REDOX TITTRATIONS_2012

8/22/2019 REDOX TITTRATIONS_2012

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REDOX TITTRATIONS

Analytical Chemistry Laboratory

2012


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FUNDAMENTAL TERMS

Reactive species in this reaction type is theelectron, which is transferred from the reductantto the oxidant

Most elements are capable of exhibiting morethan one oxidation state

It is customary to describe redox reaction inelectrochemical terms because transfer electronmay also be carried out in an electrochemicalcell


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Redox Reactions Oxidation process : loss of electron

Reduction process : gain of electron

Reducing agent is oxidized

Oxidizing agent is reduced


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NERSNT EQUATION

To relate electrochemical potentials to activities(concentration) of species in the system, we can

draw on the thermodynamics relationshipinvolving free energy change and activities,namely :

G = G0 + RT ln Q

G = -nFE - nFE = -nFE0 + RT ln Q

E = E0 - RT/nF ln QE = E0 - 0,05916/n log Q


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E0 = electrochemical potential for the reaction

when all species are in their standard state Its describe the tendency of the ion to

reductizes


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REDOX TITRATION CURVEs

To evaluate a redox titration we must know theshape of its titration curve

For redox titration, it is convenient to monitorelectrochemical potential coz we are dealingwith electron

Nernst equation relates the electrochemicalpotential to the concentrations of reactants and

products participating in a redox reaction


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Consider, for example a titration in which theanalyte in a reduced state, Ared is titrated with atitrant in an oxidized state Tox.

The titration reaction is :

A red + T ox T red + Aoxthe electrochemical potential for the reaction isthe difference between the reduction potentialsfor the reduction and oxidation half reaction;thus

Erxn = ETox/Tred EAox/Ared


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Before the equivalence point the titration

mixture consists of appreciable quantities ofboth the oxidized and reduced forms of theanalyte, but very little unreacted titrant.

The potential, therefore, is best calculated usingthe nernst equation for the analytes halfreaction

EAox/Ared = E0Aox/Ared RT/nF ln [Ared]/[Aox]


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After each addition of titrant, the reactionbetween the analyte and titrant reaches astate of equilibrium. The reactionselectrochemical potential, Erxn, therefore

is zero, andE Tox/Tred = E Aox/Ared


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After the equivalence point, the potential iseasiest quast to calculate using the Nernstequation for the titrants half reaction, sincesignificant quantities of its oxidized and

reduced forms are presentETox/Tred = E0Tox/Tred RT/nF ln [Tred]/[Tox]


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Example

Calculate the titration curve for the titration of50 mL of 0,1 M Fe2+with 0,1 M Ce4+ in a matrixof 1M HClO4. (after 5 mL, 50 mL and 60 mLtitrant added)

the reaction is

Fe 2+ + Ce 4+ Fe 3+ + Ce 3+

assume analyte and titrant react completely


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Answer

We calculate volume we need to reach the equivalent point. From

the stoichiometry we know that :

So volume Ce4+ needed were :


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Before equivalent point :

Easier for us to measure the potential from analyte half potential reaction


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Substituting these concentration into potential halfs reaction, gives us :


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Equivalent Point :

Mol of [Fe2+

] and [Ce4+] equal but so small, so we cant calculate the potentialfrom reactant or titrant halfs reaction only. We have to combine the two

Nernst Equation.

Adding together this two Nernst equation, give us :


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At the equivalent point , the titration reaction stoichiometry requires that

So the ratio of concentration become one and the log become zero, the

potential then:


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After adding 60 mL titrant : (the condition are after equivalent point),

we can calculate the potential from potential of titrant halfs reaction


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Substituting these concentration gives us :


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Evaluating the end point


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Finding the end point with visual indicator

Redox indicator : substances that do not participate inthe redox titration, but whose oxidized and reducedforms differ in color

When added to a solution containing analyte, theindicator imparts a color that depends on the solutionselectrochemical potential

Since the indicator changes color in response to the

elctrochemical potential, and not to the presence orabsence of a specific species, these compounds are calledgeneral redox indicator

Specific redox indicator : react with the presence of aspecific species


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Types of indicators used to signal end point :

MnO4-when MnO4- is used as an oxidizing titrant, the solutionremains colorless until the first drop of excess MnO4- isadded. The first tinge purple signals the end point

Starch (Specific Indicator)

forms a dark blue complex with I2 and can be used tosignals the presence of excess I2 (color change : colorlessto blue), or the completion of a reaction in which I2 isconsumed (color change : blue to colorless)

Thiocyanate (specific indicator)forms a soluble red-colored complex Fe(SCN)2+, with Fe3+


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REDOX TITRATION METHODS

Titration Involving Iodine : Iodometry and Iodimetry

Titration With Oxidizing Agent : Permanganometry,Cerimetry, potassium dichromate


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Iodimetry

Titration with I2 solution

Titration performed in neutral or mildy alkaline (pH 8) to

a weakly acid solution

Reason avoiding the pH too acid : starch as indicatortends to hydrolyze in strong acid, reducing power of

some reducing agent decreases in acid solution, iodide

produced in the reaction tends to be oxidized by dissolvedoxygen in acid solution

Indicator : Starch


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Iodometry

Add excess of Iodide (I-) to a solution of an oxidizing agent, I2

produced in an equivalent amount to the oxidizing agent

I2 present can be titrated with reducing agent such as sodium

thiosulfate

I2 + 2S2O32- 2I- + S4O6

2-

End point titration detected with starch (by disappearance of the blue

starch-I2 color)

Most titration performed in acid solution

Example : assay of potassium dichromate


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View down the starch helix,

showing iodine, inside the helix

Structure of the repeating unit of thesugar amylose.

Schematic structure of the starch-

iodine complex. The amylose chain

forms a helix around I6 unit.


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Permanganometry

Use potassium permanganate as oxidizing titrant

Acts as self indicator for end point detection


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Oxidation with Ce4+

Ce4+ + e = Ce3+ 1.7V in 1 N HClO4

yellow colorless 1.61V in 1N HNO3

1.47V in 1N HCl

1.44V in 1M H2SO4

Indicator : ferroin, diphenylamine

Preparation and standardization:

Ammonium hexanitratocerate, (NH4)2Ce(NO3)6, (primary standard grade)

Ce(HSO4)4, (NH4)4Ce(SO4)42H2O

Standardized with Sodium oxalate.


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Applications of cerimetry

(1) Menadione (2-methylnaphthoquinon: vitamin K3)

O

O

CH3

OH

OH

CH3

2 Ce(SO4)2

HCl, Zn

Reduction

(2) Iron

2FeSO4 + 2 (NH4)4Ce(SO4)4 = Fe2(SO4)3 + Ce2(SO4)3 + 4 (NH4)2SO4


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Oxidation with potassium dichromate

Cr2O72 + 14H+ + 6e = 2Cr3+ + 7H2O Eo = 1.36 V

K2Cr2O7 is a primary standard.

Indicator : diphenylamine sulphonic acid

End point colour : violet


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Ex. Redox titration ( hydroquinone vs dichromate standard

solution )

HO OH O O + 2H+ + 2e Eo= 0.700

Cr2O72 + 14H+ + 6e 2 Cr3+ + 7 H2O

Eo= 1.33

3

3

HO

OH + Cr2O72 + 8H+ 3

O

O + 2 Cr3+ + 7

H2O

E

o

= E

o

cathode E

o

anode = 1.33 0.700 =0.63 V

K = 10 nEo/0.05916 = 10 6(0.63) / 0.05916 = 10 64

redox indicator : diphenylamine

colorless to violet

Very large : quantitative : complete reaction


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Example

A solution of sodium thiosulfate was standardized by dissolving 0,1210 g of

KIO3 (mw 214 g/mol) in water, adding a large excess of KI, and acidifying with

HCl. The liberated iodine required 41,64 mL of the thiosulfate solution to

decolorize the blue starch/iodine complex. Calculate the molarity of the

Na2S2O3.

Reaction :

IO3- + 5I- + 6H+ 3 I2 + 3H2O

I2 + 2S2O3 2- 2I- + S4O6 2-


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Solution :

1 mol IO3 equivalent to 3 mol I2 equivalent to 6 mol S2O3 2-

Amount of S2O3 2- = [(0,1210 x 1000) / 214 ] x 6

= 3,392 mmol

Molarity of S2O3 2- = 3,392 : 41,64 mL = 0,0842 M


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Determining waterwith the Karl FisherReagent

The Karl Fisher reaction :

I2

+ SO2

+ 2H2

O 2HI + H2

SO4

For the determination of small amount of water, Karl Fischer(1935) proposed

a reagent prepared as an anhydrous methanolic solution containing iodine,

sulfur dioxide and anhydrous pyridine in the mole ratio 1:1:3 The reaction

with water involves the following reactions :

C5H5NI2 + C5H5NSO2 + C5H5N + H2O

2 C5H5NH I + C5H5NSO3C5H5N

+SO3 + CH3OH C5H5N(H)SO4CH3

Pyridinium sulfite can also consume water.

C5H5N+SO3

+ H2O C5H5NH+SO4H

It is always advisable to use fresh reagent because of the presence of various side reactions involving iodine. The reagent is stored in a desiccant-

protected container.

The end point can be detected either by visual( at the end point, the color

changes from dark brown to yellow) or electrometric, or photometric

(absorbance at 700nm) titration methods. The detection of water by the

coulometric technique with Karl Fischer reagent is popular.


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Pyridine free Karl Fisher reagent

In recent years, pyridine, and its objectionable odor, have been replaced in the

Karl Fisher reagent by other amines, particularly imidazole.

(1) Solvolysis 2ROH + SO2 RSO3 + ROH2

+

(2) Buffering B + RSO3 + ROH2

+ BH+SO3R

+ ROH

(3) Redox BI2 + BH+SO3R

+ B + H2O BH+SO4R

+ 2 BH+I


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HOME WORK

Derive a curve for the titration of 50 mL of 0,025 M U 4+ with 0,1 M Ce 4+ after

adding 5 mL , 25mL, and 30 mL of Ce 4+ . Assume that the solution Is 1.0 < in

H2SO4 throughout the titration ([H+] for such a solution will be about 1.0 M)

The analytical reaction is :U 4+ + 2H2O + 2 Ce4+ UO2

2+ + 2 Ce 3+ + 4H+

From the handbook :

Ce 4+ + e Ce 3+ Eo = +1.44 V

UO 2 2+ + 4H+ + 2e U 4+ + 2H2O Eo = +0,334 V

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REDOX TITTRATIONS_2012