redox titrimetry, p k mani
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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 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.
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 titratedwith an auxiliary oxidizing or reducing agent.
Ex. Pre adjustment by auxiliary reagent
Fe(II), Fe(III) Fe(II)4–
Titration
Ce4+
Chromous chloride
Jones reductor (zinc coated with zinc amalgam) Walden reductor ( solid Ag and 1M HCl)
Preoxidation : Peroxydisulfate (NH4)2S2O8
Sodium bismuthate ( NaBiO3)Hydrogen peroxide (H2O2)
Prereduction : Stannous chloride ( SnCl2)
Jones reductor : 2Zn (s) + Hg2+ Zn2+ + Zn(Hg) (s)
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 AgThe 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
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 UO2
2+ in U4+.[1]
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), Cr2+, and uranium(III), U3+, which are immediately oxidized on contact with air.[
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 Na2S2O3·5H2O
Arsenic trioxide: arsenious oxide As2O3
Sodium oxalate and oxalic acid dihydarte Na2(COO)2 , (COOH)2·2H2O
Titanium trichloride TiCl3
Potassium ferrocyanide K4Fe(CN)6 · 3H2O
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 : Dichrometry
Iodine I2 : Iodimetry, Iodometry
Potassium iodate KIO3 : Iodatimetry
Potassium bromate KBrO3 : Bromatimetry
Sodium nitrite NaNO2 :
Calcium hypochlorite Ca(ClO)2 :
Calculations
Equivalent weight = ( formula weight) / ( e– change)
Equivalents = g / eq. wt. meq = mg / eq. Wt.
Normality (N) = eq / L = meq / ml
Reaction eq. wt of reactant
Fe2+ Fe3+ + e FW Fe ÷ 1
KMnO4 + 5e Mn2+ FW KMnO4 ÷ 5
Na2S2O35H2O ½ S4O6– + e FW Na2S2O35H2O ÷ 1
Cr2O72 – + 6e 2 Cr3+ FW Cr2O7
2 – ÷ 6
MnO4– + 8H+ + 5e ⇋ Mn2 + + 4H2 O
The first partial eqn (reduction) is: MnO4– → Mn2 +
To balance atomatically, 8H+ are required:
And to balance it electrically 5e is needed on the LHS:
MnO4– + 8H+ → 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 + + eNow 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
(A) The reduction of potassium permanganate by iron(II)
sulphate in the presence of dilute H2SO4.
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 Eo = 1.51 V
violet color 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 = MnO4
2 – Eo = 0.558 V
green manganate
Oxidation with potassium dichromate
Cr2O72– + 14H+ + 6e = 2Cr3+ + 7H2O Eo = 1.36 V
K2Cr2O7 is a primary standard.
Indicator : diphenylamine sulphonic acid
Cr2O72– → Cr3+
Cr2O72– + 14H+ → 2Cr3+ + 7H2O
Cr2O72– + 14H+ + 6e = 2Cr3+ + 7H2O
To balance electrically, add 6e to the LHS
I- → I2
2I- →I2
2I-⇋ I2 + 2e
The interaction of PDC and KI in the presence of dil.H2SO4
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-⇋ I2 + 2e)
Cr2O72– + 14H+ + 6I- = 2Cr3+ + 7H2O +3I2
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
Redox potential
M ⇋ Mn+ + ne-
Metal atom
Red Ox + ne⇋ -
Reductant oxidant
nM0 anFRT
EE ln
nM0 logcn
0.0591 EE
Potentail or work done for the transformation M ⇋ Mn+ + ne-
or for the equilibrium M | Mn+
is known as Nernst Potential
If, , the potential of the system or the change of free energy for the transformation is called Standard electrode potential, (E0) of the metal.
1a nM
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.
½H2 | H+
PtH-electrode Zn | Zn+2 Zn-electrode
Pt(foil) |Pt(black)| H2 | H+ || M | Mn+ |Pt (foil)
SHEOrCalomelelectrode
LiquidJunction
Red(a=1)
Ox(a=1)
Desired half cell
coated
Complete electrochemical cellStd.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 –E0
H = E01 volt (E0
H =0)i.e., EMF = Redox potential (std) of the desired half cell
Sign convention
1. All half-cell reactions are written as OxidationRed Ox + ne⇋ -
Reductant 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 298ºKHence, Fe must liberate H2 from acid
Fe+2 Fe+3 +e- , E0= - 0.77 Volt (at 298ºK Red1 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)
Mn+2 +4H2O MnO4- + 8H+ +5e-
Red 2 Oxid 2
E0= - 1.52 Volt (at 298ºK) oxidation potential
2Cr+3 + 7H2O Cr2O7-2 + 14H+ +6e-
Red Oxid
E0= - 1.36 Volt (at 298ºK) oxidation potential
I- ½I2 + e- , 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
Chemistry 215 Copyright D Sharma 18
Familiar Redox Reactions…
Respiration
Batteries
http://www.astrazeneca.ch/
Photosynthesis
Electrolysis
http://www.sparknotes.com/
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 - µ
0Ox + RTln(aRed /aOx)
-ΔG = (µ0Red - µ
0Ox) + RTln(aRed /aOx)
but , considering the electrical work associated with transfer of n no. Of electrons,
ΔG= -nFE
RedaOxa
lnnF
RT-
nFOx
0μRed
0μE
When aOx =aRed = 1, then
nFOx
0μRed
0μE0
RedaOxa
lnnF
RT-EE 0
RedaOxa
logn
0.0591-EE, 0so
T=2980KF=96000 Coulomb
Effect of pH on red-ox potential
Hydrogen electrode: ½H2 H+ + e-
212H
0
p
Ha
log1
0.0591-EE
When , pH2
=1 atm
H0 a log 0.0591-EE
pH 0.0591EE 0 Since E0 = O (zero volt)
Redox or electrode potential, pH 0.0591E
When aH+ = 1, pH= 0, E=0
When aH+ = 0.1, pH= 1, E=0.0591 volt
At pH= 7, 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-
Mn+2 +4H2O MnO4- + 8H+ +5e-
Red 2 Oxid 2
E0= - 1.52 Volt (at 298ºK) oxidation potential
2Mn
8H-
4MnO0
a
axalog
5
0.0591-EE
pH85
0591.0
a
alog
5
0.0591-E
2Mn
-4MnO0 x
When aMnO4- = aMn+2 = aH+ = 0.1 Molar
E = -1.52 -0 + 0.096 = -1.424 Volt
i.e., MnO4- becomes weaker oxidant as pH increases
At pH 6, E = - 0.92 Volt
pH factors on the red-ox value
Halide, X- →→ ½ X2
MnO4-
At different pH, different halides are oxidisedAt pH 5-6, Only I- is oxidised to I2
at pH 3 (using CH3COOH) Br- is also Oxidised
pH<1.5 Cl- is oxidised
2
0ECl
Cl = - 1.36 Volt
Fe+2 Fe+3 +e- No hydrogen ion on the half cell reaction
32
0E
Fe
Fe Is independent of pH
But as pH increases , a new equilibrium is set upFe(OH)2 + OH- Fe(OH)3 + e-
OHa x 2Fe(OH)
Fe(OH)30
a
alog
1
0.0591-EE
Effect of pH will be observed
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
HAsO2 + 2H2O H3AsO4 +2H+ +2e- E0 = -0.56 VoltArsenious acid Arsenic acid
III V
As2O3 + 3H2O = 2H3AsO3 → HAsO2
Arsenate or arsenic acid acts as oxidising agent in acidic medium and it libertaes I2 from iodide.
Whereas at higher pH. e.g. in bicarbonate buffer medium(pH=8.2) aresenite or As2O3 is oxidised by I2
which means that the red-ox potential of the couple becomes less negative at higher pH
H2O
and falls below -0.54 Volt which is the std. potential for I-/I2
53
0E
AsAs
V54.00E2I
I
H3AsO3 + H2O + I2 → H3AsO4 + 2HI
H3AsO4 +2H+ +2I- → H3AsO3 + H2O +I2, pH
Primary standard: Arsenic (III) oxide, As2O3
Equilibrium constant of Red-ox reaction
aOxI + bRed II aRed ⇋ I + b OxII
Couple I
Couple IIMnO4
- Mn+2 Fe+2 Fe+3
bII
aI
bII
aI
Red Ox
Ox RedK
aI
aI0
IIRed
Oxlog
n0.0591
-EE bII
bII0
IIIIRed
Oxlog
n0.0591
-EE
at equilibrium of the reaction, E1=EII
bII
bII0
IIaI
aI0
IRed
Oxlog
n0.0591
-ERed
Oxlog
n0.0591
-E
KlogOxRed
RedOxlog
0.591)E-n(E
aI
bII
aI
bIII
00II
Thus, K of any redox rn can be evaluated from the std. Potential data of the involved redox couples at equilibrium,
ba KRedIIOxII
Ox Red
I
I
Value of E at the equivalence point during titration of redox system by the other is obtd as:
baaEbE
EII
0
I
0e.p.
Consider the rn. MnO4– + 8H+ + 5 Fe2+ ⇋ Mn2 + + 5Fe3+ 4H2 O
5.630.0591
) 1.52 0.771- ( 5logK
63103K xHence, the titration of Fe+2 by MnO4 can be carried out quantitatively with a very high speed of reaction.
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 e- Both 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 = -nFEo
cell
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 nFEo
cell = RT ln K= RT log 2.303 RT
Eocell
= --- ln K = -------- log K
nF nF
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