FUNDAMENTALS OF ELECTROCHEMISTRY

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REDOX REACTIONS

• FUNDAMENTALS OF ELECTROCHEMISTRY
• ELECTROCHEMISTRY IS THE BRANCH OF CHEMISTRY THAT
• DEALS WITH THE RELATIONSHIP BETWEEN ELECTRICITY
  AND
• CHEMICAL REACTIONS.
• PRODUCTION OF ELECTRICAL CURRENT BY CHEMICAL
  REACTIONS (Batteries, Fuel cells)
• CHEMICAL CHANGES PRODUCED BY ELECTRIC CURRENT
  (Electrolysis, Electroplating and refining of
  metals)
• ELECTRIC CURRENT Transfer of charge per unit
  time
• BIOELECTROCHEMISTRY study of electron transfer
  in biological regulations of organisms

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• BASIC CONCEPTS
• A redox reaction involves transfer of electrons
  from one species to another.
• Oxidation loss electrons - Reducing agent
• Reduction gain electrons - Oxidizing agent

3

• Electric charge (q) n x F
• Coulombs
• F Faraday constant 96 485.3415 C/mol of e-
• Electric current (I) is the quantity of charge
  flowing each second through the circuit. Unit
  Amperes (A)
• Eg. Calculate the mass of aluminum produced in 1
  hour by electrolysis of molten AlCl3 if the
  electrical current is 20 A.
• (Answer 33 g)

4

• Potential difference (E) between 2 points is the
  work needed when moving an electric charge from 1
  point to another, unit is Volt
• Work E. q OR J CV
• 1 Joule is the energy gained or lost when 1
  coulomb of charge moves between points whose
  potentials differ by 1V
• The greater the potential difference between 2
  points the stronger will be the push on a
  charged particle travelling between those points.
  A 12V battery pushes e- 8X harder than a 1.5V
  dry cell

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• The free energy of change, ?G, represents the
  maximum electrical work that can be done by the
  reaction on its surroundings.
• Work done on surroundings ?G - work -E.q
• ?G - nFE (-ve ?G spontaneous rxn)
• Ohms Law, I E/R
• Unit of resistance is ohms or Greek symbol O
  (omega)
• Power , P work/time
• E.q E. q E. I I2. R
• s s

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• BALANCING REDOX REACTIONS
• In an acidic medium
• In a basic medium
• Split the reaction into 2 components or half
  reactions by identifying which species are
  oxidised and which ones are reduced. Add
  electrons appropriately to match the change in
  oxidation state of each element
• Introduce H2O to balance the oxygen atoms
• Introduce H to balance the charges as well as
  the hydrogen atoms
• Multiply the half reactions so as to have an
  equal no. of e- and add the half reactions
• Whichever case, balance for the acid (H) and
  then for basic media add OH- to the side where
  there is H to eliminate it as a H2O molecule
  (this is a 11 reaction)

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STANDARD POTENTIALS
The voltmeter tells how much work is done by e-
flowing from one side to the other, ve V means
e- flow into negative terminal
LHS negative terminal Reference electrode
RHS connected to the positive terminal
Line notation for cell
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• STANDARD REDUCTION POTENTIAL (Eo)
• for each half cell is measured or setup by the
  above experiment.
• Standard means the activities ( A ) of all
  species are unity.
• The half reaction of interest is 2Ag 2e-
  2Ag(s)
• AAg 1 by definition activity of Ag (s)
  unity
• Standard Hydrogen Electrode (SHE), consists of a
  catalytic Pt surface in contact with an acid
  solution H (aq,1M), AH 1
• By convention the LHS electrode (Pt) is attached
  to the negative terminal of the potentiometer
• equilibrium rxn at SHE 2H (aq, A 1) 2e-
  H2 (g, A 1)
• half reaction always written as reduction
  reactions

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• BY INTERNATIONAL AGREEMENT THE SHE IS ARBITARILY
  ASSIGNED Eo 0.00V at 25oC.
• E0cell 0.799V
• E0cell Eored (cathode) Eored (anode) or
• E0cell RHS electrode potential LHS
  electrode potential
• E0cell E0red (reduction process) E0red
  (oxidation process)
• A positive E0 spontaneous process
• A negative E0 non spontaneous process
• Sketch the cell construction SHE ll Cd2(aq,
  A1) l Cd(s)
• The half reaction with a more positive E0 is more
  reduced and that with a less positive E0 is less
  reduced or more oxidised

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NERNST EQUATION Le Chateliers principle tells us
that conc. of reactants or products drives
rxn to the right or left respectively. The net
driving force is expressed by the NERNST
EQUATION For the half reaction a A ne- ? b
B Nernst Equation for the half cell potential, E
Q Reaction quotient
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where Eo standard reduction potential ( AA
AB 1) Ai activity of species i R gas
constant ( 8.314 J/K.mol) T Temperature (K) N
number of electrons in the half reaction F
Faraday constant (9.6485 x 104C/mol)
concentration in Kc can be replaced by
activities to account for ionic strength Ac
C ?c ?
activity co efficient measures
the deviation from
ideality , ? 1
behavior is ideal, low ionic strength
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Pure solids and liquids are omitted from Q,
because their activities are (close to)
unity. Concentration mol/L and pressure in
bars When all activities are unity Q 1, ln Q
0 then E Eo Log form of the Nernst equation, at
25oC T 298K
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NERNST EQUATION FOR A COMPLETE CELL E E - E-
Find the voltage for the cell if the right half
cell contains 0.50M AgNO3 (aq) and the left half
cell 0.010M Cd(NO3)2 (aq).
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STEP 1 Right electrode 2 Ag 2e- ?
2Ag(s) Eo 0.799V Left
electrode Cd2 2e- ? Cd(s) Eo -
0.402V STEP 2 Nernst equation for the right
electrode

0.781V STEP 3 Nernst equation for the
left electrode

- 0.461V STEP 4 Cell Voltage E E - E-
0.781 - (-0.461) 1.242V STEP 5 Net cell
reaction Eqn Right electrode Eqn Left
electrode Cd (s) 2 Ag ? Cd2 2 Ag (s)
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Note Multiplying a half reaction by a number
does not change E0 nor E To figure half cell
reactions look for the element in two different
oxidations states
Eo and the Equilibrium Constant A galvanic cell
produces electricity because the cell is not at
equilibrium. Relating E to the reaction quotient
Q Consider Right electrode aA ne- ? cC
Eo Left electrode dD ne- ? bB Eo-
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The Nernst equation will be E E - E-
E0
Q
When the cell is at equilibrium E 0
and Q K Eo can be calculated and used to find
K for 2 half reactions!
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Question 10 in Tutorial on Complexation rxns Use
the following standard-state cell potentials to
calculate the complex formation equilibrium
constant for the Zn(NH3)42 complex ion.
Zn(NH3)42 2e- ? Zn 4NH3 Eored
-1.04 V Zn2 2e- ? Zn
Eored - 0.7628 V
SOLUTION E0cell E0red (reduction process)
E0red (oxidation process) (i) Zn 4NH3 ?
Zn(NH3)42 2e- E0 1.04
V (ii) Zn2 2e- ? Zn
E0 - 0.7628 V (i) (ii) Zn2
4NH3 ? Zn(NH3)42 E0 0.28 V

10
(2)(0.28)/0.05916 2.92 x 109
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REDOX TITRATIONS

• Theory of redox titrations and common titrants.
• A redox titration is based on an oxidation-
  reduction reaction between an analyte and
  titrant.
• Environmental and biological analytes can be
  measured by redox titrations. Other analytes
  include laser and superconductor materials.
• REDOX TITRATION CURVE
• Consider the potentiometric titration of Fe(II)
  and cerium (IV).
• Titration rxn Ce4 Fe2
  Ce3 Fe3 (1)
• Each mole of Ce4 ion oxidizes 1 mole of ferrous
  ion, the titration creates a mixture of Ce4,
  Fe2, Ce3 and Fe3 ions.

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e- flow from the anode to the cathode, the
circuit measures the potential for Fe3 /Ce4
reduction at the Pt surface by e- from the
calomel electrode
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Calomel reference electrode 2Hg(l) 2 Cl-
Hg2Cl2 (s) 2e- Pt indicator electrode 2 rxns
coming to equilibrium Fe3 e- ? Fe2
Eo 0.767V (2) Ce4 e- ? Ce3
Eo 1.70V (3) Cell rxns 2Fe3
2Hg(l) 2Cl- ? Fe2 Hg2Cl2 (s)
(4) 2Ce4 2Hg(l) 2Cl- ? Ce3 Hg2Cl2
(s) (5) At equilibrium the potential
driving rxns 1 and 2 must be the same. THE CELL
RXNS ARE NOT THE SAME AS THE TITRATION RXN! The
titration reaction goes to completion and is an
oxidation of Fe2 and reduction of Ce4 Cell rxns
proceed to negligible extent. The cell is used to
measure activities, not to change them.
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HOW CELL VOLTAGE CHANGES AS Fe2 IS TITRATED WITH
Ce4 REGION 1 BEFORE THE EQUIVALENCE POINT As
each aliquot of Ce4 is added , it is consumed
(eqn 1) and creates an equal number of moles of
Ce3 and Fe3 Prior to the equivalence point
excess unreacted Fe remains in solution Since the
amounts of Fe2 and Fe3 are known, cell voltage
can be calculated from eqn 2 rather than 3 E
E - E- (calomel)
When volume titrant is half the equivalence point
( V 1/2Ve) the concentration of Fe2 and Fe3
are equal, thus E Eo For an acid-base
titration pH pKa when V 1/2Ve
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SHAPES OF TITRATION CURVES
E Eo(Ce4I Ce3) - 0.241V 1.46V
Not symmetric about the equivalence point 21
11 stoichiometry symmetric about equivalence
point same curve for diluted sample
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REGION 2 AT THE EQUIVALENCE POINT All cerium is
in the form of Ce3 Ce3 Fe3 Thus
the equilibrium form of eqn 1 Ce 4 Fe2 ? Ce3
Fe3 If a little Fe3 goes back to Fe2, an
equal no. of moles Ce4 must be made and
Ce4 Fe2 Eqns 2 and 3 are in equilibrium
at the Pt electrode, it is convenient to use both
these eqns to calculate the cell voltage At
equilibrium in a redox titration, Ecell (Eo1
Eo2)/2
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REGION 3 AFTER THE EQUIVALENCE POINT Almost
all the iron atoms are Fe3, the moles of Ce3
moles Fe3, and there is a known excess of
unreacted Ce4. We know Ce3 and Ce4 and so
we can use eqn 3 to calc E .
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FINDING THE ENDPOINT As in acid-base titration,
indicators and electrodes are commonly used to
find the endpoints of a redox titration. REDOX
INDICATORS A redox indicator is a compound that
changes colour when it goes from oxidized to a
reduced state, eg. Ferroin changes from pale blue
to red. By writing the Nernst equation we can
predict the potential range over which the
indicator will change In (oxidised) ne- ? In
(reduced)
(6)
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As with acid base indicators, the colour of In
(reduced) will be observed when In( reduced)
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In(oxidised) 1 And the colour of the
In(oxidised) will be observed when In(
reduced) 1
In(oxidised) 10 Using these 2
quotients in eqn 6, tells us the colour range
will occur over the range
Eo 1.147V, we expect the colour change to occur
1.088 1.206V wrt SHE
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STARCH IODINE COMPLEX Many analytical procedures
use redox titrations involving iodine. Starch is
used as the indicator, since it forms an intense
blue complex with iodine. Starch is not a redox
indicator because it responds to I2, not to a
change in potential. Starch is readily
biodegradable and must be freshly
prepared. ADJUSTMENT OF ANALYTE OXIDATION
STATES Before titration we adjust the oxidation
state of the analyte, eg Mn2 can be pre-oxidised
to MnO4- and then titrated with
Fe2 Pre-adjustment must be quantitative and all
excess reagent must be destroyed. Pre-oxidation
Persulphate S2O82- is a powerful oxidant that
requires Ag as a catalyst S2 O82- Ag
SO42- SO4- Ag2
Two powerful oxidants
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Excess reagent is destroyed after by boiling the
solution after oxidation is complete 2S2O82-
2 H2O boiling 4SO42- O2 4H H2O2 is
a good oxidant in basic solution and reductant in
acidic solution. The excess spontaneously
disproportionate in boiling water. H2O2
H2O O2 PRE-REDUCTION Stannous
chromous chloride, SO2 , H2S are used to
pre-reduce analytes to a lower oxidation
state. An important pre-reduction technique uses
a packed column to pre-reduce analyte to a lower
oxidation state (analyte is drawn by suction).
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Jones reductor, which contains Zn coated with Zn
amalgam. Zn is a powerful reducing agent (Eo
-0.764V) making the Jones reductor unselective,
other species eg, Cr3 are reduced and may
interfere with the titration analysis.
OXIDATION WITH POTASSIUM PERMANGANATE KMnO4 is a
strong oxidant with an intense violet colour. In
strongly acidic solutions (pHlt 1) it is reduced
to colourless Mn2 (manganous). In neutral or
alkaline the product is a brown solid MnO2 In
strongly alkaline solution, green manganate
MnO42- is produced.
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KMnO4 is not a primary standard, and can contain
traces of MnO2, thus it must be standardized with
pure Fe wire or sodium oxalate (pink end-point)
for greater accuracy, The KMnO4 solution is
unstable 4MnO4- 2H2O
4MnO2 3O2 4OH-
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OXIDATION WITH Ce4 Reduction of Ce4 (yellow) to
Ce3 (colorless) can be used in place of KMnO4.
Ce4 is used for the quantitative determination
of malonic acid as well as alcohol, ketones and
carboxylic acids. The primary standard is
prepared by dissolving the salt in 1M H2SO4 and
is stable indefinitely. OXIDATION WITH
K2Cr2O7 Powerful oxidant in acidic solution
orange dichromate iron is reduced to green
chromic ion. In 1M HCl the formal potential is
1.00V and 2M H2SO4 it is 1.11V, thus less
powerful than Ce 4 and MnO4-
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Stable primary standard that is employed to
determine Fe2 Also used in environmental
analysis of oxygen demand. COD or chemical oxygen
demand is defined as the oxygen that is
equivalent to the Cr2O72- consumed by the
oxidation of organics in water. METHODS
INVOLVING IODINE
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When a reducing analyte is titrated with iodine
to produce I- the method is called Iodimetry
(titration with I3-). In the iodimetric
determination of vitamin C-starch is added to
give an intense blue end-point. Iodometry
oxidizing analyte added to I- to produce I2
which is then titrated with thiosulfate standard,
starch is added only before the endpoint. When we
speak of using iodine as a titrant we mean a
solution of I2 plus excess I- I2 (aq) I-
I3- K 7x102 A 0.05M solution of I3- is
prepared by dissolving0.12M KI plus 0.05M I2 in
water. Reducing agent I3-
3I- Oxidizing agent 3I- I3-
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Precipitation Reactions
Gravimetric Analysis
Solid product formed
Relatively insoluble
Easy to filter
High purity
Known Chemical composition
Precipitation Conditions Particle Size
Small Particles Clog pass through filter paper
Large Particles Less surface area for attachment
of foreign particles.
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2. Particle Growth
1. Nucleation
Crystallization
Molecules form small Aggregates randomly
Addition of more molecules to a nucleus.
More solute than should be present at
equilibrium.
Supersaturated Solution
Supersaturated Solution
Nucleation faster Suspension (colloid) Formed.
Nucleation slower, larger particles formed.
Less Supersaturated Solution
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How to promote Crystal Growth
1. Raise the temperature
Increase solubility
Decrease supersaturation
2. Precipitant added slowly with vigorous
stirring.
3. Keep low concentrations of precipitant and
analyte (large solution volume).
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Homogeneous Precipitation
Precipitant generated slowly by a chemical
reaction
pH gradually increases
Large particle size
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Net ve charge on colloidal particle because of
adsorbed Ag
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Precipitation in the Presence of an Electrolyte
Consider titration of Ag with Cl- in the
presence of 0.1 M HNO3.
Colloidal particles of ppt
Surface is vely charged Adsorption of excess
Ag on surface (exposed Cl-)
Colloidal particles need enough kinetic energy to
collide and coagulate.
Addition of electrolyte (0.1 M HNO3) causes
neutralisation of the surface charges. Decrease
in ionic atmosphere (less electrostatic
repulsion)
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Digestion and Purity
Digestion
Period of standing in hot mother liquor.
Promotion of recrystallisation Crystal particle
size increases and expulsion of impurities.
Purity
Adsorbed impurities Surface-bound
Inclusions Occlusions
Within the crystal
Absorbed impurities
Inclusion Impurity ions occupying crystal
lattice sites.
Occlusion Pockets of impurities trapped
within a growing crystal.
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Coprecipitation
Adsorption, Inclusion and Occlusion
Colloidal precipitates Large surface area
BaSO4 Al(OH)3 and Fe(OH)3
How to Minimise Coprecipitation
1. Wash mother liquor, redissolve, and
reprecipitate.
2. Addition of a masking agent
Gravimetric analysis of Be2, Mg2, Ca2, or Ba2
with N-p-chlorophenylcinnamohydroxamic
acid. Impurities are Ag, Mn2, Zn2, Cd2,
Hg2, Fe2, and Ga2. Add complexing KCN.
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Ca 2 2RH ? CaR2(s)
2H Analyte
Precipitate
Mn2 6CN- ?
Mn(CN)64- Impurity Masking agent
Stays in solution
Postprecipitation
Collection of impurities on ppt during
Digestion A supersaturated impurity e.g.,
MgC2O4 on CaC2O4.
Peptization
Breaking up of charged solid particles when ppt
is washed with water.
AgCl is washed with volatile electrolyte (0.1 M
HNO3).
Other electrolytes HCl NH4NO3 and (NH4)2CO3.
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Product Composition
Difficult to weigh accurately
Hygroscopic substances
Some ppts
Variable water quantity as water of
Crystallisation.
Drying
Change final composition by ignition
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Thermogravimetric Analysis
Heating a substance and measuring its mass as a
function of temperature.
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Example
In the determination of magnesium in a sample,
0.352 g of this sample is dissolved and
precipitated as Mg(NH4)PO4.6H2O. The
precipitate is washed and filtered. The
precipitate is then ignited at for 1 hour 1100
oC and weighed as Mg2P2O7.The mass of Mg2P2O7 is
0.2168 g.
Calculate the percentage of magnesium in the
sample.
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Solution
Relative atomic mass Of Mg
The gravimetric factor is
FM of Mg2P2O7
Note 2 mol Mg2 in 1 mol Mg2P2O7.
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Mass of Mg 2 0.0471 g
13.45
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Combustion Analysis
Determination of the carbon and Hydrogen content
of organic compounds burned in excess oxygen.
H2O absorption
Prevention of entrance of atmospheric O2 and
CO2.
CO2 Absorption
Note Mass increase in each tube.
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C, H, N, and S Analyser Modern Technique
Thermal Conductivity, IR,or Coulometry for
Measuring products.
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2 mg sample in tin or silver capsule.
Capsule melts and sample is oxidised in excess of
O2.
Dynamic Flash combustion
Short burst of gaseous products
Products
Hot WO3 catalyst
Then, metallic Cu at 850 oC
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Oxygen Analysis
Pyrolysis or thermal decomposition in absence of
oxygen.
Nickelised Carbon
Gaseous products
CO formed
1075 oC
Halogen-containing compounds
CO2, H2O, N2, and HX products
HX(aq) titration with Ag coulometrically.
Silicon Compounds (SiC, Si3N4, Silicates from
rocks)
Combustion with F2 in nickel vessel
Volatile SiF4 other fluorinated products
Mass Spectrometry
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Example 1 Write a balanced equation for
the combustion of benzoic acid, C6H5CO2H, to
give CO2 and H2O. How many milligrams of CO2 and
H2O will be produced by the combustion of 4.635
mg of C6H5CO2H?
Solution
C6H5CO2H 15/2O2 FW 122.123
7CO2 3H2O 44.010
18.015
4.635 mg of C6H5CO2H
1 mole C6H5CO2H yields 7 moles CO2 and 3 moles H2O
Mass CO2 7 x 0.03795 mmol x 44.010 mg/mmol
11.69 mg CO2
Mass H2O 3 x 0.03795 mmol x 18.015 mg/mmol
11.69 mg H2O
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Example 2 A 7.290 mg mixture of cyclohexane,
C6H12 (FW 84.159), and Oxirane, C2H4O (FW
44.053) was analysed by combustion, and 21.999
mg CO2 (FW 44.010) were produced. Find the
weight of oxirane in the sample mixture.
Solution
C6H12 C2H4O 23/2O2
8CO2 8H2O
Let x mg of C6H12 and y mg of C2H4O. X
y 7.290 mg
Also,CO2 6(moles of C6H12) 2(moles of C2H4O)
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CO2
X y 7.290 mg
? x 7.290 - y
? y mass of C2H4O 0.767 mg
Therefore, Weight Oxirane
10.52
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The Precipitation Titration Curve
Reasons for calculation of titration curves
1. Understand the chemistry occurring. 2. How to
exert experimental control to influence the
quality of analytical titration.

• In precipitation titrations
• 1. Analyte concentration
• 2. Titrant concentration
• 3. Ksp magnitude

Influence the sharpness of the end point
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Titration Curve
A graph showing variation of concentration of one
reactant with added titrant.
Concentration varies over many orders of magnitude
P function pX -log10X
Consider the titration of 25.00 mL of 0.1000 M I-
with 0.05000 M Ag.
I- Ag ? AgI(s)
There is small solubility of AgI
AgI(s) ? I- Ag
Ksp AgI- 8.3 x 10-17
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I- Ag ? AgI(s) K 1/Ksp 1.2 x
1016
Ve Volume of titrant at the equivalent point
(0.02500 L)(0.1000 mol I-/L)
(Ve)(0.05000 mol Ag/L)

mol I-
mol Ag
? Ve 0.05000 L 50.00 mL
Before the Equivalence Point
Addition of 20 mL of Ag
This reaction I- Ag ? AgI(s) goes to
completion.
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Some AgI redissolves AgI(s) ? I- Ag
I- due to I- not precipitated by 20.00 mL of
Ag.
Fraction of I- reacted
Original volume of I-
Fraction of I- remaining
Total volume
Therefore,
Original Conc.
Fraction Remaining
Dilution Factor
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? I- 3.33 x 10-2 M
?
?
Ag 2.49 x 10-15 M
pAg -logAg 14.60
The Equivalence Point
All AgI is precipitated
AgI(s) ? I- Ag
Then,
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Ksp AgI- 8.3 x 10-17
And Ag I- x
? X 9.1 x 10-9 M
Ksp (x)(x) 8.3 x 10-17
? pAg -log x 8.04
At equivalence point
pAg value is independent of the original
volumes or concentrations.
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After the Equivalence Point
Note Ve 50.00 mL
Ag is in excess after the equivalence point.
Suppose that 52.00 mL is added
Volume of excess Ag
Therefore, 2.00 mL excess Ag
Total volume of solution
Original Ag Concentration
Dilution Factor
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Ag 1.30 x 10-3 M
pAg -logAg 2.89
Shape of the Titration Curve
Equivalence point point of maximum slope
Steepest slope
has maximum value
Inflection point
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Titration Curves Effect of Diluting the reactants

• 0.1000 M I- vs
• 0.05000 M Ag

2. 0.01000 M I- vs 0.005000 M Ag
3. 0.001000 M I- vs 0.0005000 M Ag
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Titrations involving 11 stoichiometry of
reactants
Equiv. Point Steepest point in titration curve
Other stoichiometric ratios 2Ag CrO42- ?
Ag2CrO4(s)
1. Curve not symmetric near equiv. point
2. Equiv. Point Not at the centre of the
steepest section of titration curve
3. Equiv. Point not an inflection point
In practice Conditions chosen such that curves
are steep enough for the steepest point to be a
good estimate of the equiv. point
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Effect of Ksp on the Titration Curve
K 1/Ksp largest
?
AgI is least soluble Sharpest change at equiv.
point
Least sharp, but steep enough for Equiv. point
location
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Titration of a Mixture
Less soluble precipitate forms first.
Titration of KI KCl solutions with AgNO3
Ksp (AgI) ltlt Ksp (AgCl)
First precipitation of AgI nearly complete before
the second (AgCl) commences.
When AgI pption is almost complete, Ag
abruptly increeases and AgCl begins to
precipitate.
Finally, when Cl- is almost completely consumed,
another abrupt change in Ag occurs.
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Titration Curve for 40.00 mL of 0.05000 M KI and
0.05000 M KCl with 0.084 M AgNO3.
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I- end point
Intersection of the steep and nearly horizontal
curves.
Note Precipitation of AgI not quite complete
when AgCl begins to precipitate.
End of steep portion better approximation of the
equivalence point.
Midpoint of the second steep section.
AgCl End Point
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The AgI end point is always slightly high for
I-/Cl- mixture than for pure I-.
1. Random experimental error both tive and
tive.
2. Coprecipitation ve error
Example Some Cl- attached to AgBr ppt and
carries down an equivalent amount of Ag.
Coprecipitation error lowers the calculated
concentration of the second precipitated halide.
High nitrate concentration to minimise
coprecipitation.
NO3- competes with Cl- for binding sites.
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Separation of Cations by Precipitation
Consider a solution of Pb2 and Hg22
Each is 0.01 M
PbI2(s) ? Pb2 2I-
Ksp 7.9 x 10 -9
Hg2I2(s) ? Hg22 2I-
Ksp 1.1 x 10 -28
Smaller Ksp Considerably Less soluble
Is separation of Hg22 from Pb2 complete?
Is selective precipitation of Hg22 with I-
feasible?
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Can we lower Hg22 to 0.010 of its original
value without precipitating Pb2?
From 0.010 M to 1.0 x 106 M?
Add enough I- to precipitate 99.990 Hg22.
Hg2I2(s) ? Hg22 2I-
Initial Concentration 0
0.010 0 Final Concentration
solid 1.0 x 10-6 x
? (1.0 x 10-6)(x)2 1.1 x 10-28
? X I- 1.0 x 10 11 M
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Will this I- 1.0 x 10 11 M precipitate 0.010
M Pb2?
Q 1.0 x 10-24
ltlt 7.9 x 109 Ksp for PbI2
Therefore, Pb2 will not precipitate.
Prediction All Hg22 will virtually precipitate
before any Pb2 precipitates on adding I-.