Redox Titration in chemical analysis Oxidation, reduction, balanced reactions

1. Oxidation-Reduction
Titration
Mr. Jopale M.K.

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. ∗ Reactions of metals or any other
organic compounds with oxygen to
give oxides are labeled as
oxidation.
∗ The removal of oxygen from metal
oxides to give the metals in their
elemental forms is labeled as reduction.
∗ In other words, oxidation is addition of
oxygen and removal of hydrogen
whereas reduction is addition of
hydrogen and removal of oxygen.
Oxidation Reduction Reaction

4. Reactions of metals or any other organic compounds with
oxygen to give oxides are labeled as oxidation.
The removal of oxygen from metal oxides to give the metals
in their elemental forms is labeled as reduction.
In other words, oxidation is addition of oxygen and removal
of hydrogen whereas reduction is addition of hydrogen and
removal of oxygen.
Oxidation Reduction Reaction

5. ∗ Reduction-Oxidation reaction
commonly known as Redox reaction
∗ Ox1
+ Red2
Red⇌ 1
+ Ox2
∗ Chemical reaction based oxidation and
reduction reaction is known as RedOx
reaction
Fe2+
+ Ce4+
→ Fe3+
+ Ce3+
(1)
Chemical reaction

6. There are two kinds of electro chemical cells,
galvanic or electrolytic.
In galvanic cells, the chemical reaction occurs
spontaneously to produce electrical energy.
In a electrolytic cell, electrical energy is used to
force the non spontaneous chemical reaction.
Galvanic cells are of importance in our further
discussion as we will be discussing the
spontaneous chemical reaction to produce
electrical energy.
Electrochemical Cells

7. If a solution containing Fe2+
is mixed with
another solution containing Ce4+
, there will be
a redox reaction situation due to their
tendency of transfer electrons. If we consider
that these two solution are kept in separate
beaker and connected by salt bridge and a
platinum wire that will become a galvanic
cell. If we connect a voltmeter between two
electrode, the potential difference of two
electrode can be directly measured.
The Fe2+
is being oxidised at the platinum wire
(the anode):
Fe2+
→ Fe3+
+ e-
The electron thus produced will flow through
the wire to the other beaker where the Ce4+
is
reduced (at the cathode).

9. The reaction involve electron transfer
Fe2+
→ Fe3+
+ e-
(2)
Ce4+
+ e-
→ Ce3+
(3)
∗ Equation (2) & (3) are called half reactions
∗ No half reaction can occur by itself
∗ There must be an electron donner (reducing agent)
and an electron accepter (oxidizing agent)
∗ Fe2+
is the reducing agent and Ce3+
is the oxidizing
agent

10. Before we start the discussion of the oxidation reduction titration
curve construction, we should understand the Nernst equation
which was introduced by German scientist, Wlater Nernst in 1889.
This equation express the relation between the potential of metal
and metal ion and the concentration of the ion in the solution.
Lets consider the following chemical reaction :-
aA + bB ⇋ cC + dD
The change in free energy is given by the equation
ΔG = ΔGo
+ 2.3RT log [C]c x [D]d / [A]a x [B]b

11. From the relation ship of the free energy
and cell potential, we can get
ΔG = -nFE
In a standard states, free energy will be
ΔGo
= -nFEo
Hence, the above equation can be written
as
-nFE = -nFEo
+ 2.3RT log [C]c [D]d / [A]a
[B]b

12. After dividing both side with –nF, we can get the
expression as
E = Eo
– 2.3 RT/ nF log [C]c [D]d / [A]a [B]b
At 25ºC, the equation can be written as :-
E = Eo
– 0.059/n log [C]c [D]d / [A]a [B]b
This equation is known as the “Nernst equation” that
correlate the electrode potential with the concentration
of the ionic species.

13. ∗ Permanganomerty
∗ Dichromatomerty
∗ Cerimetry
∗ Iodometry/ Iodimetry
∗ Bromatometry
Redox Titration
Methods

14. Redox titration is monitored by observing the change of electrode
potential. The titration curve is drawn by taking the value of this potential
against the volume of the titrant added. Unlike other titration curve the p
values are substituted by electrode potential values in the curve.
The redox reaction is rapid and the system is always in equilibrium
throughout the titration. The electrode potential of the two half reaction are
always identical.
If we consider the oxidation of Fe (II) with standard Ce (IV), then we can
write the equation as follows :-
Fe2+
+ Ce4+
Fe⇋ 3+
+ Ce3+
Redox Titration Curve

15. The electrode potential of the two half reaction will be always identical.
For iron, the electrode potential will be
E = EºFe – 0.059 log [Fe2+
] / [Fe3+
]
For cerium the electrode potential will be
E = EºCe – 0.059 log [Ce3+
] / [Ce4+
]
Therefore, either of the electrode potential could be utilised to calculate
the potential of the solution.

16. The utilisation of the either equation is based on the stage of titration.
Prior to the equivalence point, the concentration of Fe(II) and Fe(III)
are appreciable compare to Ce(IV) ion which is negligible because of
the presence of large excess of Fe(II).
Beyond the equivalence point, the concentration of Ce(IV) and Ce(III)
is readily computed from the addition and the electrode potential for
the Ce(IV) could be used.
At equivalence point, the concentration of the oxidised and reduced
forms of the two species are such that their attraction for electron are
identical. At this point, the reactant species concentration and product
species concentrations ratios are known and they are utilised to
calculate the potential of the solution.

17. # 50.0 ml of 0.05M Fe2+
is titrated with 0.1M Ce4+
in a sulphuric acid media
at all times. Calculate the potential of the inert electrode in the solution at
various intervals in the titration and plot the titration curve. Use 0.68V as
the formal potential of the Fe2+
- Fe3+
system in sulphuric acid and 1.44V
for the Ce3+
- Ce4+
system.
Initial step : After addition of 5.0 ml of Ce4+
As because the Ce4+
is too small, we are considering the iron (Fe)
electrode potential to calculate the solution potential.
[Fe3+
] = 5.0ml X 0.10M / (50.0 + 5.0) ml
= 0.5 mmol / 55.0ml
Similarly, [Fe2+
] = (50.0ml X 0.05M – 5.0 ml X 0.1M) / 55.0 ml
= 2.0 mmol / 55.0 ml

18. Substituting the values to the standard electrode potential equation, we
can get
E = EºFe – 0.059 log [Fe2+
] / [Fe3+
]
E = Eº – 0.059 log 2.0 / 0.5
E = 0.68 -0.036 = 0.64 V
Step-2 : Equivalence point
At equivalence point in the titration of Fe(II) and Ce(IV), the potential of
the solution is controlled by both the half reaction.
Eeq=E0
Ce -0.059 log [Ce3+
] / [Ce4+
]
Eeq = E0
Fe - 0.059 log [Fe2+
] / [Fe3+
]

19. Adding the two expression, we can get
2Eeq = E0
Ce + E0
Fe - 0.059 log [Ce3+
] [Fe2+
]/ [Ce4+
][Fe3+
]
At equivalence the concentration of Fe3+
= Ce3+
and Fe2+
= Ce4+
We can get ,
2Eeq = E0
Ce + E0
Fe - 0.059 log [Ce3+
] [Ce4+
]/ [Ce4+
] [Ce3+
]
Eeq = (E0
Ce + E0
Fe) / 2
Eeq = (1.44 + 0.68) / 2 = 1.06 V

20. Step-3 : After addition of 25.1 ml Ce4+
At this stage, the concentration of Fe(II) is very small and we can neglect
the value and for convenience, we will utilise the Ce(IV) electrode potential
to calculate the solution potential.
[Ce3+
] = (25.0ml X 0.1M ) / (50.0 + 25.1) ml
= 2.5 mmol / 75.1 ml
[Ce4+
] = (0.1 ml X 0.1M) / 75.1 ml
E = EºCe – 0.059 log [Ce3+
] / [Ce4+
]
∴ E = 1.44 – 0.059 log 2.5 / 0.01
= 1.30 V

21. Redox Titrations
 Equipment forEquipment for
obtaining a titrationobtaining a titration
curve for a redoxcurve for a redox
titration.titration.

22. RedOx Titration Curve
0.6
0.8
1.0
1.2
1.4
1.6
1.8
0 20 40 60 80 100 120 140 160 180 200
mL Ce4+
E,volts

23. The net balanced redox equation is the sum of the
two half-cell reactions.
It may be necessary to multiply one or both half-cells
by some coefficient so that the same number of
electrons are lost by the substance that is oxidized as
are gained by the substance reduced.
Redox Titrations

24. Balancing Redox Equations
1. Assign oxidation numbers to each atom.
2. Determine the elements that get oxidized and reduced.
3. Split the equation into half-reactions.
4. Balance all atoms in each half-reaction, except H and O.
5. Balance O atoms using H2O.
6. Balance H atoms using H+
.
7. Balance charge using electrons.
8. Sum together the two half-reactions, so that: e-
lost = e-
gained
9. If the solution is basic, add a number of OH-
ions to each side of
the equation equal to the number of H+
ions shown in the overall
equation. Note that H+
+ OH-
 H2O
Fe2+
+ MnO4-
+ H+
Mn2+
+ Fe3+
+ H2O

25. 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
Balancing simple redox reactions

26. 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:
Balancing complex redox reactions

27. 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

28. 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-

29. ∗ Many of the end point indicators used
in redox titrations are substances
that are either oxidized or reduced at
given potentials.
∗ Much like acid-base indicators, the
electrical potential at the end point
oxidizes or reduces the indicator to
one of the colored forms to signal the
end point.
Redox Titrations

30. 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 −+
×= MnOofmolesFeofMoles
Titration of unknown sample of Iron Vs
KMnO4

31. Nernst EquationNernst Equation
aOx1 +bRed2 a’Red1 + b’Ox2
Q =
[Red1]a’
[Ox2]b’
[Ox1]a
[Red2]b
E = E0
- ln QRT
nF
E0
= Standard Potential
R = Gas constant 8.314 J/K.mol
F- Faraday constant = 94485 J/V.mol
n- number of electrons

32. 250mL 250mL 250mL
Vinitial
Vfinal
End point:
Pale Permanent
Pink color
KMnO4
Vfinal- Vinital= Vused (in mL)
mL
LmLinV
LinV used
UsedKMnO
1000
1
1
)(
)(4
×=
Important requirement:
The concentration of
KMnO4 should be
known precisely.

33. ∗ If the titrant is highly colored, this
color may be used to detect end
point.
∗ 0.02 M potassium permanganate is
deep purple. A dilute solution is
pink. The product of its reduced
form (Mn2+
) colorless.
RedOx Indicators

34. Redox Indicators

35. 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
 

36. Standardization of KMnO4 solution
Potassium permanganate is not primary standard, because traces of MnO2
are invariably present.
Standardization by titration of sodium oxalate (primary standard) :
2KMnO4 + 5 Na2(COO)2 + 8H2SO4 = 2MnSO4 + K2SO4 + 5Na2SO4 + 10 CO2 + 8H2O
2KMnO4 ≡ 5 Na2(COO)2 ≡ 10 Equivalent
mw 158.03 mw 134.01
158.03 g / 5 ≡ 134.01 g / 2 ≡ 1 Eq.
31.606 g ≡ 67.005 g
1N × 1000 ml ≡ 67.005 g
x N × V ml a g
x N = ( a g × 1N × 1000 ml) / (67.005 g × V ml)

37. Applications of permanganometry
(1) H2O2
2KMnO4 + 5 H2O2 + 3H2SO4 = 2MnSO4 + K2SO4 + 5O2 + 8H2O
(2) NaNO2
2NaNO2 + H2SO4 = Na2SO4 + HNO2
2KMnO4 + 5 HNO2 + 3H2SO4 = 2MnSO4 + K2SO4 + 5HNO3 + 3H2O
(3) FeSO4
2KMnO4 + 510 FeSO4 + 8H2SO4 = 2MnSO4 + K2SO4 + 5Fe2(SO4)3 + 8H2O
(4) CaO
CaO + 2HCl = CaCl2 + H2O
CaCl2 + H2C2O4 = CaC2O4 + 2HCl (excess oxalic acid)
2KMnO4 + 5 H2C2O4 + 3H2SO4 = 2MnSO4 + K2SO4 + 10CO2 + 8H2O (back tit)
(5) Calcium gluconate
[CH2OH(CHOH)4COO]2Ca + 2HCl = CaCl + 2CH2OH9CHOH)4COOH
(NH4)2C2O4 + CaCl2 = CaC2O4 + 2 NH4Cl
CaCl2 + H2SO4 = H2C2O4 + CaSO4

38. Oxidation with Ce4+
Ce4+
+ e = Ce3+
1.7 V in 1 N HClO4
yellow colorless 1.61 V in 1N HNO3
1.47 V in 1N HCl
1.44 V in 1M HSO4
Indicator : ferroin, diphenylamine
Preparation and standardization:
Ammonium hexanitratocerate, (NH4)2Ce(NO3)6, (primary standard grade)
Ce(HSO4)4, (NH4)4Ce(SO4)4·2H2O
Standardized with Sodium oxalate.

40. Applications of cerimetry
(1) Menadione (2-methylnaphthoquinon: vitamin K3)
O
O
CH3
OH
OH
CH3
2 Ce(SO4)2
HCl, Zn
Reduction
(2) Iron
2FeSO4 + 2 (NH4)4Ce(SO4)4 = Fe2(SO4)3 + Ce2(SO4)3 + 4 (NH4)2SO4

41. Oxidation with potassium dichromate
Cr2O7
2–
+ 14H+
+ 6e = 2Cr3+
+ 7H2O Eo
= 1.36 V
K2Cr2O7 is a primary standard.
Indicator : diphenylamine sulphonic acid

42. Thank You