1. REDOX TITRATIONS
Electron Transfer Titrations
J.S.K.NAGARAJAN
ASST. PROF
OFF CAMPUS -JSS UNIVERSITY
J.S.S. COLLEGE OF PHARMACY,
OOTACAMUND – 643 001.

2. REDOX TITRATIONS
Gravimetry

3. What is a titration?
Act of adding standard solution in small quantities to the test solution till the
reaction is complete is termed titration.
What is a standard solution?
A standard solution is one whose concentration is precisely known.
What is a test solution?
A test solution is one whose concentration is to be estimated

4. REDOX TITRATIONS
Titrations Examples
Acid-base Quantification of acetic acid in vinegar
Complexometric Quantification of chloride (Cl-) in water
Precipitation Water Hardness (Calcium & Magnesium)
Redox Quantification of H2O2

5. REDOX TITRATIONS
Titration Analyte Titrant Indicator
example
Acid-base Quantification Acetic acid NaOH Phenolphthale
of acetic acid (CH3COOH) in
in avinegar
Complexometric Water Ca 2+ , Mg 2+ EDTA Eriochrome
Hardness black T
Murexide
Precipitation Quantification Chlordie AgNO3 Mohr,
of chloride Volhard,
(Cl-) in water Fajans
Redox Quantification Hydrogen peroxide KMnO4 No indicator
of hydrogen (H2O2)
peroxide
(H2O2)

6. REDOX TITRATIONS
Titrations Example Type of Titrations
Acid-base Quantification of acetic acid in
vinegar ■ Direct □ Indirect □ Back
Complexome Water Hardness (Ca 2+ & Mg2+)
tric ■ Direct □ Indirect □ Back
Precipitation Quantification of Mohr Method
Cl in Water ■ Direct □ Indirect □ Back
Fajans Method
■ Direct □ Indirect □ Back
Volhard Method
□ Direct □ Indirect ■ Back
Redox Quantification of hydrogen peroxide
(H2O2) ■ Direct □ Indirect □ Back

7. REDOX TITRATIONS
Acid/Base reactions - involves donation /acceptance of protons
S
Precipitation/ Solubility reactions : Involve donation/ acceptance of negative
charge I
REDOX TITRATION: what is being donated and accepted in a redox reaction?
M
U
L
E l e c t r o n s !
T
Consider the reaction taking place in a disposable battery: A
2Zn + 3MnO2 Mn3O4 + 2ZnO N
E
How can you tell that electrons are being donated and accepted? Which
species is donating electron( s) and which is accepting electron (s)?
O
U
Tr a n s f e r l e a d s t o - S
Increase in ON of element = O X I D A T I O N L
Decrease in ON of element = R E D U C T I O N
Y

8. REDOX TITRATIONS
Oxidation Reduction
Loss of electrons Gain of electrons
Gain in oxygen Loss of oxygen
0 1
Cl e Cl
Sodium is oxidized Chlorine is reduced
LEO SAYS GER

9. REDOX TITRATIONS
REDUCTION
OXIDATION: Old definition: Removal of oxygen from
a compound
Old Definition:
Combination of substance with WO3 (s) + 3H2(g) W(s) + 3H2O(g)
oxygen
C (s) + O2(g) CO2(g) Current definition: Gain of Electrons is
Reduction (GER)
Current definition:
Cl + e- Cl
-
Loss of Electrons is Oxidation (LEO)
Na Na + + e-
Negative charge represents electron
richness
ONE NEGATIVE CHARGE MEANS RICH
Positive charge represents electron
BY ONE ELECTRON
deficiency
Positive OS reflects the tendency atom to loose electrons
Negative OS reflects the tendency atom to gain electrons

10. REDOX
TITRATIONS:
•Direct Titrations
•Indirect Titrations
•Back Titrations

11. REDOX TITRATIONS
OBJECTIVES
Define oxidation & reduction in terms of loss or gain of O2/Electrons electrons
OBJECTIVES
State the characteristics of a redox reaction and identify the oxidizing agent
and reducing agent.
Many of reactions may not even involve oxygen
Redox- electrons are transferred between reactants
Mg + S→ Mg2+ + S2- (MgS)
Magnesium atom (has zero charge) changes to a Mg ion by losing 2 electrons,
and is oxidized to Mg2+
The sulfur atom (which has no charge) is changed to a sulfide ion by gaining 2
electrons, and is reduced to S2-
Active metals: Lose electrons easily- easily oxidized- strong reducing agents
Active nonmetals: Gain electrons easily-easily reduced-strong oxidizing agents

12. REDOX TITRATIONS
Losing electrons is oxidation & substance that
loses the electrons is called the reducing agent
Gaining electrons is reduction, & substance that
gains the electrons is called the oxidizing agent
Mg is the
reducing
Mg(s) + S(s) → MgS(s)
agent Mg is oxidized: loses e-, becomes a Mg2+ ion
S is the oxidizing agent S is reduced: gains e- = S2- ion
0 1
Na Na e Sodium is oxidized – it is the reducing agent
0 1
Cl e Cl Chlorine is reduced – is the oxidizing agent

13. REDOX TITRATIONS
 The reaction of a metal and non-metal
 All the electrons must be accounted for!
Mg + S 2+ + 2-
→ Mg S

14. REDOX TITRATIONS
In water, oxygen is highly
electronegative, so:
It is easy to see the
loss and gain of • Oxygen gains electrons (is
reduced & is oxidizing
electrons in ionic agent)
compounds, but
what about
covalent • Hydrogen loses electrons (is
oxidized & is reducing
compounds?
agent)

15. REDOX TITRATIONS
Not All Reactions are Redox Reactions
Reactions - no change in ON are NOT redox reactions.
Examples:
1 5 2 1 1 1 1 1 5 2
Ag N O 3 ( aq ) Na Cl ( aq ) Ag Cl ( s ) Na N O 3 ( aq )
1 2 1 1 6 2 1 6 2 1 2
2 Na O H ( aq ) H 2 S O 4 ( aq ) Na 2 S O 4 ( aq ) H 2 O (l )
Not all oxidation processes that use oxygen involve burning:
Elemental iron slowly oxidizes to iron (III) oxide- “rust”
Bleaching stains in fabrics
H2O2 - releases oxygen when decomposes

16. REDOX TITRATIONS
Corrosion
Damage done to metal is costly to prevent and repair.
Iron- Metal, corrodes by being oxidized to ions of iron by oxygen.
Corrosion is faster in presence of salts & acids, because these materials
make electrically conductive solutions that make electron transfer easy
L u c kily, n o t a l l m e t a ls c o r rode e a s ily
Gold & platinum are noble metals - are resistant to losing their electrons by corrosion.
Other metals lose their electrons easily, but protected by coating on surface, such as Al.
Iron has an oxide coating, but is not tightly packed, so water & air can penetrate easily

17. Oxidation Numbers
OBJECTIVES
Determine: Oxidation number of an atom of any element in a
pure substance.
Define: Oxidation & Reduction in terms of a change in ON, &
identify atoms being oxidized or reduced in redox reactions.
An “oxidation number” is positive /negative number assigned
to an atom to indicate its degree of oxidation or reduction.
Generally, a bonded atom’s oxidation number
is the charge it would have if the electrons in
the bond were assigned to the atom of the
more electronegative element

18. Rules for Assigning Oxidation Numbers
1. ON of any uncombined element = 0
2.ON of a monatomic ion equals its charge.
0 0 1 1
2 Na Cl 2 2 Na Cl
3. ON of oxygen in compounds is -2, except in peroxides, such as
H2O2 where it is -1.
4. ON of hydrogen in compounds is +1, except in metal hydrides,
like NaH, where it is -1.
5. Sum of ON of atoms in compound must equal 0
1 2
2 2 1
H 2 O Ca (O H ) 2
2(+1) + (-2) = 0 (+2) + 2(-2) + 2(+1) = 0
H O Ca O H
6. Sum of ON in formula of polyatomic ion is equal to its ionic charge.
thus
? 2 X + 3(-2) = -1 thus ? 2 X + 4(-2) = -2
S O4
2
X =+6
N O3 N O X = +5 S O

19. Rules for Oxidation States (OS)
• The charge the atom would have in molecule (or ionic compound) if electrons were
completely transferred.
1 ON of elements in their standard states is zero. Eg: Na, Be, K, Pb, H2, O2, P4 = 0
2.OS for monatomic ions are the same as their charge.
– Eg: Li+, Li = +1; Fe3+, Fe = +3; O2-, O = -2
3. Oxygen is assigned an OS of -2 in its covalent compounds except as a peroxide.
4. ON of hydrogen is +1 except when it is bonded to metals in binary compounds. In
these cases, its ON is –1.
5. Group IA metals are +1, IIA metals are +2 and fluorine is always –1.
6. Sum of ON of all atoms in a molecule /ion is equal to charge on molecule /ion.

20. Balancing Redox Equations
OBJECTIVES
Describe how ON are used to identify redox reactions.
OBJECTIVES
Balance a redox equation using the ON-change method.
OBJECTIVES
Balance a redox equation by breaking the equation into oxidation
and reduction half-reactions, and then using the half-reaction method.

21. In general, all chemical reactions can be assigned to one of two
classes:
1) oxidation-reduction, in which electrons are transferred:
• Single-replacement, combination, decomposition, &
combustion
2) Second class has no electron transfer, & includes all others:
• Double-replacement and acid-base reactions
In an electrical storm, N & O react to form NO
N2(g) + O2(g) → 2NO(g) YES!
•Is this a redox reaction?
•If the ON of an element in a reacting species changes, then that
element has undergone either oxidation or reduction; therefore,
the reaction as a whole must be a redox.

22. Balancing Redox Equations
• It is essential to write a correctly balanced equation that
represents what happens in a chemical reaction
– Fortunately, two systematic methods are available, and are based on
the fact that the total electrons gained in reduction equals the total
lost in oxidation. The two methods:
1. Use oxidation-number changes
2. Use half-reactions

23. Using Oxidation-Number Changes
• Sort of like chemical bookkeeping, you compare the increases
and decreases in oxidation numbers.
– start with the skeleton equation
– Step 1: assign oxidation numbers to all atoms; write above their symbols
– Step 2: identify which are oxidized/reduced
– Step 3: use bracket lines to connect them
– Step 4: use coefficients to equalize
– Step 5: make sure they are balanced for both atoms and charge.

24. Using half-reactions
Half-reaction is an equation showing -oxidation or reduction that
takes place they are then balanced separately, and finally combined
• Step 1: write unbalanced equation in ionic form
• Step 2: write separate half-reaction equations for oxidation &
reduction
• Step 3: balance the atoms in the half-reactions
• Step 4: add enough electrons to one side of each half-reaction to
balance the charges
• Step 5: multiply each half-reaction by a number to make the
electrons equal in both
• Step 6: add the balanced half-reactions to show an overall
equation
• Step 7: add the spectator ions and balance the equation

25. p- block
s- block
d- block
f- block

26. Group 1A Group 2A
Has 1e- in the Has 2e- in the
outermost shell outermost shell
Tend to loose Tend to loose
1e- OS = +1 2e-OS = +2
Alkali metals Alkaline-earth
metals

27. Electronegativity Increases
p- block
Electro-
negativity
decreses
Electronegativity increases as we more left to right along a period.
Electronegativity decrease as move top to bottom down a group.

28. Group 3A Group 4A Group 5A
3e- in outermost shell 4e- in outermost shell 5e- in outermost shell
Tend to loose 3e- loose 4e-/gain 4e- loose 5e- /gain 3e-
OS=+3 Oxidation state Oxidation state -3, +5
-4,-3,-2,-1,0,+1,+2,+3,+4

29. Group 6A Group 7A
6e- in outermost shell 7e- in outermost shell Group 8A
Tend to gain 2e- Tend to gain 1e- 8e- in outermost shell
Oxidation state -1 Tend to gain/loose 0 e-
Oxidation state -2
Oxidation state -- 0
Group number - 8
Group number – 8 Inert elements/Noble
chalcogens Halogens
gases

30. Balancing simple redox reactions
Cu (s) + Ag +(aq) Ag(s) + Cu2+(aq)
Step 1: Pick out similar species from the equation
Cu(s) Cu2+(aq)
Ag +(aq) Ag (S)
Step 2: Balance the equations individually for charges and number of atoms
Cu0(S) Cu2+(aq) + 2e-
Ag +(aq) + e Ag (S)

31. Balancing simple redox reactions
Cu0(S) Cu2+(aq) + 2e-
Cu0(S) becomes Cu 2+(aq) by loosing 2 electrons -Cu0(S) oxidized to Cu2+(aq) -
oxidizing half reaction.
Ag +(aq) + e- Ag (S)
Ag+(aq) becomes Ag 0 (S) by gaining 1 electron -Ag+(aq) reduced to Ag(S) is the
reducing half reaction.
Final Balancing act: Making the number of electrons equal in both half reactions
[Cu0(S) Cu2+(aq) + 2e-] 1
[Ag +(aq) + e- Ag (S)] 2
So we have, Cu0(S) Cu2+(aq) + 2e-
2Ag +(aq) + 2e- 2Ag (S)
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

32. Balancing complex redox reactions
Fe+2(aq) + MnO4-(aq) Mn+2(aq) + Fe+3(aq)
Oxidizing half: Fe+2(aq) Fe+3(aq) + 1e-
Reducing half: MnO4-(aq) Mn+2(aq)
Balancing
Balancing atoms: MnO4-(aq)+ Mn+2(aq) + 4H2O oxygens:
Balancing hydrogens: MnO4-(aq)+8H+ Mn+2(aq) + 4H2O
Reaction happening in an acidic medium
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

33. Balancing complex redox reactions
Final Balancing act:
Making the number of electrons equal in both half reaction
[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-
5Fe2++MnO4-(aq)+8H+ 5Fe3+ +Mn+2(aq) + 4H2O
5 Fe 2+ ions oxidized by 1 MnO4- ion to 5 Fe3+ions.
Conversely 1 MnO4- is reduced by 5 Fe2+ ions to Mn2+.

34. Choosing a Balancing Method
1.The oxidation number change 2.The half-reaction method works
method works well if the oxidized best for reactions taking place in
and reduced species appear only acidic or alkaline solution.
once on each side of the equation,
and there are no acids or bases.

35. Displacement Reaction a.k.a Single Hydrogen Displacement Reaction
Replacement
A + BC AC + B
Sr+2H2O Sr(OH)2 +H2
Hydrogen Displacement
TiCl4 + 2Mg Ti + 2MgCl2
Metal Displacement
Cl2 + 2KBr 2KCl + Br2 The Activity Series fo
Halogen Displacement
M + BC AC + B
M is metal, BC is acid or H2O
B is H2
Ca + 2H2O Ca(OH)2 + H2

36. Types of Oxidation-Reduction
Reactions
0 +1 +2 0
+4 0 0 +2
0 -1 -1 0

37. C3H8O + CrO3 + H2SO4
Cr2(SO4)3 + C3H6O + H2O
Balancing equations using
oxidation numbers

38. Review: balancing chemical equations
Balance the following chemical reaction:
CuCl2 + Al Cu + AlCl3
• Balanced equations by “inspection”.
• Balancing equations-equal numbers of atoms on each
side of the equation
• Balance equations using oxidation #s.
• Relies on the idea that -number of electrons lost by-
element must be equal to number gained by different
element.
• Total gain in oxidation numbers -- equal to total lost.

39. Using Oxidation Numbers
total +2 -2 0 0 +3 -3
oxidation #
+2 -1+
CuCl Al
0 Cu +
0AlCl
+3 -1
2 3
• Notice: Cu has gained 2e– (oxidation # by 2)
• Notice: Al has lost 3e– (oxidation # by 3)
• But, number of e– gained must equal e– lost
• Multiply Cu by 3, Al by 2: change is 6 for both
-6 +6
change
+6 0 0 +6
total +2 -2 0 0 +3 -3
oxidation # +2 -1 0 0 +3 -1
3CuCl2 + 2Al 3Cu + 2AlCl3

40. Steps to balancing equations
Write the skeleton equation
Assign oxidation numbers to all atoms
Identify which atoms change oxidation number
Make the number of atoms that change oxidation
number the same on both sides by inserting
temporary coefficients
Compute the total change in oxidation number
Make total increase in oxidation number equal the
total decrease by multiplication using appropriate
factors
Balance the remainder by inspection.
Do not change what has been balanced.
Compounds with elements that have changed in one
case but not in another are considered twice.

41. Example 1
Balance the following equation:
-2 x 3 = -6
+6
change
total +6 0 +6 +4
ox. # +2+6 -8 0 +6 +18-24 +4 -4 +2 -2
H2SO4 + Al
+1+6 -2 0Al2(SO4)3 ++6 -2+ H2O -2
+3 SO2 +4 +1 -2
3H2SO4+3 2 3 6
Step 6:7:Identify whichremainderin oxidation
Step 4: Make the number of atoms that change
Step3:Make the total increase done ox. us
Step
1: Write equation: already by inspection.
Balance the atoms change for #
Step 5: Compute to +4) andthat have oxidation
Step 2: Assigncompoundschange in+3) already
oxidationS (+6the totalsame on tobynot
Al (0 both sides by
Note: equal oxidationdecrease
number only the totalnumbers
number the
inserting temporary coefficients
number balanced need to be balanced here.
been
multiplication using appropriate factors

42. Balance the following equation:
+2 x 4 = +8
-8
0 +5 +2 -3
0 +1+5 -6 +1+5-6 +2+10-12 -3 +4+5 -6 +2 -2
0 +1+5-2 +1+5-2 +2 +5 -2 -3 +1+5 -2 +1 -2
4Zn+ HNO3+ 9 HNO3 4Zn(NO3)2 + NH4NO3+ H2O
3
Step 7: Balance the remainder by inspection.
Step 6: Make the total increase in oxidation
Step 4: Make the number of atoms that change
Note:Identify which oxidation numbers# us
number equal totaltotal thatin oxidation number
Step only compounds decreasedone
Write equation: already not already
the change have by
Step 3: Step 2: Assign atoms change ox. for by
1: number the same on both sides
oxidation
Step 5: Compute need to be balanced here.
been balanced +2) appropriate factors
inserting Zn (0 to coefficients to -3)
multiplication using and N (+5
temporary

43. Balance the following equation:
-5 x 2 = -10
+2 x 5 = +10
+7 +4 +2 +6
1 7 -8 2 6 -8 2 6 -8 2 6 -8 2 6 -8 6 18 -24 2-2
1 7 -2 2 6 -2 1 6 -2 1 6 -2 2 6 -2 3 6 -2 1-2
2KMnO4+2FeSO4+ H2SO4
8 K2SO4+2MnSO4+ 5Fe2(SO4)3+8H2O
10
Step 6: Make the total increase in oxidation
Step 7: Balance the remainder by inspection.
Step 4: Make the number of atoms that change
number equal the total decrease by
Step 3: Identify which atoms change ox. #
Step 1: Write Assign oxidation numbers us
Note:Step 2: equation: already done for
oxidation number the same on both sides by
only compounds that have not already
Step 5: Compute total appropriate factorsnumber
multiplication using change Feoxidation
Mn (+7 to +2) and in (+2 to +3)
inserting temporary coefficients
been balanced need to be balanced here.

44. Balance the following equation:
-2 x 3 = -6
+6 oxidation numbers
Step 2: Assign
+6 0 +6 +4
1 7 -8 2 6 -8 0 +6 +18 -24 +4 -4 +2 -2
1 7 -2 1 6 -2 0 +3 +6 -2 +4 -2 +2 -2
KMnO4+ H2C2O4+ H2SO4
2 CO2+ K2SO4+3 MnSO4+ H2O
3
Step 3: Identify which atoms change ox. #
Step 4: Make (+6 the+4)increaseby oxidation
Step 6: Make the to remainder(0 tothat change
7: Balance number of Al in inspection.
S the total and atoms +3)
Step 5: Compute thethe same inhaveby sides by
oxidationequal total change on both already
Note: only compoundsdecrease not number
that oxidation
number number total already done for us
Step 1: Write equation:
inserting temporary coefficients factors
been balanced need to be balanced here.
multiplication using appropriate

45. Oxidizing agents Reducing agents
Examples: Examples:
permanganate (MnO4-), chromate Active metals
(CrO42-), and dichromate (Cr2O72-)
ions, sodium hypochlorite (bleach) sodium, magnesium,
nitric acid (HNO3), aluminum & zinc,
perchloric acid (HClO4), and NaH, CaH2 and LiAlH4.
sulfuric acid (H2SO4)
Oxidizing agents used Reducing agents used as
as redox titrants redox titrants
Sodium thiosulfate
I3- (iodimetry) -most common
KMnO4, pot. permanganate -Stable in oxygen
-reducing agents are oxidized by
K2Cr2O7, potassium dichromate dissolved oxygen.
-Stronger reducing agents are
Cerium(IV) solutions
required, must work in oxygen-
Titrations that create/consume I2 free environment

46. Redox Titrations- POTASSIUM PERMANGANATE
• POWERFUL OXIDISING AGENT
• 1ST Introduced by Margueritte for Iron
• Acidic conditions – Strong Oxidising Agent.
• Sulfuric Acid – No action on permanganate in dilute solution.
• But with HCl – Chlorine liberates.
• MnO4- + 8H++5e= Mn2++4H2O
• With alkaline condition other reactions
• Cant use as Primary standard
• Standard solution and stability
• Permanganate ion used often because - its own indicator.
• MnO4- is purple, Mn+2 is colorless. When reaction solution
remains clear, MnO4- is gone.

47. Redox Titrations- POTASSIUM PERMANGANATE
• Standard solution and stability
• Permanganate ion used often because - its own indicator.
• MnO4- is purple, Mn+2 is colorless.
• When reaction solution remains clear, MnO4- is gone.
• Aqueous solution of MnO4- are not entirely stable because the
ions tend to oxidize water:
• 4MnO4- + 2H2O 4MnO2(s)+3O2(g)+4OH-
• Decomposition reaction-catalysed by light, heat,acids, MnO2
• Stored in dark glass bottles & Kept away high temp.
• Not a Primary Standard.
• Standardised = Std. Oxalic acid
• 2MnO4- + 5H2C2O4 +6H 2Mn2+ +10CO2+8H2O

48. Redox Titrations- POTASSIUM PERMANGANATE
BACK TITRATION TECHNIQUE: Eg. Glycerol Determination
• C3H8O3 + 14MnO4 - 20 OH- +3CO3
• Ba2+ is added to ppt manganate, to mask its green color.
• Acidify with sulfuric acid & back titrate xs St. MnO4- = std. Oxalic acid.
2MnO4- + 5H2C2O4 + 6H ------ 2Mn2+ +10CO2+8H2O
Potassium mangante(VII) -in acid reduced to manganese(II)
MnO4- + 8H+ + 5e- Mn2+ + 4H2O

49. Titration of unknown sample of Iron Vs KMnO4:
The unknown sample of iron contains, iron in Fe2+ oxidation state. So we are basically
doing a redox titration of Fe2+ Vs KMnO4
5Fe2++MnO4-(aq)+8H+ 5Fe3+ +Mn+2(aq) + 4H2O
Problem with KMnO4
Unfortunately, the permanganate solution, once prepared, begins to decompose by the
following reaction:
4 MnO4-(aq) + 2 H2O(l)  4 MnO2(s) + 3 O2(g) + 4 OH-(aq)
So we need another solution whose concentration is precisely known to be able to find
the precise concentration of KMnO4 solution.

50. Vinitial
KMnO4 Titration of Oxalic acid Vs KMnO4
Vfinal
Primary secondary
standard standard
End point:
Pale Permanent
Pink color
250mL 250mL
250mL
16 H+(aq) + 2 MnO4-(aq) + 5 C2O4-2(aq) 2 Mn+2(aq) +10CO2(g) + 8 H2O(l)

51. Redox Titrations- POTASSIUM DICHROMATE
Slightly Weaker Oxidising agents than MnO4
Its reactions are slow(Back titration)
It does not oxidise Chloride
Primary Std. Its solution need not be standardised (pure grade)
Titrations are carried out in acidic condition
In basic solution, Cr2O72- converted to yellow chromate ion CrO4 2-
(oxidising power is nil)
Potassium dichromate(VI) oxidising agent -reduced to chromium(III)
Cr2O7- + 14H+ + 6e- 2Cr3+ + 7H2O
Potassium dichromate acts as oxidizing agent in acidic medium only:
The neutral aqueous solution of Potassium dichromate is 1:1 equilibrium mixture of
dichromate and chromate, a consequence of hydrolysis of dichromate ions.
Cr2O7 2– + H2O = 2 CrO42– + 2H+
Orange yellow
Chromate ions are weaker oxidizing agent than dichromate. Thus oxidizing strength of
dichromate is reduced in neutral solution. The above hydrolysis reaction however can be
reversed by adding acid to the solution and this explains the necessity of acidic medium
for the reaction.

52. Redox titrations -Iodine
• I2, or I3- --Good oxidizing agent - used as a titrant in iodimetry
– I2 not very soluble in water. Dissolved in aqs. KI solution.
– In excess I-, forms I3-, very soluble, dark red
• I2 in Equations should replaced by I3-, with addition I- on other side
• Stability:
Iodine solutions lack stability
Volatility of I, Looses of Iodine from an open vessel occur in short time.
Air oxidation of Iodide in M of Iodine solution.
4I- + O2(g) + 4H+ 2I2 + 2H2O
This reaction is promoted by Acids, Heat, Light and Nitrogen oxides.
Stored in closed dark glass stoppered bottles and from elevated temp.
• I3- usually standardized against As2O3
• Iodine is volatile, solution used immediately after standardization
• Starch indicator detects appearance of excess I3-

53. Titration involing with Iodine
IODIMETRIC IODOMETRIC
DIRECT METHOD INDIRECT METHOD
Analysing strong reducing species Analysing strong oxidising species
Titration of I2 produced by analyte against
Titration of analyte with std. I2 soln. (I3-)
sod. thiosulfate
Add Xs of I- to soln of analyte. I2 is
Titration in acidic/weakly alkaline solution produced in amount equivalent to oxi.
agent
Red. Power of red. Agents is increased in
Lib I2 is titrated against Sod. Thio sulfate
neutral solution
pH is maintained neutral by adding Cr2O7 2- + 6I-(Xs)---14H + 2Cr3+ +3I2 + 7H2O
NaHCO3 I2+2S2O3 2- --- 2I- + S4O6 2-
Sodium thio sulfate is universal titrant for Iodine
AsO3 3- + I2 + H2O ---- AsO4 3- +2I- + 2H+ in neutral/acidic soltuion.
In neutral solution, at low acid:
equilibirium is shited to right Titrations occur in acidic solutions
Acidity promotes oxidising agent- iodide
Potential of As(V)/As(III) couple is
reaction
decreased sufficiently that As(III) will
reduce I2 and increase reducing power of Lib I2 is titrated against Sod. Thio sulfate
AsO3 3-
AsO4 3- + 2H+ +2e-------AsO33- + H2O How? Equiliibrium is shitted to right.
MnO4 - + 8H+ + 5e ---- Mn2+ +4H2O
H3AsO4 + 2H+ + 2e ---- H3AsO3 +4H2O

54. Titration involing with Iodine
IODIMETRIC IODOMETRIC
pH Effect: pH Effect:
Very Little influence on electrode potential of Increase of acidity promotes oxidising
I2/I couple because H+ does not participate in
agent – iodide reaction
the half reaction. Keep the solution neutral.
Complexing agent: Fe3+/Fe2+ system Precipitating agent:
Fe 3+ +e --- Fe 2+ E0=0.77V Determination of Cu 2+
I3-+2e -----3I- E0=0.536V Cu 2+ + e -------Cu+
E0 Fe 3+ /Fe+ is hr than E0 I /I- system
2 To force the reaction ----oxidation of I-by
Under normal condition I 2 could not oxidise Cu2+ ---- you have to increase E of
Fe 2+ into Fe 3+ Cu2+/Cu+ system to be >0.53V
Addition of complexing agenta as F- or EDTA By adding SCN/I- reagent to ppt
that form stable complex with Fe3+ shit to Cu+ ----- [Cu+] --- E of Cu2+/Cu+
lowerr E (0.53v) that allows oxidation of Fe2+
with I2 Indicator: Starch
Indicator: Starch Sources of error in Iodometry:
Sources of error in Iodimetry: Decomposition of thiosulfate solution
Due to I2: lack of stability (vol, air oxi of Iodide) Premature addition ofstarch & its decomposition
Due to Starch: Due to I2: lack of stability (vol, air oxi of Iodide)
Aq. Starch decomposes –few days- bacterial action. Determination of oxidising agents,
Decomposition products: Glucose, 2MnO4, Cr2O7, BrO3, IO3.
Boric acid/formamide-as preservative 2Fe3+ , H3AsO4
Direct titrations Titrate with liberated I2 against sod. Thio sulfate
Back titrations techniques.

55. Titration involing with Iodine
IODIMETRIC IODOMETRIC
Direct titration with only 1 reaction: Not direct titration because 2 reactions:
analyte +titrant (I2)→product (iodide I-) analyte + I- →I2
unknown known I2 + std. thiosulfate → product
a) A reducing analyte b) One reaction a) An oxidizing analyte b) Two reactions
c) Standard solution: Iodine (I2) c) Standard solution: Sodium thisoufate
Analytical applications:
Species analyzed : (reducing analytes), Species analyzed : (oxidizing analytes) HOCl,
SO2, H2S, Zn2+, Cd2+, Hg2+,Pb2+ Cysteine, Br2, IO3- , IO4- , O2, H2O2, O3,
glutathione, mercaptoethanol, Glucose NO2- , Cu 2+, MnO4-, MnO2
(and other reducing sugars)
Species Reaction
Species Oxidation reaction
SO2 SO2 + H2O < == > H2SO3 HOCl HOCl+H++3I- <=>Cl-+I3-+ H2O
H2SO3 + H2O< == > SO24- + 4H+ + 2e- Br2 Br2 + 3 I- < == > 2 Br - + I3-
IO3- 2 IO3-+16 I-+12 H+<=>6 I3- +6 H2O
H2S H2S < == > S(s) + 2H+ + 2e- IO4- 2 IO4-+22 I-+16 H+<=>8 I3-+ 8 H2O
Zn2+, Cd2+-- M2+ + H2S MS(s) + 2H+ O2 O2+4 Mn(OH)2+2H2O<=>4 Mn(OH)3
Hg2+, Pb2+ -- MS(s) < == > M2+ + S + 2e- 2Mn(OH)3+6H++6I-<=>2Mn2++2I3-+6H2O
H2O2 H2O2+3I-+2H+<=>I3-+2H2O
Cysteine, glutathione, 2RSH<=>RSSR+2H++2e- O O3+3I-+2H+<=>O2+ I3-+H2O
mercaptoethanol NO2 - 2HNO2+2H++3I-<=>2NO+I3-+2H2O
Aldehydes H2CO+3OH-<=>HCO2-+2H2O+2e- S2O 8 2- S O 2-+3I-<=>2SO 2- +I -
2 8 4 3
Glucose (and other reducing sugar) O Cu2+ 2 Cu2+ + 5 I - < == > 2 CuI(s) + I3-
MnO4- 2MnO4-+16H++15 I-<=>2Mn2++5I3-+8
RCH + 3OH- < == > HCO2-+2H2O + 2e- H 2O
Ascorbic acid MnO2 MnO2(s)+4H+ +3I-<=>Mn2++I3-+2H2O

56. Example: Quantification of Ascorbic Acid (Vitamin C)
C6H8O6 + I2 → CçH6O6 + 2I- + 2H+
Iodine rapidly oxidizes ascorbic acid, C6H8O6 , to produce
dehydroascorbic acid, C6H6O6 .
Ascorbic acid Dehydroascorbic acid

57. Redox Titrations- Cerimetry
Strong Oxidizing agent
Eg: Determination of Hydrogen peroxide (direct), Glycerol (back titration)
Cerric solution in H2SO4 does not oxidize chloride and can be used to titrate
HCl solutions of analytes
Its solultion need not be standardised
Standard solution and stability:
Salt is dissolved in H2SO4 – to prevent pptn of basic salts.
Solution in sulfuric acid is indefinitely stable.
Solution in HNO3 undergoes photo-chemical decomposition but slowly.
Ceric salts hydrolyses to ceric hydroxide if not in acid.

58. Introduction
• Sodium Thiosulfate (Na2S2O3) is used in Iodometry to quantify
the amount of iodine in solution, in the form of Triiodide (I3-)
• Since Sodium Thiosulfate is rarely/never available as a primary
standard, Na2S2O3 solutions must be standardized between
preparation and use
• To standardize an Na2S2O3 solution, KIO3 is usually employed as
the primary standard
• Two chemical equilibrium equations are involved in the process:
(1) IO3- + 8I- + 6H+  3I3- + 3H2O
(2) I3- + 2S2O3 2-  3I- + S4O6 2-

59. Preparation of
Primary Standard
The primary standard (KIO3) is accurately weighed, dissolved in dH2O,
and chemically treated to obtain I3- in solution, according to Eq. (1):
(1) IO3- + 8I- + 6H+  3I3- + 3H2O
A slight excess of I- is supplied The H+ can be supplied by adding
by dissolving solid KI… a non-reactive acid, such as H2SO4…
In this manner, 3 moles of I3- can be obtained in the solution
for every mole of KIO3 dissolved…

60. Standardization of Sodium
Thiosulfate Solution
The standardization of Na2S2O3 occurs according to Eq. (2):
(2) I3- + 2S2O3 2-  3I- + S4O6 2- Starch indicator is
used to accentuate
An exact amount of the treated the endpoint…
primary standard soln. is placed
in an Erlenmeyer flask… The unstandardized thiosulfate soln.
is placed in the buret…
Thus, 2 moles of S2O3 2- are required to neutralize each mole of
I3-…

61. CERIMETRY
VITAMIN c PERMANGANOMETRY
FERROUS SULFATE
FERROUS FUMARATE AMMONIUM CHLORIDE
FERROUS GLUCONATE FAS
NIFIDIPINE HYDROGEN PEROXIDE
ACETAMENOPHEN OXALIC ACID
VITAMIN C
IODOMETRY REDOX DIIODOHYDROXY
TITRATIONS QUINOLINE
BENZALKONIUM Cl DIMERCAPROL
CAPTOPRIL GLYCERYLMONOSTERATE
CEPHLORIDINE GUAPHENESIN
CETRIMIDE MANNITOL
PHENINDIONE POVIDONE IODINE
SODIUM METABISULFITE SODIUM THIOSULFATE
BROMOMETRY
SODIUM DIATRIAZOATE IODINE
IRON
CHLOROCRESOL
CHLOROXYLENOL
PHENOL