REDOX TITTRATIONS

8/10/2019 REDOX TITTRATIONS

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REDOKS TITTRATIONS

Aliya Nur Hasanah

Laboratorium kimia analitis

2013

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TITRASI YANG DIDASARKAN PADA REAKSIOKSIDASI DAN PENGURANGAN ANTARAANALYTE DAN TITRANT


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Redoks reaksi:

Istilah dasar Proses oksidasi: kehilangan elektron proses penurunan: keuntungan dari

elektron agen pereduksi teroksidasi

Agen oksidasi berkurang


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NERSNT EQUATION

It is customary to describe redox reaction inelectrochemical terms because transfer electron mayalso be carried out in an electrochemical cellNernst Equation

To relate electrochemical potentials to activities(concentration) of species in the system, we candraw on the thermodynamics relationship involvingfree energy change and activities, namely :

G = G + RT ln Q 0 G = -nFE - nFE = -nFE + RT ln Q 0

RT/nF ln QE = E0

E = E0 0,05916/n log Q


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E = electrochemical potential for the reaction 0

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, A is titrated with aredtitrant in an oxidized state Tox

The titration reaction is :

A + T Tredthe electrochemical potential for the reaction is

ox red + Aox

the 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

electrochemical potential, Eof equilibrium. The reactions

, thereforerxn

is zero, andE = ETox/Tred 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 = E Tox/Tred RT/nF ln [Tred]/[Tox]0


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Example

Calculate the titration curve for the titration of 50 mL of 0,1 M Fe2+

with 0,1 M Ce4+ in a matrix of 1M HClO4. (after 5 mL, 50 mL and 60mL titrant added).

E Feo 3+/Fe2+ = +0,767 Volt. E Ceo 4+/Ce 3+ = 1,70 Volt

the reaction isFe 2+ + Ce 4+ Fe

assume analyte and titrant react completely

3+ + Ce 3+


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Answer

We calculate volume we need to reach the equivalent point. Fromthe 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 [Fe ] and [Ce4+] equal but so small, so we cant calculate the potential2+from 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, thepotential 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 in theredox titration, but whose oxidized and reduced formsdiffer in color

When added to a solutionindicator imparts a color that depends on the solutions

containing analyte, the

electrochemical 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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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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2- - 2-

Iodometry

Add excess of Iodide (I-) to a solution of an oxidizing agent, I 2

produced in an equivalent amount to the oxidizing agent

I

thiosulfate

2 present can be titrated with reducing agent such as sodium

I2

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

+ 2S O2 3 2I + S O4 6

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 I unit.6


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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, (NH ) Ce(NO ) , (primary standard grade)4

2 3 6

Ce(HSO ) , (NH ) Ce(SO ) 2H O 44 4 4 4 4 2

Standardized with Sodium oxalate.


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Applications of cerimetry

(1) Menadione (2-methylnaphthoquinon: vitamin K )3

O

O

CH3

OH

OH

CH3

2 Ce(SO )4 2

HCl, Zn

Reduction

(2) Iron

2FeSO4 + 2 (NH ) Ce(SO )4 4 4 4 = Fe (SO )2

4 3 + Ce (SO ) + 4 (NH ) SO 24 3 4 2 4


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Oxidation with potassium dichromate

Cr O2 72 + 14H + 6e = 2Cr +3+ + 7H O E = 1.36 V 2o

K Cr O2 2 7 is a primary standard.

Indicator : diphenylamine sulphonic acid

End point colour : violet


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Ex. Redox titration

solution )

( hydroquinone vs dichromate standard

HO OH O O + 2H+ + 2e E = 0.700o

Cr O2 7E = 1.33o

2 + 14H+ + 6e 2 Cr3+ + 7 H O2

3

3 HO OH + Cr O2

72 + 8H+ 3 O O + 2 Cr3+ + 7 H O2

E = E

o o

0.63 Vcathode E

o

anode = 1.33 0.700 =

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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)

Determining water with the Karl Fisher Reagent

The Karl Fisher reaction :

I2

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

+ SO + 2H O2 2

2HI + H SO2 4

a reagent prepared as an anhydrous methanolic solution containing iodine,sulfur dioxide and anhydrous pyridine in the mole ratio 1:1:3 The reactionwith water involves the following reactions :

C H NI + C H NSO + C H N + H O 55 2 5 5 2 5 5 2

2 C H NHI + C H NSO5 5 5 5 3C H N SO5 5

+3 + CH OH3

Pyridinium sulfite can also consume water.

C H N(H)SO CH5 5 4 3

C H N SO5 5+

3 + H O2

It is always advisable to use fresh reagent because of the presence of

C H NH SO H5 5+

4

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

(absorbance at 700 nm

coulometric technique with Karl Fischer reagent is popular.

titration methods. The detection of water by the


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You wish to standardize the solution of KMnO4 0,010 M against standard Na2C2O4(Mr = 134). If you want to use between 30 and 45 mL of the reagent for thestandardiization, what range of weights of the primary standard

should you take?

HOME WORK


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HOME WORK

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

adding 5 mL , 25mL, and 30 mL of Ce 4+with 0,1 M Ce 4+

. Assume that the solution Is 1.0 M in

after

H SO throughout the titration ( [H+] for such a solution will be about 1.0 M) 24

The analytical reaction is :U 4+ + 2H O + 2 Ce 4+2 UO22+ + 2 Ce 3+ + 4H+

From the handbook :

Ce 4+ + e Ce2+ + 4H + 2e U

+

4+

3+ o

+ 2H O E = +0,334 V o

E = +1.44 V

UO 2 2

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REDOX TITTRATIONS