Oxidation-Reduction Titration Dr. A.K.M. Shafiqul Islam August 24, 2007

Oxidation-Reduction Oxidation-Reduction TitrationTitration

Dr. A.K.M. Shafiqul IslamDr. A.K.M. Shafiqul Islam

August 24, 2007August 24, 2007

Oxidation Reduction ReactionOxidation Reduction Reaction Reactions of metals or any other organic

compounds with oxygen to give oxides are labeled as oxidation.

The removal of oxygen from metal oxides to give the metals in their elemental forms is labeled as reduction.

In other words, oxidation is addition of oxygen and removal of hydrogen whereas reduction is addition of hydrogen and removal of oxygen.

Reactions of metals or any other organic compounds with oxygen to give oxides are labeled as oxidation.

The removal of oxygen from metal oxides to give the metals in their elemental forms is labeled as reduction.

In other words, oxidation is addition of oxygen and removal of hydrogen whereas reduction is addition of hydrogen and removal of oxygen.

Oxidation Reduction ReactionOxidation Reduction Reaction

Chemical reactionChemical reaction

Reduction-Oxidation reaction commonly Reduction-Oxidation reaction commonly known as RedOx reactionknown as RedOx reaction

OxOx11 + Red + Red22 ⇌⇌ Red Red11 + Ox + Ox22

Chemical reaction based oxidation and Chemical reaction based oxidation and reduction reaction is known as RedOx reduction reaction is known as RedOx reactionreaction

FeFe2+2+ + Ce + Ce4+4+ → Fe → Fe3+3+ + Ce + Ce3+3+ (1)(1)

There are two kinds of electro chemical cells, galvanic or electrolytic.

In galvanic cells, the chemical reaction occurs spontaneously to produce electrical energy.

In a electrolytic cell, electrical energy is used to force the non spontaneous chemical reaction.

Galvanic cells are of importance in our further discussion as we will be discussing the spontaneous chemical reaction to produce electrical energy.

Electrochemical CellsElectrochemical Cells

If a solution containing Fe2+ is mixed with another solution containing Ce4+, there will be a redox reaction situation due to their tendency of transfer electrons. If we consider that these two solution are kept in separate beaker and connected by salt bridge and a platinum wire that will become a galvanic cell. If we connect a voltmeter between two electrode, the potential difference of two electrode can be directly measured.

The Fe2+ is being oxidised at the platinum wire (the anode):

Fe2+ → Fe3+ + e-

The electron thus produced will flow through the wire to the other beaker where the Ce4+ is reduced (at the cathode).

Ce4+ + e- → Ce3+

The reaction involve electron transferThe reaction involve electron transferFeFe2+2+ → Fe → Fe3+ 3+ + e+ e- - (2)(2)CeCe4+4+ + e + e-- → Ce → Ce3+3+ (3)(3)

Equation (2) & (3) are called half reactions Equation (2) & (3) are called half reactions

No half reaction can occur by itselfNo half reaction can occur by itself

There must be an electron donner (reducing agent) There must be an electron donner (reducing agent) and an electron accepter (oxidizing agent)and an electron accepter (oxidizing agent)

FeFe2+2+ is the reducing agent and Ce is the reducing agent and Ce3+3+ is the oxidizing is the oxidizing agentagent

Before we start the discussion of the oxidation reduction titration curve construction, we should understand the Nernst equation which was introduced by German scientist, Wlater Nernst in 1889.

This equation express the relation between the potential of metal and metal ion and the concentration of the ion in the solution.

Lets consider the following chemical reaction :-

aA + bB ⇋ cC + dD

The change in free energy is given by the equation

ΔG = ΔGo + 2.3RT log [C]c x [D]d / [A]a x [B]b

From the relation ship of the free energy and cell potential, we can get

ΔG = -nFE

In a standard states, free energy will be

ΔGo = -nFEo

Hence, the above equation can be written as

-nFE = -nFEo + 2.3RT log [C]c [D]d / [A]a [B]b

After dividing both side with –nF, we can get the expression as

E = Eo – 2.3 RT/ nF log [C]c [D]d / [A]a [B]b

At 25ºC, the equation can be written as :-

E = Eo – 0.059/n log [C]c [D]d / [A]a [B]b

This equation is known as the “Nernst equation” that correlate the electrode potential with the concentration of the ionic species.

Redox titration is monitored by observing the change of electrode potential. The titration curve is drawn by taking the value of this potential against the volume of the titrant added. Unlike other titration curve the p values are substituted by electrode potential values in the curve.

The redox reaction is rapid and the system is always in equilibrium throughout the titration. The electrode potential of the two half reaction are always identical.

If we consider the oxidation of Fe (II) with standard Ce (IV), then we can write the equation as follows :-

Fe2+ + Ce4+ Fe⇋ 3+ + Ce3+

Redox Titration CurveRedox Titration Curve

The electrode potential of the two half reaction will be always identical.

For iron, the electrode potential will be

E = EºFe – 0.059 log [Fe2+] / [Fe3+]

For cerium the electrode potential will be

E = EºCe – 0.059 log [Ce3+] / [Ce4+]

Therefore, either of the electrode potential could be utilised to calculate the potential of the solution.

The utilisation of the either equation is based on the stage of titration. Prior to the equivalence point, the concentration of Fe(II) and Fe(III) are appreciable compare to Ce(IV) ion which is negligible because of the presence of large excess of Fe(II).

Beyond the equivalence point, the concentration of Ce(IV) and Ce(III) is readily computed from the addition and the electrode potential for the Ce(IV) could be used.

At equivalence point, the concentration of the oxidised and reduced forms of the two species are such that their attraction for electron are identical. At this point, the reactant species concentration and product species concentrations ratios are known and they are utilised to calculate the potential of the solution.

# 50.0 ml of 0.05M Fe2+ is titrated with 0.1M Ce4+ in a sulphuric acid media at all times. Calculate the potential of the inert electrode in the solution at various intervals in the titration and plot the titration curve. Use 0.68V as the formal potential of the Fe2+- Fe3+ system in sulphuric acid and 1.44V for the Ce3+- Ce4+ system.

Initial step : After addition of 5.0 ml of Ce4+

As because the Ce4+ is too small, we are considering the iron (Fe) electrode potential to calculate the solution potential.

[Fe3+] = 5.0ml X 0.10M / (50.0 + 5.0) ml

= 0.5 mmol / 55.0ml

Similarly, [Fe2+] = (50.0ml X 0.05M – 5.0 ml X 0.1M) / 55.0 ml

= 2.0 mmol / 55.0 ml

Substituting the values to the standard electrode potential equation, we can get

E = EºFe – 0.059 log [Fe2+] / [Fe3+]

E = Eº – 0.059 log 2.0 / 0.5

E = 0.68 -0.036 = 0.64 V

Step-2 : Equivalence point

At equivalence point in the titration of Fe(II) and Ce(IV), the potential of the solution is controlled by both the half reaction.

Eeq=E0Ce -0.059 log [Ce3+] / [Ce4+]

Eeq = E0Fe - 0.059 log [Fe2+] / [Fe3+]

Adding the two expression, we can get

2Eeq = E0Ce + E0 Fe - 0.059 log [Ce3+] [Fe2+]/ [Ce4+][Fe3+]

At equivalence the concentration of Fe3+ = Ce3+ and Fe2+ = Ce4+

We can get ,

2Eeq = E0Ce + E0 Fe - 0.059 log [Ce3+] [Ce4+]/ [Ce4+] [Ce3+]

Eeq = (E0Ce + E0 Fe) / 2

Eeq = (1.44 + 0.68) / 2 = 1.06 V

Step-3 : After addition of 25.1 ml Ce4+

At this stage, the concentration of Fe(II) is very small and we can neglect the value and for convenience, we will utilise the Ce(IV) electrode potential to calculate the solution potential.

[Ce3+] = (25.0ml X 0.1M ) / (50.0 + 25.1) ml

= 2.5 mmol / 75.1 ml

[Ce4+] = (0.1 ml X 0.1M) / 75.1 ml

E = EºCe – 0.059 log [Ce3+] / [Ce4+]

E = 1.44 – 0.059 log 2.5 / 0.01

= 1.30 V

Redox TitrationsRedox Titrations

Equipment for Equipment for obtaining a obtaining a titration curve for titration curve for a redox titration.a redox titration.

RedOx Titration Curve RedOx Titration Curve

0.6

0.8

1.0

1.2

1.4

1.6

1.8

0 20 40 60 80 100 120 140 160 180 200

mL Ce4+

E, v

olt

s


The net balanced redox equation is the The net balanced redox equation is the sum of the two half-cell reactions. sum of the two half-cell reactions.

It may be necessary to multiply one or It may be necessary to multiply one or both half-cells by some coefficient so both half-cells by some coefficient so that the same number of electrons are that the same number of electrons are lost by the substance that is oxidized as lost by the substance that is oxidized as are gained by the substance reduced.are gained by the substance reduced.

Cu0(S) Cu2+(aq) + 2e-

2Ag +(aq) + 2e- 2Ag (S)

Cu0(S) + 2Ag +(aq) + 2e- Cu2+(aq) + 2Ag (S) + 2e-

Cu0(S) + 2Ag +(aq) Cu2+(aq) + 2Ag (S)

Number of e-s involved in the overall reaction is 2

Balancing simple redox reactionsBalancing simple redox reactions

Fe+2(aq) + MnO4-(aq) Mn+2(aq) + Fe+3(aq)

Fe+2(aq) Fe+3(aq) + 1e-

MnO4-(aq) Mn+2(aq)

Oxidizing half:

Reducing half:

Balancing atoms:

MnO4-(aq)+ Mn+2(aq) + 4H2O

Balancing oxygens:

Balancing complex redox reactionsBalancing complex redox reactions

Balancing complex redox reactions

MnO4-(aq)+8H+ Mn+2(aq) + 4H2O

Balancing hydrogens:

Oxidation numbers: Mn = +7, O = -2 Mn = +2

Balancing electrons:The left side of the equation has 5 less electrons than the right side

MnO4-(aq)+8H++ 5e- Mn+2(aq) + 4H2O

Reducing Half

Reaction happening in an acidic medium

Balancing complex redox reactionsFinal Balancing act:Making the number of electrons equal in both half reactions

[Fe+2(aq) Fe+3(aq) + 1e- ]× 5

[MnO4-(aq)+8H++ 5e- Mn+2(aq) + 4H2O]×1

5Fe+2(aq) 5Fe+3(aq) + 5e- MnO4

-(aq)+8H++ 5e- Mn+2(aq) + 4H2O

5Fe2++MnO4-(aq)+8H++ 5e-

5Fe3+ +Mn+2(aq) + 4H2O + 5e-


Many of the end point indicators Many of the end point indicators used in redox titrations are used in redox titrations are substances that are either oxidized substances that are either oxidized or reduced at given potentials. or reduced at given potentials.

Much like acid-base indicators, the Much like acid-base indicators, the electrical potential at the end point electrical potential at the end point oxidizes or reduces the indicator to oxidizes or reduces the indicator to one of the colored forms to signal the one of the colored forms to signal the end point.end point.

RedOx IndicatorsRedOx Indicators

If the titrant is highly colored, this If the titrant is highly colored, this color may be used to detect end color may be used to detect end point. point.

0.02 M potassium permanganate is 0.02 M potassium permanganate is deep purple. A dilute solution is pink. deep purple. A dilute solution is pink. The product of its reduced form The product of its reduced form (Mn(Mn2+2+) colorless.) colorless.

The unknown sample of iron contains, iron in Fe2+

oxidation state. So we are basically doing a redoxtitration of Fe2+ Vs KMnO4

5Fe2++MnO4-(aq)+8H+ 5Fe3+ +Mn+2(aq) + 4H2O

Titration of unknown sample of Iron Vs KMnO4

250mL 250mL 250mL

Vinitial

Vfinal

End point:Pale PermanentPink color

KMnO4

Vfinal- Vinital= Vused (in mL)

Important requirement:The concentration of KMnO4 should be known precisely.

Redox IndicatorsRedox Indicators

Titrations Involving IodineTitrations Involving Iodine

One of the most common redox One of the most common redox titrations involve either using iodine titrations involve either using iodine (I(I22) as a mild oxidizing agent or ) as a mild oxidizing agent or iodide (Iiodide (I--) as a mild reducing agent.) as a mild reducing agent.

Iodine as oxidizing agent:Iodine as oxidizing agent:

II22 + 2 e + 2 e -- = 2 I = 2 I --

When iodine is used as the titrant the When iodine is used as the titrant the method is known as iodimetry.method is known as iodimetry.


II22 is not very soluble in water (only about is not very soluble in water (only about1.3 x 10 1.3 x 10 -3-3 mol/L). It solubility is increased mol/L). It solubility is increased

in the presence of excess iodide by the in the presence of excess iodide by the formationformation

of the triiodide (Iof the triiodide (I33--) species, ) species,

II22 + I + I -- = I = I33 –– So it is really the triiodide species, though So it is really the triiodide species, though

it will commonly be referred to as iodine it will commonly be referred to as iodine that is involved in the chemical that is involved in the chemical reactions.reactions.

Due to the difficulty is maintaining the Due to the difficulty is maintaining the concentration of concentration of II22, (limited solubility in , (limited solubility in water, appreciable vapor pressure of the water, appreciable vapor pressure of the II22.) iodometric methods are more .) iodometric methods are more commonly used. The amount of commonly used. The amount of II2 2

produced by the action of the oxidizing produced by the action of the oxidizing analyte on excess iodide is usually titrated analyte on excess iodide is usually titrated with standardized sodium thiosulfate, with standardized sodium thiosulfate, NaNa22SS22OO33..



Iodide as reducing agent:Iodide as reducing agent:

2 I2 I-- = I = I22 + 2 e + 2 e --

When iodine is produced by the When iodine is produced by the addition of an oxidizing analyte to an addition of an oxidizing analyte to an excess of iodide, the method is as excess of iodide, the method is as iodometry.iodometry.


Thiosulfate (SThiosulfate (S22OO33-2-2) is commonly used ) is commonly used

in titration reactions involving iodine, in titration reactions involving iodine, both for iodimetric and iodometric both for iodimetric and iodometric methods. The iodimetric methods methods. The iodimetric methods generally involve an excess of generally involve an excess of standard Istandard I22 (as I (as I33

- -) followed by back ) followed by back titration with standard thiosulfate.titration with standard thiosulfate.


In the iodometric methods, an excess In the iodometric methods, an excess of iodide is added to the sample of of iodide is added to the sample of an oxidizing analyte and a an oxidizing analyte and a stoichiometric amount of iodine (Istoichiometric amount of iodine (I22 or or II33

-- ) is produced. This iodine is ) is produced. This iodine is titrated with a standard solution of titrated with a standard solution of thiosulfate. This reaction is shown thiosulfate. This reaction is shown on the following slide.on the following slide.



The reaction with iodine needs to The reaction with iodine needs to occur in a solution whose pH < 9 to occur in a solution whose pH < 9 to prevent side reactions which produce prevent side reactions which produce iodates (IOiodates (IO33

--). Generally acetic acid ). Generally acetic acid is added to the analyte mixture is added to the analyte mixture before titration to assure the proper before titration to assure the proper pH. In some cases, appropriate pH pH. In some cases, appropriate pH buffers may also be added.buffers may also be added.


The titrant solution of thiosulfate The titrant solution of thiosulfate cannot be prepared directly. It is cannot be prepared directly. It is made to an approximate concentration made to an approximate concentration and then standardized with a primary and then standardized with a primary standard oxidizing agent by standard oxidizing agent by iodometry. The thiosulfate solution is iodometry. The thiosulfate solution is unstable if the pH < 5, undergoing the unstable if the pH < 5, undergoing the following disproportionation reaction.following disproportionation reaction.

SS22OO33-2-2 + 2 H + 2 H++ < == > H < == > H22SOSO33 + S + S


This disproportionation reaction is This disproportionation reaction is prevented by using freshly boiled prevented by using freshly boiled deionized water as the solvent and deionized water as the solvent and adding a small amount of NaOH. adding a small amount of NaOH. Although the thiosulfate needs to be Although the thiosulfate needs to be stored in a basic solution, as stored in a basic solution, as mentioned earlier, its reaction as a mentioned earlier, its reaction as a reductant titrant needs to occur in an reductant titrant needs to occur in an acid solution.acid solution.

Chapter 16 – Iodine MethodsChapter 16 – Iodine Methods

The indicator for both iodimetric and The indicator for both iodimetric and iodometric titrations is a starch iodometric titrations is a starch solution; in the presence of iodine, it solution; in the presence of iodine, it shows a blue or purple color. The shows a blue or purple color. The starch indicator should not be added starch indicator should not be added when Iwhen I22 is in large excess because is in large excess because the desorption of Ithe desorption of I22 from the starch from the starch molecule is not reversible.molecule is not reversible.

Starch IndicatorStarch Indicator

The The repeating repeating amyloseamylose unit in unit in the starch the starch molecule. molecule. Starch is a Starch is a polymer of polymer of amylose.amylose.

Starch IndicatorStarch Indicator

The starch-iodine The starch-iodine complex where complex where the sugar chain the sugar chain forms a helix forms a helix about about II66 unitsunits..

Iodimetric titrations: Titrations with Iodimetric titrations: Titrations with standard iodine (actually Istandard iodine (actually I33

))

Species analyzedSpecies analyzed Oxidation reactionOxidation reactionSOSO22 SOSO22 + H + H22O < == > HO < == > H22SOSO33

HH22SOSO33 + H + H22O < == > SOO < == > SO224-4- + 4H + 4H++ + +

2e2e--

HH22SS HH22S < == > S(s) + 2HS < == > S(s) + 2H++ + 2e- + 2e-

ZnZn2+2+, Cd, Cd2+2+, , MM2+2+ + H + H22S S MS(s) + 2H MS(s) + 2H++

HgHg2+2+, Pb, Pb2+2+ MS(s) < == > MMS(s) < == > M2+2+ + S + 2e + S + 2e--

Iodimetric titrations: Titrations with standard iodine Iodimetric titrations: Titrations with standard iodine (actually I(actually I33

))

Species analyzedSpecies analyzed Oxidation reactionOxidation reaction

Cysteine, glutathione,Cysteine, glutathione, 2RSH < == > RSSR + 2H2RSH < == > RSSR + 2H++ + + 2e2e--

mercaptoethanolmercaptoethanol

AldehydesAldehydes HH22CO + 3OHCO + 3OH-- < == > HCO < == > HCO22-- + 2H + 2H22O + O +

2e 2e--

Glucose (and other reducing sugar)Glucose (and other reducing sugar) OO

RCH + 3OHRCH + 3OH-- < == > HCO < == > HCO22-- + 2H + 2H22O + 2eO + 2e--

Ascorbic acid Ascorbic acid See next slideSee next slide(or vitamin C)(or vitamin C)

Ascorbic acid (or vitamin C)Ascorbic acid (or vitamin C)

Iodometric titrations: Titrations of iodine (actually Iodometric titrations: Titrations of iodine (actually II33

) produced by the analyte) produced by the analyte

Species analyzedSpecies analyzed ReactionReaction

HOClHOCl HOCl + HHOCl + H++ + 3I + 3I- - < == > Cl < == > Cl -- + I + I33-- + H + H22OO

BrBr22 BrBr22 + 3 I + 3 I-- < == > 2 Br < == > 2 Br -- + I + I33--

IOIO33-- 2 IO2 IO33

-- + 16 I + 16 I -- + 12 H + 12 H++ < == > 6 I < == > 6 I33- - + 6 H + 6 H22OO

IOIO44-- 2 IO2 IO44

-- + 22 I + 22 I -- + 16 H + 16 H++ < == > 8 I < == > 8 I33-- + 8 H + 8 H22OO

OO22 OO22 + 4 Mn(OH) + 4 Mn(OH)22 + 2 H + 2 H22O < == > 4 Mn(OH)O < == > 4 Mn(OH)33

2 Mn(OH)2 Mn(OH)33 + 6 H + 6 H++ + 6 I + 6 I -- < == > 2 Mn < == > 2 Mn2+2+ + 2 I + 2 I33--

+ 6 H+ 6 H22OO

HH22OO22 HH22OO22 + 3 I + 3 I -- + 2H + 2H++ < == > I < == > I33-- + 2 H + 2 H22OO

Iodometric titrations: Titrations of iodine (actually Iodometric titrations: Titrations of iodine (actually II33

) produced by the analyte) produced by the analyte

Species analyzedSpecies analyzed ReactionReaction

OO33 OO33 + 3 I + 3 I -- + 2 H + 2 H++ < == > O < == > O22 + I + I33-- + H + H22OO

NONO22-- 2 HNO2 HNO22 + 2 H + 2 H ++ + 3 I + 3 I -- < == > 2 NO + I < == > 2 NO + I33

-- + 2 + 2 HH22OO

SS22OO882-2- S S22OO88

2-2- + 3 I + 3 I -- < == > 2 SO < == > 2 SO442-2- + I + I33

--

CuCu2+2+ 2 Cu2 Cu2+2+ + 5 I + 5 I - - < == > 2 CuI( < == > 2 CuI(ss) + I) + I33--

MnOMnO44-- 2 MnO 2 MnO44

-- + 16 H + 16 H++ + 15 I + 15 I -- < == > 2 Mn < == > 2 Mn2+2+ + 5 I + 5 I33--

+ 8 H+ 8 H22OO

MnOMnO22 MnOMnO22((ss) + 4 H ) + 4 H ++ + 3 I + 3 I -- < == > Mn < == > Mn2+2+ + I + I33-- + 2 + 2

HH22O O

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Oxidation-Reduction Titration Dr. A.K.M. Shafiqul Islam August 24, 2007