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