This experiment determines the percentage of iron in a sample through a redox titration between the iron sample and potassium permanganate (KMnO4). The reaction involves the oxidation of Fe2+ ions in the iron sample by MnO4- ions from the KMnO4 solution. For every 5 moles of Fe2+ ions oxidized, 1 mole of MnO4- ions is reduced. The moles of MnO4- ions used in the titration allows calculation of the moles of Fe2+ ions and hence the mass of iron in the original sample. An initial titration of KMnO4 solution against oxalic acid, a primary standard, is required to determine the exact molar

1. REDOX TITRATION
Experiment #3
OXIDATION- REDUCTION TITRATION:
IRON & PERMANGANATE

2. What are we doing in
this experiment?
Determine the % of Iron, in a sample by
performing a redox titration between a
solution of the iron sample and potassium
permanganate (KMnO4).

3. What is redox titration?
A TITRATION WHICH DEALS WITH A
REACTION INVOLVING OXIDATION AND
REDUCTION OF CERTAIN CHEMICAL
SPECIES.
What is a titration?
The act of adding standard solution in small
quantities to the test solution till the reaction
is complete is termed titration.

4. 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

5. What is oxidation?
Old definition:
Combination of substance with oxygen
C (s) + O2(g) CO2(g)
Current definition:
Loss of Electrons is Oxidation (LEO)
Na Na+ + e-
Positive charge represents electron deficiency
ONE POSITIVE CHARGE MEANS DEFICIENT BY ONE ELECTRON

6. What is reduction?
Old definition:
Removal of oxygen from a compound
WO3 (s) + 3H2(g) W(s) + 3H2O(g)
Current definition:
Gain of Electrons is Reduction (GER)
Cl + e- Cl -
Negative charge represents electron richness
ONE NEGATIVE CHARGE MEANS RICH BY ONE ELECTRON

7. OXIDATION-REDUCTION
Oxidation and reduction go hand in hand.
In a reaction, if there is an atom undergoing
oxidation, there is probably another atom
undergoing reduction.
When there is an atom that donates electrons,
there is always an atom that accepts electrons.
Electron transfer happens from one atom to
another.

8. How to keep track of electron
transfer?
Oxidation number or oxidation state (OS):
Usually a positive, zero or a negative number (an integer)
A positive OS reflects the tendency atom to loose electrons
A negative OS reflects the tendency atom to gain electrons

9. Rules for assigning OS
The sum of the oxidation numbers of all of the atoms
in a molecule or ion must be equal in sign and value
to the charge on the molecule or ion.
KMnO4 SO4
2-
Potassium Permanganate Sulfate anion
OS of K + OS of Mn +
4(OS of O) = 0 OS of S + 4(OS of O) = -2
NH4
+
Ammonium cation
OS of N + 4(OS of H) = +1

10. Also, in an element, such as S8 or O2 , this rule
requires that all atoms must have an oxidation
number of 0.
In binary compounds (those consisting of only two
different elements), the element with greater
electronegativity is assigned a negative OS equal
to its charge as a simple monatomic ion.
NaCl
Na+ Cl-
MgS
Mg2+ S2-

11. When it is bonded directly to a non-metal atom,
the hydrogen atom has an OS of +1. (When bonded
to a metal atom, hydrogen has an OS of -1.)
Except for substances termed peroxides or
superoxides, the OS of oxygen in its compounds is -
2. In peroxides, oxygen has an oxidation number of
-1, and in superoxides, it has an oxidation number
of -½ .
H2O HCl
2H+ O2-
NH4
+
N3- 4(H+) H+ Cl-
Hydrogen peroxide: H2O2= 2H+ 2O-
Potassium superoxide: KO2= K+ 2O -1/2

12. Please Remember !!
In a periodic table,
Vertical columns are called GROUPS
Horizontal rows are called PERIODS
Electronegativity increases as we more left to right
along a period.
Electronegativity decrease as we move top to bottom
down a group.

13. d- block
p- block
s-
block
f- block

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

15. d- block
p- block
s- block
f- block

16. p - block
Electronegativity Increases
Electro-
negativity
decreses

17. Group 3A
Tend to loose 3e-
OS = +3
Has 3e- in the
outermost shell

18. Group 4A
Can either loose 4e-
Or gain 4e-
Has 4e- in the
outermost shell
Exhibits variable
Oxidation state
-4,-3,-2,-1,0,+1,+2,+3,+4

19. Group 5A
Has 5e- in the
outermost shell
Can either loose 5e-
Or gain 3e-
Oxidation state
-3,+5

20. Group 6A
Has 6e- in the
outermost shell
Tend to gain 2e-
Oxidation state
-2
Group number - 8
Chalcogens

21. Group 7A
Has 7e- in the
outermost shell
Tend to gain 1e-
Oxidation state
-1
Group number - 8
Halogens

22. Group 8A
Has 8e- in the
outermost shell
Tend to gain/loose 0 e-
Oxidation state
0
Group number - 8
Inert elements
or
Noble gases

23. Sample problem
Find the OS of each Cr in K2Cr2O7
Let the OS of each Cr be = x
OS of K = +1 (Remember K belongs to Gp. 1A)
OS of O = -2 (Remember O belongs to Gp. 6A)
Net charge on the neutral K2Cr2O7 molecule = 0
So we have,
2(OS of K) + 2 ( OS of Cr) + 7 (OS of O)= 0
2(+1) + 2 ( x) + 7 (-2)= 0
2+ 2 ( x) +(-14)= 0

24. 2+ 2 ( x) +(-14)= 0
2 ( x) +(-12) = 0
2 ( x) = (12)
x = 6
Find the OS of each C in C2O4
2-
Let the OS of each C be = x
OS of O = -2 (Remember O belongs to Gp. 6A)
So we have,
2(OS of C) + 4 ( OS of O) = -2
2(x) + 4 ( -2) = -2
2 ( x) +(-8)= -2

25. 2 ( x) +(-8)= -2
2 ( x) = +6
( x) = +3
Find the OS of N in NH4+
Let the OS of each N be = x
OS of H = +1 (Remember H belongs to Gp. 1A)
So we have,
(OS of N) + 4 ( OS of H) = +1
(x) + 4 ( +1) = +1
( x) +(4)= +1
( x) = -3

26. 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)

27. Balancing simple redox reactions
Cu0(S) Cu2+(aq) + 2e-
Ag +(aq) + e- Ag (S)
Cu0(S) becomes Cu 2+ (aq) by loosing 2 electrons.
So Cu0(S) getting oxidized to Cu2+(aq) is the
oxidizing half reaction.
Ag+(aq) becomes Ag 0 (S) by gaining 1 electron.
So Ag+(aq) getting reduced to Ag (S) is the
reducing half reaction.
LEO-GER

28. Balancing simple redox reactions
Final Balancing act:
[Cu0(S) Cu2+(aq) + 2e-] × 1
[Ag +(aq) + e- Ag (S)]× 2
Making the number of electrons equal in both
half reactions
So we have,
Cu0(S) Cu2+(aq) + 2e-
2Ag +(aq) + 2e- 2Ag (S)

29. Cu0(S) Cu2+(aq) + 2e-
2Ag +(aq) + 2e- 2Ag (S)
Balancing simple redox reactions
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

30. Balancing complex redox reactions
Fe+2(aq) + MnO4
-(aq) Mn+2(aq) + Fe+3(aq)
Fe+2(aq) Fe+3(aq) + 1e-
MnO4
-(aq) Mn+2(aq)
Oxidizing half:
Reducing half:
Balancing atoms:
MnO4
-(aq)+ Mn+2(aq) + 4H2O
Balancing
oxygens:

31. Balancing complex redox reactions
MnO4
-(aq)+8H+ Mn+2(aq) + 4H2O
Balancing hydrogens:
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
Reducing Half
Reaction happening in an acidic medium

32. Balancing complex redox reactions
Final Balancing act:
Making the number of electrons equal in both half reactions
[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-

33. Balancing complex redox reactions
5Fe2++MnO4
-(aq)+8H+ 5Fe3+ +Mn+2(aq) + 4H2O
5 Fe 2+ ions are oxidized by 1 MnO4
- ion to 5 Fe3+
ions. Conversely 1 MnO4
- is reduced by 5 Fe2+ ions
to Mn2+.
If we talk in terms of moles:
5 moles of Fe 2+ ions are oxidized by 1mole of
MnO4
- ions to 5 moles of Fe3+ ions. Conversely
1 mole of MnO4
- ions is reduced by 5 moles of
Fe2+ ions to 1 mole of Mn2+ ions.

34. Conclusion from the balanced
chemical equation
For one mole of MnO4
- to completely react
With Fe2+, you will need 5 moles of Fe2+ ions.
So if the moles of MnO4
- used up in the reaction
is known, then the moles of Fe2+ involved in the
reaction will be 5 times the moles of MnO4
-
Mathematically written:
]
[
5 4
2 


 MnO
of
moles
Fe
of
Moles

35. How does this relationship concern our experiment?
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
]
[
5 4
2 


 MnO
of
moles
Fe
of
Moles

36. 250mL 250mL 250mL
Vinitial
Vfinal
End point:
Pale Permanent
Pink color
KMnO4
Vfinal- Vinital= Vused (in mL)
mL
L
mL
in
V
L
in
V used
Used
KMnO
1000
1
1
)
(
)
(
4


Important requirement:
The concentration of
KMnO4 should be
known precisely.

37. )
(
4
4
4
L
in
V
MnO
of
Molarity
MnO
of
Moles Used
KMnO




]
[
5 4
2 


 MnO
of
moles
Fe
of
Moles





 2
2
2
2
1
85
.
55
Fe
of
moles
Fe
of
mole
Fe
of
g
Fe
of
Grams
%
100
%
2



grams
in
sample
of
mass
Fe
of
grams
sample
in
Fe

38. 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.

39. Titration of Oxalic acid Vs KMnO4
Primary
standard
Secondary
standard
16 H+(aq) + 2 MnO4
-(aq) + 5 C2O4
-2(aq)  2 Mn+2(aq) +
10 CO2(g)
+ 8 H2O(l)
5 C2O4
2- ions are oxidized by 2 MnO4
- ions to 10 CO2
molecules. Conversely 2 MnO4
- is reduced by 5 C2O4
2-
ions to 2Mn2+ ions.

40. Titration of Oxalic acid Vs KMnO4
16 H+(aq) + 2 MnO4
-(aq) + 5 C2O4
-2(aq) 2 Mn+2(aq)
+10CO2(g) + 8 H2O(l)
If we talk in terms of moles:
5 moles of C2O4
2- ions are oxidized by 2 moles MnO4
-
ions to 10 moles of CO2 molecules. Conversely 2 moles
of MnO4
- is reduced by 5 moles of C2O4
2- ions to
2 moles of Mn2+ ions.

41. Conclusion from the balanced
chemical equation
For 5 moles of C2O4
2- ions to be completely oxidized
by MnO4
- we will need 2 moles of MnO4
- ions.
Conversely for 2 moles of MnO4
- to be completely
reduced by C2O4
2-, we will need 5 moles of C2O4
2- ions


 4
2
4
2
2
5 MnO
of
moles
O
C
of
Moles



 4
2
4
2
5
2
1 MnO
of
moles
O
C
of
Moles

42. 250mL 250mL 250mL
Vinitial
Vfinal
End point:
Pale Permanent
Pink color
KMnO4
Vfinal- Vinital= Vused (in mL)
mL
L
mL
in
V
L
in
V used
Used
KMnO
1000
1
1
)
(
)
(
4


Important requirement:
The concentration of
KMnO4 should be
known precisely.
0.15 g OXALIC ACID
+ 100 mL of 0.9 M
H2SO4.Heated to 80C

43. )
(
.
)
(
2
4
2
mol
g
acid
Oxalic
of
Wt
Mol
g
in
acid
oxalic
of
Weight
O
C
of
Moles 

4
2
2
2
4
2
2
4
2
4
4
1
1
5
2
O
C
Na
mol
O
C
mol
Ox
of
Moles
O
C
mol
MnO
mol
MnO
mol







)
(
]
[
4
4
4
L
in
V
MnO
mol
MnO
Used
KMnO




44. When preparing 0.9 M H2SO4
1.Wear SAFETY GOGGLES AND GLOVES
2.Use graduated cylinder to dispense the acid
from the bottle
3. Please have about 100 mL of water in
500 mL volumetric flask, before adding
acid in to it.
4. Add acid to the flask slowly in small
aliquots.