Zn2+(aq) + 2NH3 (aq) + 2Cl-(aq) g Zn(NH3)2Cl2(s)
Overall Cell reaction:
2MnO2 (s) + 2NH4Cl(aq) + Zn(s) g Zn(NH3)2Cl2(s) + H2O (l) + Mn2O3(s)
Ecell = 1.5 V
Uses: common household items, such as portable radios, toys, flashlights,
Advantage; Inexpensive, safe, available in many sizes
Disadvantages: At high current drain, NH3(g) builds up causing drop in voltage, short shelf life because
zinc anode reacts with the acidic NH4+ ions.
DRY CELL OR LeClanche cell
Invented by George Leclanche, a French Chemist.
Acid version:
Zinc inner case that acts as the anode and a carbon rod in contact with
a moist paste of solid MnO2 , solid NH4Cl, and carbon that acts as the cathode. As battery
wear down, Conc. of Zn+2 and NH3 (aq) increases thereby decreasing the voltage.
Half reactions:
E°Cell = 1.5 V
Anode:
Zn(s) g Zn+2(aq) + 2eCathode: 2NH4+(aq) + MnO2(s) + 2e- g Mn2O3(s) + 2NH3(aq) + H2O(l)
Advantage:
Inexpensive, safe, many sizes
Disadvantage:
High current drain, NH3(g) build up, short
shelf life
Alkaline Battery
Alkaline Battery
The alkaline battery is an improved dry cell. The half-reactions are similar, but the electrolyte is a basic KOH paste,
which eliminates the buildup of gases and maintains the Zn electrode.
Anode (oxidation)
Zn(s) + 2OH- (aq) g ZnO(s) + H2O (l) + 2eCathode (reduction).
2MnO2 (s) + 2H2O (l) + 2e- g Mn(OH)2(s) + 2OH-(aq)
Overall Cell reaction:
2MnO2 (s) + H2O (l) + Zn(s) g ZnO(s) + Mn(OH)2(s)
Ecell = 1.5 V
Uses: Same as for dry cell.
Advantages: No voltage drop and longer shell life than dry cell because of
alkaline electrolyte; sale ,amu sizes.
Disadvantages; More expensive than common dry cell.
ALKALINE BATTERY
Leclanche Battery: Alkaline Version
In alkaline version; solid NH4Cl is replaced with KOH or NaOH. This makes cell last
longer mainly because the zinc anode corrodes less rapidly under basic conditions versus
acidic conditions.
Half reactions: E°Cell = 1.5 V
Anode:
Zn(s) + 2OH-(aq) g ZnO(s) + H2O(l) + 2eCathode:
MnO2 (s) + H2O(l) + 2e- g MnO3 (s) + 2OH-(aq)
Nernst equation: E = E° - [(0.592/n)log Q], Q is constant !!
Advantage:
No voltage drop, longer shelf life.
Disadvantage:
More expensive
Alkaline Batteries
MERCURY BUTTON CELL
Mercury and Silver batteries are similar.
Like the alkaline dry cell, both of these batteries use zinc in a basic medium as the anode.
The solid reactants are each compressed with KOH, and moist paper acts as a salt bridge.
Half reactions:
Anode:
Cathode (Hg):
Cathode (Ag):
E°Cell = 1.6 V
Zn(s) + 2OH-(aq) g ZnO(s) + H2O(l) + 2eHgO (s) + 2H2O(l) + 2e- g Hg(s) + 2OH-(aq)
Ag2O (s) + H2O(l) + 2e- g 2Ag(s) + 2OH-(aq)
Advantage:
Small, large potential, silver is
nontoxic.
Disadvantage:
Mercury is toxic, silver is
expensive.
LEAD STORAGE BATTERY
Lead-Acid Battery. A typical 12-V lead-acid battery has six cells
connected in series, each of which delivers about 2 V. Each cell
contains two lead grids packed with the electrode material: the anode is
spongy Pb, and the cathode is powered PbO2. The grids are immersed
in an electrolyte solution of 4.5 M H2SO4. Fiberglass sheets between the
grids prevents shorting by accidental physical contact. When the cell
discharges, it generates electrical energy as a voltaic cell.
Half reactions: E°Cell = 2.0 V
Anode: Pb(s) + SO42- g PbSO4 (s) +2 eE° = 0.356
Cathode (Hg): PbO2 (s) + SO42- + 4H+ + 2e- g
PbSO4 (s) + 2 H2O
Net: PbO2
(s) +
Pb(s) + 2H2SO4 g PbSO4 (s) + 2 H2O
E° = 1.685V
E°Cell = 2.0 V
Note hat both half-reaction produce Pb2+ ion, one through oxidation of Pb, the
other through reduction of PbO2. At both electrodes, the Pb2+ react with SO42- to
form PbSO4(s)
NICKEL CADMIUM BATTERY
Battery for the Technological Age
Rechargeable, lightweight “ni-cad” are used for variety of cordless appliances. Main advantage is
that the oxidizing and reducing agent can be regenerated easily when recharged. These produce
constant potential.
Half reactions: E°Cell = 1.4 V
Anode:
Cd(s) + 2OH-(aq) g Cd(OH)2 (s) + 2e-
Cathode:
2Ni(OH) (s) + 2H2O(l) + 2e- g Ni(OH)2 (s) + 2 OH-(aq)
FUEL CELLS
FUEL CELLS; BATTERIES
Fuel Cell also an electrochemical device for converting chemical energy
into electricity.
In contrast to storage battery, fuel cell does not need to involve a reversible reaction since
the reactant are supplied to the cell as needed from an external source. This technology
has been used in the Gemini, Apollo and Space Shuttle program.
Half reactions: E°Cell = 0.9 V
Anode:
2H2 (g) + 4OH-(aq) g 4H2O(l) + 4e-
Cathode:
O2 (g) + 2H2O(l) + 4e- g
4OH-(aq)
Advantage:
Clean, portable and product is water. Efficient
(75%) contrast to 20-25% car, 35-40% from coal
electrical plant
Disadvantage:
Cannot store electrical energy, needs continuous
flow of reactant, Electrodes are short lived and
expensive.
Rust: Fe2O3 • X H2O
Anode: Fe(s) g Fe+2 + 2eE° = 0.44 V
Cathode:
O2 (g) + 4H+ + 4e- g 2H2O (l)
E° = 1.23 V
Net: Fe+2 will further oxidized to Fe2O3 • X H2O
Corrosion
Not all spontaneous redox reaction are beneficial.
Natural redox process that oxidizes metal to their oxides and sulfides runs billions
of dollars annually. Rust for example is not the direct product from reaction
between iron and oxygen but arises through a complex electrochemical process.
CONDITIONS FOR CORROSION
Conditions for Iron Oxidation:
Iron will oxidize in acidic medium
SO2 g H2SO4 g H+ + HSO4+
Anions improve conductivity for oxidation.
Cl- from seawater or NaCl (snow melting) enhances rusting
Conditions for Prevention:
Iron will not rust in dry air; moisture must be present
Iron will not rust in air-free water; oxygen must be present
Iron rusts most rapidly in ionic solution and low pH (high H+)
The loss of iron and deposit of rust occur at different placm on object
Iron rust faster in contact with a less active metal (Cu)
Iron rust slower in contact with a more active metal (Zn)
IRON CORROSION CHEMISTRY
Most common and economically destructive form of
corrosion is the rusting of iron. Rust is not a direct
product of the reaction between iron and oxygen but
arises through complex electrochemical process. The
features of a voltaic cell can help explain this process.
Iron will not rust in dry air; moisture must be present.
Iron will not rust in air-free water; oxygen must be present
Iron rusts most rapidly in ionic solutions and at low pH (High H+)
The loss of iron and the depositing of rust
often occur at different places on the same
object.
Iron rust faster in contact with a less active
metal (such as Cu) and more slowly in
contact with a more active metal (such as
Zn).
CORROSION AND…
CORROSION PREVENTION
PREVENTING CORROSION
• Coating to keep out air and water.
• Galvanizing - Putting on a zinc coat
• Has a lower reduction potential, so it is more easily
oxidized.
• Alloying with metals that form oxide coats.
• Cathodic Protection - Attaching large pieces of an
active metal like magnesium that get oxidized instead.
Corrosion Prevention