Red-Ox titrimetry
Pabitra Kumar Mani , Assoc. Prof., Ph.D. ACSS, BCKV
Linus Pauling (1901-1994)

His work in chemical bonding, X-ray
crystallography, and related areas had a
tremendous impact ...
Leo the Lion!
• LEO the lion says GER
– Loss of electrons is oxidation, gain of electrons is
reduction
Pre adjustment of analyte oxidation state
It is necessary to adjust the oxidation state of the analyte to one that can be ...
Jones reductor : 2Zn (s) + Hg2+ → Zn2+ + Zn(Hg) (s)
The Walden reductor is a reduction column filled with metallic
silver which can be used to reduce a metal ion in aqueous s...
Reagents used in redox titration
Reducing agents

Ferrous salts :
ammonium iron(II) sulfate hexahydrate (Mohr’s salt)
FeSO...
Reagents used in redox titration
Oxidizing agents
Potassium permanganate KMnO4 :

Permanganometry

Ceric sulfate / Ceric a...
Calculations



Equivalent weight = ( formula weight) / ( e– change)
Equivalents = g / eq. wt.

meq = mg / eq. Wt.

Norm...
(A) The reduction of potassium permanganate by iron(II)
sulphate in the presence of dilute H2SO4.
The first partial eqn (r...
Permanganate titration


Oxidation with permanganate : Reduction of permanaganate
KMnO4 : Powerful oxidant that the most...
Oxidation with potassium dichromate
Cr2O72– + 14H+ + 6e = 2Cr3+ + 7H2O

Eo = 1.36 V

K2Cr2O7 is a primary standard.
Indica...
Iodimetry and iodometry



Iodimetry : a reducing analyte is titrated directly with iodine.
Iodometry : an oxidizing an...
Redox potential
M ⇋ Mn+ + ne-

Potentail or work done for the
transformation M ⇋ Mn+ + ne-

Metal atom

Red

Reductant

⇋ ...
Pt(foil)

|Pt(black)| H | H ||
coated
+

2

| Ox
M
Red
M

n+

|Pt (foil)

SHE
Liquid
(a=1) (a=1)
Or
Junction
Desired
Calom...
Sign convention
1. All half-cell reactions are written as Oxidation
Red ⇋ Ox + neReductant

oxidant

The potential may be ...
Mn+2 +4H2O 
Red 2

MnO4- + 8H+ +5eOxid 2

E0= - 1.52 Volt (at 298ºK)
2Cr+3 + 7H2O 
Red

Cr2O7-2 + 14H+ +6eOxid

0

E = -...
Familiar Redox Reactions…
Photosynthesis

Respiration

http://www.astrazeneca.ch/

Batteries

Chemistry 215 Copyright D
Sh...
Derivation of the EMF equation
For the general rn.,

Red

⇋ Ox + ne-

The chemical potential are given by

µRed = µ0Red+ R...
Effect of pH on red-ox potential
Hydrogen electrode:

½H2  H+ + e-

a +
0.0591
E =E log H 1
1
(pH2 ) 2
0

When ,

E = E 0...
Mn+2 +4H2O 
Red 2

MnO4- + 8H+ +5eOxid 2

E0= - 1.52 Volt (at 298ºK)

a
x( a H + )
0.0591
MnO 4E =E log
5
a Mn +2

oxidat...
Halide, X-

→ ½ X2
MnO4-

At different pH, different halides are oxidised
At pH 5-6, Only I- is oxidised to I2
at pH 3 (us...
Fe+2
E0

Fe+ 2


Fe+ 3

No hydrogen ion on the half cell
reaction
Is independent of pH

Fe+3 +e-

But as pH increases , a...
III

V

HAsO2 + 2H2O  H3AsO4 +2H+ +2eArsenious acid

E0 = -0.56 Volt

Arsenic acid

Arsenate or arsenic acid acts as oxid...
Primary standard: Arsenic (III) oxide, As2O3
Equilibrium constant of Red-ox reaction
Couple I

[ Red I ] a [ Ox II ] b
K=
[ Ox I ] a [ Red II ] b

aOxI + bRed II ⇋ aRe...
Value of E at the equivalence point during titration of redox
system by the other is obtd as:

E e.p.
Consider the rn.

bE...
Free energy, Cell EMF, and Equilibrium Constants
The change in Gibbs free energy, ΔG is a measure of the spontaneity of a ...
Redox titrimetry, P K MANI
Redox titrimetry, P K MANI
Redox titrimetry, P K MANI
Redox titrimetry, P K MANI
Redox titrimetry, P K MANI
Redox titrimetry, P K MANI
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Redox titrimetry, P K MANI

  1. 1. Red-Ox titrimetry Pabitra Kumar Mani , Assoc. Prof., Ph.D. ACSS, BCKV
  2. 2. Linus Pauling (1901-1994) His work in chemical bonding, X-ray crystallography, and related areas had a tremendous impact on chemistry, physics, and biology. He is the only person to receive two unshared Nobel prizes: for chemistry(1954) and for his efforts to ban nuclear weapons, the peace prize (1962). This photo of Pauling tossing an orange into the air is symbolic of his work and importance of being able to determine concentrations of ascorbic acid at all levels in fruits and commercial vitamin preparations. Redox titrations with iodine are widely used to determine ascorbic acid.
  3. 3. Leo the Lion! • LEO the lion says GER – Loss of electrons is oxidation, gain of electrons is reduction
  4. 4. Pre adjustment of analyte oxidation state It is necessary to adjust the oxidation state of the analyte to one that can be titrated with an auxiliary oxidizing or reducing agent. Ex. Pre adjustment by auxiliary reagent Fe(II), Fe(III) Fe(II) – 4 Titration Ce4+ Preoxidation : Peroxydisulfate (NH4)2S2O8 Sodium bismuthate ( NaBiO3) Hydrogen peroxide (H2O2) Prereduction : Stannous chloride ( SnCl2) Chromous chloride Jones reductor (zinc coated with zinc amalgam) Walden reductor ( solid Ag and 1M HCl)
  5. 5. Jones reductor : 2Zn (s) + Hg2+ → Zn2+ + Zn(Hg) (s)
  6. 6. The Walden reductor is a reduction column filled with metallic silver which can be used to reduce a metal ion in aqueous solution to a lower oxidation state. It can be used e.g. to reduce UO22+ in U4+. [1] A wire of copper is contacted with a solution of silver nitrate. Dentritic crystals of silver immediately forms on the copper wire according to the following redox reaction: Cu + 2 Ag+ —> Cu2+ + 2 Ag The silver crystals are then removed from the copper wire, washed with pure water to remove the copper nitrate and the excess of silver nitrate and packed in a small glass column A Jones reductor is a device which can be used to reduce a metal ion in aqueous solution to a very low oxidation state. The active component is a zinc/mercury amalgam. It can be used to prepare solutions of ions, such as chromium(II), Cr 2+, and uranium(III), U3+, which are immediately oxidized on contact with air.[
  7. 7. Reagents used in redox titration Reducing agents Ferrous salts : ammonium iron(II) sulfate hexahydrate (Mohr’s salt) FeSO4(NH4)2SO4· 6H2O iron(II) ethylene diamine sulfate (Oesper’s salt) FeC2H4(NH3)2(SO4)2· 4H2O Sodium thiosulfate pentahydrate Arsenic trioxide: arsenious oxide Na2S2O3·5H2O As2O3 Sodium oxalate and oxalic acid dihydarte Na2(COO)2 , (COOH)2·2H2O Titanium trichloride TiCl3
  8. 8. Reagents used in redox titration Oxidizing agents Potassium permanganate KMnO4 : Permanganometry Ceric sulfate / Ceric ammonium sulfate Ce(SO4)2·2(NH4)2SO4· 4H2O : Cerimetry Potassium dichromate K2Cr2O7 : Iodine I2 Potassium iodate Dichrometry : Iodimetry, Iodometry KIO3 Potassium bromate KBrO3 Sodium nitrite NaNO2 : : Iodatimetry : Bromatimetry
  9. 9. Calculations  Equivalent weight = ( formula weight) / ( e– change) Equivalents = g / eq. wt. meq = mg / eq. Wt. Normality (N) = eq / L = meq / ml Reaction Fe2+ → Fe3+ + e eq. wt of reactant FW Fe ÷ 1 KMnO4 + 5e → Mn2+ FW KMnO 4 ÷ 5 Na2S2O35H2O → ½ S4O6– + e FW Na2S2O35H2O ÷ 1 Cr2O72 – + 6e → 2 Cr3+ FW Cr2O72 – ÷ 6
  10. 10. (A) The reduction of potassium permanganate by iron(II) sulphate in the presence of dilute H2SO4. The first partial eqn (reduction) is: MnO4– → Mn2 + To balance atomatically, 8H+ are required: MnO4– + 8H+ → Mn2 + + 4H2 O And to balance it electrically 5e is needed on the LHS: MnO4– + 8H+ + 5e ⇋ Mn2 + + 4H2 O The second partial eqn (oxidation) is: Fe2+ → Fe3 + And to balance it electrically 1e must be added to the RHS or subtracted from the LHS : Fe2+ ⇋ Fe3 + + e Now the gain and loss of electron must be equal. One permanganate ion utilises 5 electrons, and one iron(II) ion liberates 1 electron; hence two partial equations must apply in the ratio of 1:5 MnO4– + 8H+ + 5e ⇋ Mn2 + + 4H2 O 5 (Fe2+ ⇋ Fe3 + + e) MnO4– + 8H+ + 5 Fe2+ ⇋ Mn2 + + 5Fe3+ 4H2 O
  11. 11. Permanganate titration  Oxidation with permanganate : Reduction of permanaganate KMnO4 : Powerful oxidant that the most widely used. In strongly acidic solutions (1M H2SO4 or HCl, pH ≤ 1) MnO4– + 8H+ + 5e = Mn2 + + 4H2 O violet color Eo = 1.51 V colorless manganous KMnO4 is a self-indicator. In feebly acidic, neutral, or alkaline solutions MnO4– + 4H+ + 3e = MnO2 (s) + 2H2 O Eo = 1.695 V brown manganese dioxide solid In very strongly alkaline solution (2M NaOH) MnO4– + e = MnO42 – green manganate Eo = 0.558 V
  12. 12. Oxidation with potassium dichromate Cr2O72– + 14H+ + 6e = 2Cr3+ + 7H2O Eo = 1.36 V K2Cr2O7 is a primary standard. Indicator : diphenylamine sulphonic acid The interaction of PDC and KI in the presence of dil.H2SO4 Cr2O72– → Cr3+ Cr2O72– + 14H+ → 2Cr3+ + 7H2O To balance electrically, add 6e to the LHS Cr2O72– + 14H+ + 6e = 2Cr3+ + 7H2O I - → I2 2I- →I2 2I-⇋ I 2 + 2e One dichromate ion uses 6e, and 2 iodine ions liberate 2e; hence the two partial eqns apply in the ratio of 1:3 Cr2O72– + 14H+ + 6e = 2Cr3+ + 7H2O 3(2I-⇋ I 2 + 2e) Cr2O72– + 14H+ + 6I- = 2Cr3+ + 7H2O +3I 2
  13. 13. Iodimetry and iodometry  Iodimetry : a reducing analyte is titrated directly with iodine. Iodometry : an oxidizing analyte is added to excess iodide to produce iodine, which is then titrated with standard thiosulfate solution. Its solubility is enhanced by complexation with iodide. I2 + I– = I3– K = 7 ×102
  14. 14. Redox potential M ⇋ Mn+ + ne- Potentail or work done for the transformation M ⇋ Mn+ + ne- Metal atom Red Reductant ⇋ Ox + neoxidant RT E = E0 + lna M n + nF 0.0591 E = E0 + logcM n + n or for the equilibrium is known as Nernst Potential M |M n+ If, a M n + = 1, the potential of the system or the change of free energy for the transformation is called Standard electrode potential, (E0) of the metal. SHE (Std. Hydrogen electrode): H2 gas at 1 atm pressure (pH2= 1 atm) bubbling through an acid solution having aH+ =1 (1N strong acid) in contact with a Pt foil. Pt H-electrode ½H2 |H + Zn | Zn +2 Zn-electrode
  15. 15. Pt(foil) |Pt(black)| H | H || coated + 2 | Ox M Red M n+ |Pt (foil) SHE Liquid (a=1) (a=1) Or Junction Desired Calomel half cell electrode Complete electrochemical cell Std.potential of the electrode is 0 (zero) volt at 298º K. The potentials of all other electrodes or half cells are referred to the value of SHE as 0 volt. The redox potential is a more general term than electrode potential since like the oxidation reduction rns in soln, the rn is taking place at the electrodes are also the oxidation– reduction rn. The red-ox potential values are determined by measuring the EMF of the cell constituted by combining the desired half cell or std electrode with the help of a potentiometer. EMF= E volt= E01 –E0H = E01 volt (E0H =0) i.e., EMF = Redox potential (std) of the desired half cell
  16. 16. Sign convention 1. All half-cell reactions are written as Oxidation Red ⇋ Ox + neReductant oxidant The potential may be called simple “Oxidation potential” 2. Any reductant which has a stronger reducing power than H2 i.e., which can liberate H2 from H+ (from acid/H2O) will have a positive value of “E” 3. Any oxidant having stronger oxidizing power than H+ will have a negative value of “ E” Fe  Fe+2 +2e- , E0= + 0.44 Volt (at 298ºK Hence, Fe must liberate H2 from acid Fe+2 Red1  Fe+3 +e- , E0= - 0.77 Volt (at 298ºK Oxid1 Fe+2 can’t be oxidised to Fe+3 by boiling with acid only without using any other stronger oxidant than Fe+3,(HNO3,bromine water)
  17. 17. Mn+2 +4H2O  Red 2 MnO4- + 8H+ +5eOxid 2 E0= - 1.52 Volt (at 298ºK) 2Cr+3 + 7H2O  Red Cr2O7-2 + 14H+ +6eOxid 0 E = - 1.36 Volt (at 298ºK) I-  oxidation potential ½I2 + e- , oxidation potential E0= - 0.54 Volt (at 298ºK) More negative the value of oxidation potential of any redox couple the stronger is the oxidising power than the oxidant of that couple. Thus the oxidant form of any couple having higher negative oxidation potential value will react with the reductant (Red) form of another couple having lower negative or higher +ve value of oxidation potential. Ox II + Red I → Red II + Ox I Ox II > OxI Red I > Red II
  18. 18. Familiar Redox Reactions… Photosynthesis Respiration http://www.astrazeneca.ch/ Batteries Chemistry 215 Copyright D Sharma Electrolysis 18 http://www.sparknotes.com/
  19. 19. Derivation of the EMF equation For the general rn., Red ⇋ Ox + ne- The chemical potential are given by µRed = µ0Red+ RTlnaRed µOx = µ0Ox+ RTlnaOx µRed - µOx = µ0Red - µ0Ox + RTln(aRed /aOx) -ΔG = (µ0Red - µ0Ox) + RTln(aRed /aOx) but , considering the electrical work associated with transfer of n no. Of electrons, ΔG= -nFE  μ0  − μ0    Red Ox  - RT ln aOx E= nF nF a Red RT a Ox ∴E = E ln nF a Red 0 When aOx =aRed = 1, then μ0  − μ0   Ox  0  Red E = nF a 0.0591 so, E = E log Ox n a Red 0 T=2980K F=96000 Coulomb
  20. 20. Effect of pH on red-ox potential Hydrogen electrode: ½H2  H+ + e- a + 0.0591 E =E log H 1 1 (pH2 ) 2 0 When , E = E 0 - 0.0591 log a H + E = E 0 + 0.0591 pH 0 Since E = O (zero volt) Redox or electrode potential, When When At aH+ = 1, aH+ = 0.1, pH= 7, pH2=1 atm E = 0.0591 pH pH= 0, pH= 1, E=0 E=0.0591 volt E = 0.4137 volt For other redeox couples, the effect of pH on the redox potential value is observed if the redox half reaction contains H+ or OH-
  21. 21. Mn+2 +4H2O  Red 2 MnO4- + 8H+ +5eOxid 2 E0= - 1.52 Volt (at 298ºK) a x( a H + ) 0.0591 MnO 4E =E log 5 a Mn +2 oxidation potential 8 0 a 0.0591 0.0591 MnO 4=E log + x8pH 5 a Mn +2 5 0 When aMnO - = 4 E = -1.52 -0 pH factors on the red-ox value aMn+2 = aH+ = 0.1 Molar + 0.096 = -1.424 Volt i.e., MnO4- becomes weaker oxidant as pH increases At pH 6, E = - 0.92 Volt
  22. 22. Halide, X- → ½ X2 MnO4- At different pH, different halides are oxidised At pH 5-6, Only I- is oxidised to I2 at pH 3 (using CH3COOH) Br- is also Oxidised Cl- is oxidised pH<1.5 E0 Cl − = - 1.36 Volt Cl2
  23. 23. Fe+2 E0 Fe+ 2  Fe+ 3 No hydrogen ion on the half cell reaction Is independent of pH Fe+3 +e- But as pH increases , a new equilibrium is set up Fe(OH)2 + OH-  Fe(OH)3 + eEffect of pH will be observed a Fe(OH)3 0.0591 E =E log 1 a Fe(OH) 2 x a 0 OH − Value of red-ox potential will depend on pH of the medium. As the pH of the medium increases the reducing property of Fe+2 increases i.e., red-ox potential value of Fe+2  Fe+3 becomes less negative at higher pH
  24. 24. III V HAsO2 + 2H2O  H3AsO4 +2H+ +2eArsenious acid E0 = -0.56 Volt Arsenic acid Arsenate or arsenic acid acts as oxidising agent in acidic medium and it libertaes I2 from iodide. H3AsO4 +2H+ +2I- → H3AsO3 + H2O +I2, pH Whereas at higher pH. e.g. in bicarbonate buffer medium(pH=8.2) aresenite or As2O3 is oxidised by I2 H3AsO3 + H2O + I2 → H3AsO4 + 2HI which means that the red-ox potential of the couple becomes E0 + 3 As less negative at higher = 2H AsO → HAsO As O + 3H O pH +5 2 3 2 3 3 2 As H2O and falls below -0.54 Volt which is the std. potential for E0 I − = − 0.54V I-/I2 I2
  25. 25. Primary standard: Arsenic (III) oxide, As2O3
  26. 26. Equilibrium constant of Red-ox reaction Couple I [ Red I ] a [ Ox II ] b K= [ Ox I ] a [ Red II ] b aOxI + bRed II ⇋ aRed I + b OxII +3 MnO4- EI = EI 0 Fe+2 Fe Mn+2 Couple II [ Ox I ]a 0.0591 log n [ Red I ]a at equilibrium of the reaction, E1=EII E II = E II 0 [ Ox II ]b 0.0591 log n [ Red II ] b [ Ox I ] a 0 0.0591 EI log n [ RedI ] a [ Ox II ] b 0 0.0591 = E II log n [ RedII ] b [ Ox II ] b [ RedI ] a = log K n(E II 0 - E 0 I ) = log b a 0.591 [ Red II ] [ Ox I ] Thus, K of any redox rn can be evaluated [ Red I ] [ OxII] = a +b K = from the std. Potential data of the involved [ Ox I ] [ RedII ] redox couples at equilibrium,
  27. 27. Value of E at the equivalence point during titration of redox system by the other is obtd as: E e.p. Consider the rn. bE 0 I + aE 0 II = a+b MnO4– + 8H+ + 5 Fe2+ ⇋ Mn2 + + 5Fe3+ 4H2 O 5 ( - 0.771 + 1.52 ) logK = = 63.5 0.0591 ∴ K = 3 x10 63 Hence, the titration of Fe+2 by MnO4 can be carried out quantitatively with a very high speed of reaction.
  28. 28. Free energy, Cell EMF, and Equilibrium Constants The change in Gibbs free energy, ΔG is a measure of the spontaneity of a process that occurs at constant temperature and pressure. Since the emf of a redox reaction indicates whether the reaction is spontaneous, we would expect some relationship to exist between emf and free energy change. This relationship is given by the following equation. ΔG = -nFEcell In this equation n is the number of moles of electrons transferred in the reaction and F is Faraday's constant, named after Michael Faraday. Faraday's constant is the quantity of electrical charge on one mole of electrons. This quantity of charge is called a faraday, F: 1 F = 96,500 C/mol e- or 96,500 J/V-mol eBoth n and F are positive quantities. Thus, a positive value of E, the cell potential, leads to a negative value of ΔG. So keep in mind that a positive value of E and a negative value of ΔG both indicate that a reaction is spontaneous. When both the reactants and the products are in their standard states, the equation above can be modified to give: ΔGo = -nFEocell The measurement of cell potentials gives us another way to obtain equilibrium constants. We can take the equation above and an equation relating free energy and the equilibrium constant, that we discussed in thermodynamics and combine them as shown below. ΔGo = -nFEocell ΔGo = -RT ln K nFEocell = RT ln K= RT log 2.303 RT Eocell = --- ln K = -------- log K
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