redox titrations [compatibility mode]

Redox titrations

Dr.Jehad M Diab, faculty of pharmacy Damascus University

Pharma.analytical chemistry II

REDOX titrationsA volumetric method of analysis whichrelies on oxidation or reduction of theanalyte using redox indicators orpotentiometry.This unit covers:changes in solution potential during a

titrationbasic calculationsmethods of sample preparation andcommon titrantsExample applications

Dr.Jehad diab

Dr.Jehad diab

Determination of equivalence point potential:Fe2+ Ce4+ = Fe3+ Ce3+

E0Fe3+= 0.771 v , E0

Ce4+=1.70 v

Eeq = E0Fe3+ - 0.0592/1 log [Fe2+] / [Fe3+] (1)

Eeq = E0Ce4+ - 0.0592/1 log [Ce3+] / [Ce4+] (2)

1+2:2Eeq = E0

Ce4+ + E0Fe3+ - 0.0592/1 log [Fe2+][Ce3+] / [Fe3+] [Ce4+]

At the equivalence point:[Fe3+] =[Ce3+ ][Fe2+]=[Ce4+ ]Rearrangement:2Eeq = E0

Ce4+ + E0Fe3+ - 0.0592/1 log [Ce4+][Ce3+] / [Ce3+] [Ce4+]

Eeq = (E0Ce4+ + E0

Fe3+) /2= 1.70+0.771/2= 1.24 volt

Dr.Jehad diab

2Eeq = E0Ce4+ + E0

Fe3+ - 0.0592/1 log [Fe2/ [Fe3+]-0.0592/1 log [Ce3+] / [Ce4+] 2Eeq = E0

Ce4+ + E0Fe3+ - 0.0592/1(log [Fe2]/ [Fe3+]+ log

[Ce3+] / [Ce4+]2Eeq = E0

Ce4+ + E0Fe3+ - 0.0592/1( (log [Fe2]-log [Fe3+]+

log[Ce3+]-log[Ce4+]2Eeq = E0

Ce4+ + E0Fe3+ - 0.0592/1(log [Fe2]+log[Ce3+]-log

[Fe3+]-log[Ce4+]2Eeq = E0

Ce4+ + E0Fe3+ - 0.0592/1 log ([Fe2+][Ce3+] / [Fe3+] [Ce4+])

draft

Dr.Jehad diab

Dr.Jehad diab

E eq =(n1e0ox+n2e0

red) / n1+n2

e eq= 0.77+0.059/1 × log [Fe3+] / [Fe2+] (1)e eq=1.51+0.059/5 × log [Mno4

-][H+]8/[Mn2+] (2) × 5

1+2:

6eeq= 0.77+5 × 1.51+0.059 × log [Fe3+][MnO4- ][H+]8/[Fe2+][Mn2+]

[Fe2+]=5[MnO4-] , [Fe3+]=5[Mn2+] ,

[Fe3+]/[Fe2+]= [Mn2+] / [MnO4-]

6eeq = 0.77+5×1.51+.059 log [Mn2+][MnO4-][H+]8/[MnO4

-][Mn2+]

Determine of Eeq for the following reaction:

Mno4- +5 Fe2+ 8H+ = Mn2+ + 5Fe3++ 4H2O

A more complex example Dr.Jehad diab

6eeq = 0.77+5×1.51+.059 log [Mn2+][MnO4-][H+]8/[MnO4

-][Mn2+]

6eeq=0.77+5×1.51+0.059 log[H+]8

eeq=(0.77+5×1.51)/6 + 0.059/6 log[H+]8

eeq=(0.77+5×1.51)/6 - 0.079 pH

If [H+]=1M (pH= 0):

eeq=(0.77+5×1.51)/6 + 0.059/6 log[1]8 =1.39 v

So, the potential at equivalence point is dependent on [H+]. If [H+]=1M we can calculate from the following equation:

Dr.J.Diabeeq=(n1e0ox + n2e0

Red ) / n1+n2

Volume of reagent,ml

E,vo

lt

E,vo

ltDr.Jehad diab

Complete titration curve

(or volume of titrant) Dr.Jehad diab

Ce4+ + Fe2+ Ce3++ Fe3+ 100 ml 0.1 M Fe2+ with 0.1 M Ce4+ solution

Dr.Jehad diab

Dr.Jehad diab

Dr.Jehad diab

Ce4+ + Fe2+ Ce3++ Fe3+

Addition of 50 ml of Ce4+:

vEFe 771.0

501001.05050100

1.0501.0100

log1059.0771.0

Addition of 20 ml of Ce4+:

vEFe 735.0

201001.02020100

1.0201.0100

log1059.0771.0

Ce4+ (0.1 M) + Fe2+ (100 ml 0.1 M) Ce3++ Fe3+

Dr.Jehad diab

Dr.Jehad diab

Dr.Jehad diab

E eq =n1e0ox+n2e0

red / n1+n2

Fe2+ Ce4+ = Fe3+ Ce3+ , e0Fe3+/Fe2+= 0.771 v , e0

Ce4+/Ce3+=1.70 v

Dr.Jehad diab

Dr.Jehad diab

Ce4+ + Fe2+ Ce3++ Fe3+

Dr.Jehad diab

Dr.Jehad diab

10ml x0.1M=210ml x [Ce4+]200 x0.05= 210 x[Ce3+]

Ce4+ + Fe2+ Ce3++ Fe3+

Dr.Jehad diab

E = e0 – 0.059 log [Ce3+]/ [Ce4+]

volt

Ce4+ + Fe2+ Ce3++ Fe3+

1.70 -1.64 v

1.66 V

Or E =1.70-0.059log 100%/10%=1.64 v

Dr.Jehad diab

(200 x0.05/ 210)

(10ml x0.1M/210ml)

Homework

You are titrating 50 ml 0.1M of Co2+ solution wuth0.1 M Ce4+ titrant.E0

Co=0.85 v ,E0Ce=1.70.

What is the potential for the titration system after addition of:a. 0 ml of Ce4+

b. 25 ml of Ce4+

c. 50 ml of Ce4+

d. 75 ml of Ce4+

Dr.Jehad diab

Ce4+ + Co2+ Ce3++ Co3+

Equivalent point volume: CCe4+ VCe4+ = CCo2+ V Co2+

0.1 x VCe4+ = 50 x0.1 ==> VCe4+ = 50 ml a) Addition of 0 ml of Ce4+ ,absence of Co3+ ,E can

not be calculatedb) Addition of 25 ml of Ce4+ before the equivalent

point; excess of Co2+

E= 0.85+0.059/1xlog([25x0.1)/75/(50 x0.1-25x0.1)/75=0.85vc)Addition of 50 ml of Ce4+;at the equivalent point:E= (e0

Co+ e0Ce)/2= (0.85 +1.70)/2= 1.275 v

Dr.Jehad diab

c)Addition of 75 ml of Ce4+;after the equivalent point excess of Ce4+

E= e0Ce+0.059/1xlog[Ce4+]/[Ce3+]

E =1.70+0.059/1xlog(75x0.1-50x0.1)/125 /(50x0.1)/125= 1.682 v

Dr.Jehad diab

Self indicators: KMnO4 (purple) → Mn2+ (colorless) I2 (yellow) → 2I- (colorless)Ce4+(yellow) → Ce3+(colorless)

Dr.Jehad diab

(True)

Redox indicatorsSpecific indicators:Starch:Starch +I2 <--> blue complexIt is an easy to detect and rapid indicator. This explains why iodine is a common titrant even though it is a weak oxidant.KSCN:Fe3++ SCN- ( Indicator) → FeSCN2+(red complex)Determination of Fe3+ with Ti3+ in presence of SCN-

Fe3+ with Ti3+ --> Fe2+ with Ti4+

when [Fe3+] decreases then color of red complex disappears which indicates the end point.

Dr.Jehad diab

____

1

E=E0ind - 0.059/n (color of reducing form)

E=E0ind + 0.059/n (color of oxidizing form) Dr.Jehad diab

Dr.Jehad diab

general

(Ferroin)

1.150v

Dr.Jehad diab

Dr.Jehad diab

Ox

Potential range for color change is: e0-0.059/n - e0+0.59/n:From 0.80-0.59/1 to 0.80 +0.59/1=(0.741→ 0.859 v)

Dr.Jehad diab

Dr.Jehad diab

√

√√

√

Potential IndicatorEnd point is determined by measuring the potential ofindicator electrode against reference electrode and plottingthe potential against the volume of titrant.

Dr.Jehad diab

(Pt for E( mv) and glass electrode for pH)

Potentiometric titrationIt is possible to monitor the course of a titration using potentiometric measurements. The Pt electrode, for example, is appropriate for monitoring an redox titration and determining an end point in lieuof an indicator. The procedure has been called a potentiometrictitration. The end point occurs when the easuredpotential undergoes a sharp change—when all the oxd. or red. in the titration vessel is reacted

Dr.Jehad diab

المعایرة الكمونیة

Fig. 13.10. Typical pH meter.

The potential scale is calibrated in pH units (59.16 mV/pH at 25o C).

A temperature adjustment feature changes the slope by 2.303RT/F.

©Gary Christian, Analytical Chemistry, 6th Ed. (Wiley)

Dr.Jehad diab

Dr.Jehad diab

العومل المؤكسدة والمرجعة المساعدة

Dr.Jehad diab

Dr.Jehad diab

Dr.Jehad diab

العوامل المرجعة المساعدة

Dr.Jehad diab Reducing column

Auxiliary reducing agent

nThan that of Walden

Dr.Jehad diab

Fe2++ Fe3+ (reductor ) → Fe2+ , Cr3+ +e →Cr2+

Fe2++ Fe3+ (reductor ) → Fe2+ , Cr3+ not reduced

Dr.Jehad diab

Auxiliary oxidizing agents

Sodium bismuthate-NaBiO3 بیزموتات الصودیوم

Dr.Jehad diab

Ammonium peroxydisulphate- (NH4)2S2O8

,ammonium persulfateفوق كبریتات األمونیوم

Dr.Jehad diab

Oxidizes Co2+, Fe2+, Mn2+

Dr.Jehad diab

H2O2,Na2O2

stability standardizationeo voltReduction٭indicator٭٭

productOxidizing

agent

(b)Mno4-

AS2O3 , Na2C2O4

, Fe1.51Mn2+KMnO4

(a)(1)KBrO31.44Br-KBrO3

Potassium

bromate

(a)(2)Na2C2O4, AS2O3 ,

Fe

1.44 (H2SO4)

1.70(HClO4)

Ce3+Ce4+

Some common oxidants Dr.Jehad diab

Potassium permanganate

Cerium(iv)

(a)(3)K2Cr2O71.33Cr3+K2Cr2O7

(b)starchAS2O31.60IO3-H5IO6

(a)(4)KIO31.24ICl2-KIO3

(c)starchBaS2O3.H2O , AS2O3

0.536I-I2

Dr.Jehad diab

Some common oxidants

Potassium dichromate

Periodic acid

Potassium iodate

Iodine

*(1) α- Naphthoflavone; (2) Ferroin;(3) diphenyl amine sulfonicacid;(4) disapperance of I2 from chlorofom.

**(a) stable ;(b) moderately stable ;(c) unstable

Dr.Jehad diab

E0 =1.33 v

Cr2O7-2 +14H+ +6e-2Cr3++ 7H2O

Dr.Jehad diab

Primary applications of Cr2O72-

- Determination Fe2+

- Indirect determination of oxidizing agents;A known excess of fe2+ is added to the sample which is oxidant such as MnO4

- and the excess of fe2+ is back titrated with Cr2O7

2 -

- Ethanol (C2H5OH) Reactions:6Fe2+ + Cr2O7

2- + 14H+ → 6Fe3+ + 2Cr3+ + 7H2O

3C2H5OH + 2Cr2O72- + 16H+ → 4Cr3+ + 3CH3COOH + 11H2O

Dr.Jehad diab

Moles of Fe2+ = 6 moles of Cr2O7-2

W Fe/ Mw Fe = 6 CCr2O7-2 x VCr2O7-2 (L)0.2464/56 = 6 CCr2O7-2 x 39.31/1000CCr2O7-2 = 0.01871 M

Dr.Jehad diab

of titrant

Dr.Jehad diab

Dr.Jehad diab

MnO4- at pH < 1(H2SO4 is used ,HCl can not be used

because MnO4- oxidize 2Cl- to Cl2)

MnO4- + 8H+ + 5 e- → Mn2+ + 4 H2O (E0 = 1.51 V strong acidic med)

MnO4- + 2H2O +3e- →MnO2 +4OH- (E0=0.59 v weak basic med)

MnO4- + 4H+ +3e- →MnO2 +2H2O (E0=1.69 v weak acidic to neutral )

MnO4- + e- →MnO4

-2 (Eo=0.56 v strong basic med)

Standardization of MnO4-

Dr.Jehad diab

-Application of KMnO4 in Redox Titrations

MnO4- + 8 H+ + 5 e- Mn2+ + 4 H2O

Dr.Jehad diab

H2O2 → O2 +2H+ + 2e-H2O2

Ti3+ Ti3+ +H2O→ TiO2 +2H+ + e

H3AsO3 H3AsO3 + H2O →H3AsO4+ 2H++ 2e

Mo3+ Mo3+ +4H2O →MoO42- + 8H+ +3e

Common titrantsOxidizing titrants

Cerium (IV) (Ce4+),E0=1.44 v and 1.70 v in 1M H2SO4 and 1M HClO4 respectively

- Commonly used in place of KMnO4- Works best in acidic solution- Can be used in most applications in

previous table - Used to analyze some organic compounds- Color change not distinct to be its own

indicator ,Ferroin indicator is usedYellow colorless

Dr.Jehad diab

Common titrantsOxidizing titrantsI2 (Iodimetry)I2 + 2e- →2I-I2 + I- → I3- (I- added to increase solubility of I2)I3- + 2 e- ------> 3 I- , E0 =0.536 VWeek oxidizing agentsUnstable Needed to be re-standardizedI2 + 2S2O3

2- → S4O62- + 2I-

( end point: disappearance of blue color)

Less popular than Ce4+, MnO4- ,Cr2O7

2-

used for the determination of strong reductants

starch

Dr.Jehad diab

Applications of Iodine in Redox Titrations(Iodimetry)

I3- + 2 e- → 3 I- , I2 + 2e- →2I-Iodimetric titrations carried out in weak acidic to weak basic medium because in strong basic I2 + OH- → OI-+H+

And 3OI-disproportionate to IO3- +2I- this resulted in error by

disturbing the stoichiometry of the reaction. In strong acidic medium starch decomposes.H2S + I2 → S + 2I- + 2H+

H3AsO3 +I2+ H2O →H3AsO4 +2I- +2H++ 2eSO3

2- + I2+H2O →SO42- + 2I- + 2H+

Sn2+ + I2 → Sn4+ + 2 I-AsO3

3- + I2 + H2O →AsO43- +2 I- +3H+

N2H4 + 2I2 → N2 +4H+ +4I- Dr.Jehad diab

المقیاس الیودي

Iodimetric determination of hydroquinone by back titration

I2Excess of standard I2 is added and excess is back titrated with Na2S2O3

I2 + 2S2O32- → S4O6

2- + 2I-

Dr.Jehad diab

Dr.Jehad diab

C6H8O6 + I3- C6H6O6 + 3I- + 2H+

→ +2H+ + 2e

Dr.Jehad diab

Oxidizing titrants

Potassium iodate:KIO3

IO3- + 5I- (excess) + 6H+(0.1-1M) → 3I2 + 3H2O

IO3- + 2I2 + 10Cl- + 6H+(3M) → 5ICl2- + 3H2O

IO3- + 2I- + 6Cl- + 6H+(3M) → 3ICl2- + 3H2O

In intermediate step iodate is converted to I2;I2 In presence of CHCl3 resulted in violet color. in final step iodine converted to ICl2-

and violet color disappear which match the end point.

Dr.Jehad diab

یودات البوتاسیوم

Application:

Determination of iodine(I2) and iodide(I-) in an aqueous mixture

the iodine in an aliquot of a mixture is determinedby standard sodium thiosulfate solution inpresence of starch as indicator . the concentrationof iodine plus iodide is then determined withstandard potassium iodate solution in strong HClin presence of chloroform as indicator.I2 + 2S2O3

2- → S4O62- + 2I-

IO3- + 2I2 + 10Cl- + 6H+(3M) → 5ICl2- + 3H2O

IO3- + 2I- + 6Cl- + 6H+(3M) → 3ICl2- + 3H2O

Dr.Jehad diab

BrO3- (standard soln.) + 5Br- (excess) + 6H+ →3Br2 + 3H2O (BrO3 ≡3Br2 ≡6e-)

Bromometry

Determination of phenol (back and substitute titration):

C6H5OH + 3Br2 (Excess) →C6H2Br3OH 3Br2 + 6I-→ 3I2 +6Br- , 3 I2 + 6S2O3

2- → 3S4O62- + 6I-

3Br2 ≡ 3I2 ≡ 6S2O32-

C6H5OH ≡3Br2 ≡6e- Dr.Jehad diab

(Iodometry)

Dr.Jehad diab

O2

I2

المقیاس الیودوي

Dr.Jehad diab

Iodometry

4S4O62- is

S4O62-

Dr.Jehad diab

2Cu2+ + 4I- → 2CuI + I22Ce4+ + 2I- →2 Ce3+ +I22MnO4

- + 10I- + 2H+ →5 I2+ 2 Mn2+ 8H2OH2O2 +2I- +2H+ → I2 +2H2OIO3

- + 5I- + 6H+ → 3I2 +3H2OHNO2 +2I- →I2 + 2NO+H2O Fe3+ + 2I- → Fe2++ I2

Application of Iodide in Redox Titrations that Produce I3- ,triiodide (Iodometry)

I2 + 2S2O32- → S4O6

2- + 2I-

Iodometric titrations carried out in strong acidic medium

Dr.Jehad diab

starch

Reducing titrants Na2S2O3Dr.Jehad diab

Reactions:

2S2O32- + I3- → S4O6

2- + 3I-

Dr.Jehad diab

MCC

08147.061

04164.0214121.0

Dr.Jehad diab

Dr.Jehad diab

Application :determination of strong oxidizing agents as MnO4

- ,Cr2O7 2+, Ce4+ ….

iron(II) ammonium sulfate (Mohr's salt)

NSO

I2 + SO2 + H2O → 2HI + SO3

C5H5N.I2 + C5H5N.SO2 + C5H5N+H2O →

2C5H5N.HI +C5H5N.SO3Dr.Jehad diab

Dr.Jehad diab

The titration’s end point is signaled when the solution changes from the yellow color of the products to the brown color of the Karl Fisher Reagent.

Example: A 1.0120 g mineral sample was crushed and dissolved in acid solution. This sample was passed through a Jones reactor (Fe3+ converted to Fe2+). Titration of the Fe(II) required 23.29 ml of 0.01992 M KMnO4. What is the %Fe in the sample?

%8320.120120.1

10012990.02%

12990.0215

01992.002329.0562

4

2

Fe

gwFe

wFe

MnOof molesFe of moles

Dr.Jehad diab

2+

2+

Example:A 0.1165 g of primary standard Cu was dissolved and then treated with an excess of kI. Reaction:

2Cu2+ + 4I- CU2I2(s) +I2Calculate the normality of Na2S2O3 soln. if 36.24 ml were needed titrate the librated I2I2 + 2S2O3

2- → S4O62- + 2I- ,

10125.0

03624.054.63

1165.0

N

N

2Cu2+ ≡ I2 ≡2e-

Dr.Jehad diab

The amount of ascorbic acid, C6H8O6, in orange juice wasdetermined by oxidizing the ascorbic acid todehydroascorbic acid, C6H6O6, with a known excess of I3–,and back titrating the excess I3– with Na2S2O3. A 5.00-mLsample of filtered orange juice was treated with 50.00 mLof excess 0.01023 M I3–. After the oxidation was complete,13.82 mL of 0.07203 M Na2S2O3 was needed to reach thestarch indicator end point. Report the concentration ofascorbic acid in milligrams per 100 mL.

Example:

C6H8O6 + I3- C6H6O6 + 3I- + 2H+

I3- + 2S2O32- 3I- + S4O6

2-(I3- =I2)

Dr.Jehad diab

C6H8O6 + I3- + C6H6O6 + 3I- + 2H+

I3- + 2S2O32- 3I- + S4O6

2-

Moles (I3-)total= moles (I3-)ascorbic acid +(mole I3-)back titrationMV(I3-) = WmgC6H8O6/Fw + 0.5(MV)S2O3

2-

50 ×0.01023= W/176.13+ 0.5 × 13.82 × 0.07203W= (50*0.01023-0.5 × 13.82 × 0.07203) × 176.13=2.43mgC6H8O6 in 5 ml sample or(2.43 ×100) /5 =48.60 mg/100 ml orange juice.

Or: 50 × 0.01023=5×M + 0.5 ×13.82 × 0.07203M=0.0028C(g/l)= 0.0028 × 176.13=0.493C(mg/100ml)= (0.486 × 1000)/10=48.60 mg/100 ml

Dr.Jehad diab

150 یكافئ ما المسحوق من یحل ،ثم وتسحق مضغوطة 20 توزن:تمرین والماء الكبریت حمض من مؤلف مزیج في األسكوربي حمض من مغ

0.1بمحلول المحلول ،یعایر M المصروف الرباعي،فكان السیریوم من ھو األسكوربي حمض من الواحدة المضغوطة محتوى أن علماً .مل 17

.غ 176.13 األسكوربي لحمض الجزیئي الوزن ،ومغ 500 من الفعلي المحتوى ھو وما ، األسكوربي لحمض المئویة النسبة ھي ما

. الواحدة المصغوطة في األسكوربي حمضC6H8O6 + 2Ce4+→ C6H6O6 + 2Ce3+ + 2H+

mgW

W OHC

7.1492

13.1761.017

1.01713.176

2 686

2113.176/

44

686686 Ce

OHC

Ce

OHC

MVW

molesmoles

Dr.Jehad diab

: النسبة المئویة

80.99150

1007.149%

idAscorbicac

:محتوى المضغوطة الفعلي من حمض األسكوربي

mg499100

80.99500

Dr.Jehad diab

Problem: Calculate the normality of the solution produced by dissolving 2.064 g of primary standard K2Cr2O7 in sufficient water to give 500 ml.N=2.064 g / (294/6)=0.421g/0.500L=0.0842 NProblem: calculate the molarity of the I2 solution that is 0.04N with respect to the following reaction:I2 + H2S →2I- +2H++SM=0.04/2=0.02 MProblem: calculate the molarity of the KIO3solution that is 0.04 N with respect to the following reaction:IO3

-+2I-+6H++6Cl- →3ICl2-+3H2OM=0.04/4=0.01 M Dr.Jehad diab

The End