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

REDOX REACTIONSREDOX REACTIONS

BASIC CONCEPTS

A redox reaction involves transfer of electrons from one species to another.

Oxidation: loss electrons - Reducing agent

Reduction : gain electrons - Oxidizing agent

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)

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

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)

Ohm’s Law, I = E/R

Unit of resistance is ohms or Greek symbol Ω (omega)

Power , P = work/time = E.q = E. q = E. I = I2. R s s

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 1:1 reaction)

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

Line notation for cell

RHS connected to the positive terminal

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

BY INTERNATIONAL AGREEMENT THE SHE IS ARBITARILY ASSIGNED Eo = 0.00V at 25oC.

E0cell = + 0.799V

E0cell = Eo

red (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, A=1) 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

NERNST EQUATION

Le Chatelier`s 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

aA

bBo

A

A

nF

RTEE ln

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] γc γ = activity co efficient – measures

the deviation from ideality , γ = 1

behavior is ideal, low ionic strength

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

aA

bBo

A

A

n

VEE log

05916.0

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).

NERNST EQUATION FOR A COMPLETE CELL

E = E+ - E-

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)

2]50.0[

1log

2

05916.0799.0

VE

]010.0[

1log

2

05916.0402.0

VE

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-

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

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!

aA

cCo

A

A

n

VE log

05916.0 )log

05916.0(

dD

bBo

A

A

n

VE

bB

aA

dD

cCoo

AA

AA

n

VEE log

05916.0

Qn

VE o log

05916.0

1005916.0/0nE

K

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 + 4NH⇌ 3 Eo

red = -1.04 V

Zn2+ + 2e- Zn⇌ Eored = - 0.7628 V

SOLUTION: E0cell = E0

red (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)4

2+ E0 = + 0.28 V

= 10 (2)(0.28)/0.05916 = 2.92 x 10910

05916.0/0nE

K

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.

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

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.

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)

241.0][

][log05916.0767.0

3

2

Fe

FeE

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

SHAPES OF TITRATION CURVES

E ~ Eo(Ce4+I Ce3+) - 0.241V = 1.46V

1:1 stoichiometry symmetric about equivalence point & same curve for diluted sample

Not symmetric about the equivalence point – 2:1

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 + Eo

2)/2

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 .

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)

)]([

)]([log

05916.0

oxidisedIn

reducedIn

n

VEE o (6)

As with acid base indicators, the colour of In (reduced) will be observed when:

[In( reduced)] 10

[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

voltsn

EE o )05916.0

(

Eo= 1.147V, we expect the colour change to occur ~ 1.088 – 1.206V wrt SHE

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+ SO4

2- + SO4- + Ag2+ Two powerful oxidants

Excess reagent is destroyed after by boiling the solution after oxidation is complete

2S2O82- + 2 H2O boiling 4SO4

2- + 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).

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 (pH< 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.

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-

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

-

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 Cr2O7

2- consumed by the oxidation of organics in water.

METHODS INVOLVING IODINE

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-

Precipitation ReactionsPrecipitation 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.

Crystallization1. Nucleation2. Particle

Growth

Molecules form smallAggregates randomly

Addition of more molecules to a nucleus.

Supersaturated Solution: More solute than should be present at equilibrium.

Supersaturated Solution: Nucleation faster; Suspension (colloid) Formed.

Less Supersaturated Solution: Nucleation slower, larger particles formed.

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).

Homogeneous Precipitation

Precipitant generated slowly by a chemical reaction

C

O

H2N NH2

+ 3H2OHeat

CO2+ 2NH4

+ + 2OH-

pH gradually increasesC

O

H OH

+ OH- HCO2- + H2O

Formate

Formic Acid

3HCO2- + Fe3+ Fe(HCO2)3

.nH2O(s)

Fe(III)formate

Large particle size

Net +ve charge on colloidal particle because of adsorbed Ag+

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 chargedAdsorption 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)

Digestion and Purity

Digestion: Period of standing in hot mother liquor.Promotion of recrystallisationCrystal particle size increases and expulsion of impurities.

Purity:

Adsorbed impurities: Surface-bound

Absorbed impurities: Within the crystal Inclusions &Occlusions

Inclusion: Impurity ions occupying crystal lattice sites.

Occlusion: Pockets of impurities trapped within

a growing crystal.

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.

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 impuritye.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.

Product CompositionProduct Composition

Hygroscopic substances: Difficult to weigh accurately

Some ppts: Variable water quantity as water of Crystallisation.

Drying

Change final composition by ignition:

Fe(HCO2)3.nH2O

850 oC

(1 Hour)Fe2O3 + CO2(g) + xH2O(g)

2Mg(NH4)PO4.6H2O Mg2P2O7 + 2NH3 + 13H2O

1100oC

(1 Hour)

Thermogravimetric AnalysisThermogravimetric Analysis

CO2CaO2C

H2O 200oC

OH OH

CO2CaO2C

OH OH

O

Ca

O

O

300oC

500oCCaCO3

700oCCaO

Calciumcarbonate

Calciumoxide

Calcium salicylate monohydrate

Heating a substance and measuring its mass as a function of temperature.

2Mg(NH4)PO4.6H2O Mg2P2O7 + 2NH3 + 13H2O

1100oC

(1 Hour)

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.

Solution:

The gravimetric factor is:

Grams of Mg in analyte

Grams of Mg2P2O7=

2 x (24.305)

222.553

Relative atomic massOf Mg

FM of Mg2P2O7

Grams of Mg in analyte = Grams of Mg2P2O7 formed2 x (24.3050)

222.553

553.222

)3050.24(22168.0

xg

Note: 2 mol Mg2+ in 1 mol Mg2P2O7.

Mass of Mg 2+ = 0.0471 g

% Mg =Mass of Mg2+

sample Mass(100) = 0.0474

0.352(100)

= 13.45 %

Combustion Analysis

Determination of the carbon and Hydrogen content of organic compounds burned in excess oxygen.

H2O absorption

CO2 AbsorptionPrevention of entrance of

atmospheric O2 and CO2.

Note: Mass increasein each tube.

C, H, N, and S Analyser: Modern Technique

Thermal Conductivity, IR,or Coulometry for Measuring products.

2 mg sample in tin or silver capsule.

Capsule melts and sample is oxidised in excess of O2.

C, H,N, S1050 oC; O2 CO2(g) + H2O(g) + N2(g) + SO2(g) + SO3(g)

(95 % SO2)

Products Hot WO3 catalyst: CarbonHeat

CrO3 Cat. CO2

Then, metallic Cu at 850 oC:

Cu + SO3850 oC

SO2 + CuO(s)

2Cu + O2850 oC

2CuO(s)

Dynamic Flash combustion: Short burst of gaseous products

Oxygen Analysis:

Pyrolysis or thermal decomposition in absence of oxygen.

Gaseous products: Nickelised Carbon1075 oC

CO formed

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 MassSpectrometry

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 + 3H2O44.010 18.015

4.635 mg of C6H5CO2H = mmolmmolmg

mg03795.0

/123.122

634.4

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

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)

mmolmg

mgyx

/010.44

999.21

053.442

161.846

mmolmg

mgyx

/010.44

999.21

053.442

161.846

X + y = 7.290 mg x = 7.290 - y

CO2

mmolmg

mgyy

/010.44

999.21

053.442

161.84

290.76

y = mass of C2H4O = 0.767 mg

Therefore, % Weight Oxirane = )100(294.7

767.0

mg

mg

= 10.52 %

The Precipitation Titration CurveThe 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

Titration CurveTitration CurveA graph showing variation of concentration of one reactant with added titrant.

Concentration varies over many orders of magnitude

P function: pX = -log10[X]

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 = [Ag+][I-] = 8.3 x 10-17

I- + Ag+ AgI(s) K =1/Ksp = 1.2 x 1016

Ve = Volume of titrant at the equivalent point:

Before the Equivalence Point:

Addition of 20 mL of Ag+:

This reaction: I- + Ag+ AgI(s) goes to completion.

Ve = 0.05000 L = 50.00 mL

(0.02500 L)(0.1000 mol I-/L) (Ve)(0.05000 mol Ag+/L)=

mol I- mol Ag+

I

KAg sp [I-] due to I- not precipitated by

20.00 mL of Ag+.

Fraction of I- reacted: )00.50(

)00.20(

mL

mL

Fraction of I- remaining: )00.50(

)00.30(

mL

mL

Some AgI redissolves: AgI(s) I- + Ag+

Therefore,

mL

mLM

mL

mLI

00.45

00.25)1000.0(

00.50

00.30][

FractionRemaining

OriginalConc.

DilutionFactor

Original volumeof I-

Totalvolume

[I-] = 3.33 x 10-2 M

I

KAg sp

2

17

1033.3

103.8

x

xAg

[Ag+] = 2.49 x 10-15 M

pAg+ = -log[Ag+] = 14.60

The Equivalence PointThe Equivalence Point::

All AgI is precipitated

AgI(s) I- + Ag+Then,

Ksp = [Ag+][I-] = 8.3 x 10-17

And [Ag+] = [I-] = x

Ksp = (x)(x) = 8.3 x 10-17 X = 9.1 x 10-9 M

pAg+ = -log x = 8.04

At equivalence point:

pAg+ value is independent of the original volumes or concentrations.

After the Equivalence PointAfter the Equivalence Point:

[Ag+] is in excess after the equivalence point.

Note: Ve = 50.00 mL

Suppose that 52.00 mL is added:

Therefore, 2.00 mL excess Ag+

mL

mLMAg

00.77

00.2)05000.0(][

Original Ag+

Concentration Dilution Factor

Volume of excessAg+

Total volumeof solution

[Ag+] = 1.30 x 10-3 M

pAg+ = -log[Ag+] = 2.89

Shape of the Titration Curve:

Steepest slope:dx

dyhas maximum value

Equivalence point: point of maximum slope

Inflection point: 02

2

dx

yd

Titration Curves: Effect of Diluting the reactantsTitration Curves: Effect of Diluting the reactants

1. 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+

Titrations involving 1:1 stoichiometry of reactantsTitrations involving 1:1 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 steepenough for the steepest point to be a good estimate of the equiv. point

Effect of KEffect of Kspsp on the Titration Curve on the Titration Curve

AgI is least solubleSharpest change at

equiv. point

Least sharp, but steep enough for Equiv.

point location

K = 1/Ksp largest

Titration of a MixtureTitration of a Mixture

Less soluble precipitate forms first.

Titration of KI & KCl solutions with AgNO3

Ksp (AgI) << 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.

Titration Curve for 40.00 mL of 0.05000 M KI and 0.05000 M KCl with 0.084 M AgNO3.

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.

AgCl End Point: Midpoint of the second steep section.

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

High nitrate concentration to minimise coprecipitation.

Example: Some Cl- attached to AgBr ppt and carries down an equivalent amount of Ag+.

NO3- competes with Cl- for binding sites.

Coprecipitation error lowers the calculated concentration of the second precipitated halide.

Separation of Cations by PrecipitationSeparation of Cations by Precipitation

Consider a solution of Pb2+ and Hg22+: Each is 0.01 M

PbI2(s) Pb⇌ 2+ + 2I-

Hg2I2(s) Hg⇌ 22+ + 2I-

Ksp = 7.9 x 10 -9

Ksp = 1.1 x 10 -28

Smaller Ksp

ConsiderablyLess soluble

Is separation of Hg22+ from Pb2+ “complete”?

Is selective precipitation of Hg22+ with I- feasible?

Can we lower [Hg22+ ] to 0.010 % of its original value

without precipitating Pb2+?

From 0.010 M to 1.0 x 10–6 M?

Add enough I- to precipitate 99.990 % Hg22+.

Hg2I2(s) Hg⇌ 22+ + 2I-

Initial Concentration: 0 0.010 0 Final Concentration: solid 1.0 x 10-6 x

spKIHg 222 ]][[ (1.0 x 10-6)(x)2 = 1.1 x 10-28

X = [IX = [I--] = 1.0 x 10 ] = 1.0 x 10 –11–11 M M

Will this [I [I--] = 1.0 x 10 ] = 1.0 x 10 –11–11 M M precipitate 0.010 M Pb2+?

21122 )100.1)(010.0(]][[ xIPbQ

Q = 1.0 x 10-24 << 7.9 x 10–9 = Ksp for PbI2

Therefore, Pb2+ will not precipitate.

Prediction: All Hg22+ will virtually precipitate before any

Pb2+ precipitates on adding I-.