This document discusses oxidation-reduction (redox) reactions and electrochemistry. 1. It explains how to identify redox reactions by checking if the oxidation number (O.N.) of any species changes in the reaction. An example reaction between permanganate and oxalic acid is given. 2. Balancing redox reactions is important, and the document outlines the step-by-step process for balancing both acidic and basic redox reactions. 3. Electrochemical cells are described as either galvanic cells that generate potential or electrolytic cells that consume potential. The standard hydrogen electrode is used as a reference electrode with a standard potential of 0 V.

1. 1
Oxidation Reduction
Equilibria and Titrations

2. 2
Identification of a Redox Reaction
It is a good practice to indicate the O.N. of each
species in a chemical reaction in order to check
if it is a Redox reaction or not. If the O.N. of any
species changes, then it is a definite indication
of a Redox reaction. Example,
2 KMnO4 + 5 H2C2O4 + 6 HCl D 2 MnCl2 + 2KCl + 10 CO2 + 8 H2O
It is observed that in the left-hand part of the
equation, manganese has an O.N. equals 7 and
carbon has an O.N. equals 3. In the right-hand
part, the O.N. of manganese is 2 and that of
carbon is 4. Therefore, permanganate is
reduced while oxalic acid is oxidized.

3. 3
Balancing Redox Reactions
Balanced chemical equations are the basis of
any reasonable quantitative calculations.
Therefore, it is very important to learn
balancing chemical equations, especially
Redox equations that can be manipulated
through definite steps. For example,
dichromate reacts with iron yielding Fe3+
and Cr3+
Cr2O7
2- + Fe2+ D Cr3+ + Fe3+
To balance this equation, the following steps
should be followed:

4. 4
1. Split the equation into two half reactions
Fe2+ D Fe3+
Cr2 O7
2- D Cr3+
2. Proceed with each half reaction separately
starting with mass balance.
Let us start with first half reaction
Fe2+ D Fe3+
One mole of Fe2+ yields one mole of Fe3+ which
is balanced.

5. 5
3. Balance the charges on both sides.
Fe2+ D Fe3+
It is clear that an electron (e) should be
added to the right side in order for the
charges to be equal
Fe2+ D Fe3+ + e
This is a straightforward process but
now consider the second half reaction,
which contains species that are not
mass balanced

6. 6
Cr2O7
2- D Cr3+
1. Adjust the number of moles of chromium on both
sides.
Cr2O7
2- D 2 Cr3+
2. For each oxygen atom place an H2O on the other
side
Cr2O7
2- D 2 Cr3+ + 7 H2O
3. Adjust the number of hydrogen atoms on both sides
of the equation by adding H+
14 H+ + Cr2O7
2- D 2 Cr3+ + 7 H2O
4. The charges on both sides should be balanced at
this point. This can be done by addition of 6
electrons to the left side of the equation
6 e + 14 H+ + Cr2O7
2- D 2Cr3+ + 7H2O

7. 7
5. This is the case if the reaction is carried out in
acidic solution. The combination of the two half
reactions necessitates the multiplication of the half
reaction involving the iron by a factor of 6 in order
to cancel the electrons involved
6 Fe2+ D 6 Fe3+ + 6 e
6 e + 14 H+ + Cr2O7
2- D 2 Cr3+ + 7 H2O
_________________________________________
6 Fe2+ + 14 H+ + Cr2O7
2- D 6 Fe3+ + 2Cr3+ + 7 H2O
This is the balanced equation assuming acidic
conditions.
In basic solutions, balancing Redox equations requires
extra effort where all previous steps are needed in
addition to other steps.

8. 8
Example
C2O4
2- + MnO4
- D Mn4+ + CO2
First, proceed as in case of acidic solution.
1. Split the equation into two half reactions
C2O4
2- D CO2
MnO4
- D Mn4+
2. Balance the first half reaction by adjusting the
number of atoms on both sides (mass balance)
C2O4
2- D 2 CO2

9. 9
3. Adjust the charges on both sides (charge balance)
C2O4
2- = 2 CO2 + 2 e
The first half reaction is completely balanced.
Now, turn to balance the second half reaction
MnO4
- D Mn4+
1. Mass balance shows 1 mole of Mn on both sides,
therefore Mn is adjusted.
2. Adjust the oxygens on both sides by placing an H2O
for each oxygen atom present.
MnO4
- D Mn4+ + 4 H2O

10. 10
3. Adjust the number of hydrogen atoms on both sides
by additon of H+
8 H+ + MnO4
- D Mn4+ + 4 H2O
4. Adjusting the charges on both sides gives
3 e + 8 H+ + MnO4
- D Mn4+ + 4 H2O
Now, watch carefully.
5. Add an OH- on both sides for each H+ present
8 OH- + 3 e + 8 H+ + MnO4
- D Mn4+ + 4 H2O + 8 OH-
6. Combine the OH- and H+ to form H2O
3 e + 8 H2O + MnO4
- D Mn4+ + 4 H2O + 8 OH-

11. 11
7. Adjust the number of H2O molecules on both sides
3 e + 4 H2O + MnO4
- D Mn4+ + 8 OH-
8. Combine the two half reactions
3 (C2 O4
2- D 2 CO2 + 2 e)
2 ( 3 e + 4 H2O + MnO4
- D Mn4+ + 8 OH-)
______________________________________________
3 C2O4
2- + 8 H2O + 2 MnO4
- D 6 CO2 + 2 Mn4+ + 16 OH-
The first half reaction was multiplied by 3 and the
second was multiplied by 2 in order to cancel the
electrons on both sides.

12. 12
Electrochemical Cells
There are two types of electrochemical cells in the
concept whether the cell generates potential
(called a galvanic cell) or consumes potential
(called an electrolytic cell).
A cell is simply constructed from two electrodes
immersed in solution. The electrode at which
reduction occurs is called the cathode while that
at which oxidation occurs is called the anode.
In galvanic cells, the cathode is positive (+) while
the anode is negative (-). The signs of the anode
and cathode are the opposite in electrolytic cells.

13. 13
The cells we will study in this chapter are of the
first type where electrons generated by one half
reaction will be consumed by the other half
reaction forcing the current to flow.

14. 14
The Standard Hydrogen Electrode
(SHE)
The standard hydrogen electrode is an
electrode with an arbitrarily assigned of
zero. It is also known as the normal
hydrogen electrode (NHE), and is used
in combination with other half cells at
standard state in order to determine the
standard electrode potential of the
other half cell.

15. 15
Hydrogen electrode is based on the
redox half cell:
2H+
(aq) + 2e → H2(g)
This redox reaction occurs at platinized
platinum electrode. The electrode is
dipped in an acidic solution and pure
hydrogen gas is bubbled through it. The
concentration of both the reduced form
and oxidized form is maintained at unity.
That implies that the pressure of hydrogen
gas is 1 bar and the activity of hydrogen
ions in the solution is unity.

16. 16
Hydrogen electrode
2H+ + 2e- = H2
Eo = 0.000 V. when PH2 = 1 atm, [H+] = 1 M, 298K

17. 17
Determination of Standard Electrode Potential
The standard electrode potential of a half reaction
can be determined using a conventional two
electrode cell but replacing one electrode with a
standard electrode like the standard hydrogen
electrode (SHE) for which an arbitrary potential
zero is assigned. Therefore, the electrode
potential is the reading of the voltmeter. For
example, when a 1.00 M Cu2+ solution is placed
in contact with a Cu wire and the cell is
completed with a SHE at standard conditions of
temperature and pressure, the potential will read
0.342 V. This is due to reaction:
Cu2+ + 2e D Cu (s) Eo = 0.342 V

18. 18
Cu2+ + 2e = Cu(s)
H2 = 2H+ + 2e

19. 19
On the other hand, when Zn2+ solution is
placed in contact with a Zn electrode in an
electrochemical cell with a SHE as the
second electrode, the potential will read -
0.762 V. The reaction is:
Zn2+ + 2e D Zn (s) Eo = - 0.762 V
From these results we can predict that Cu2+ is a
better oxidizing agent since the standard
electrode potential is more positive than that
for Zn2+.

20. 20

21. 21
The more positive the Eo, the better oxidizing agent is the oxidized form (e.g., MnO4
-).
The more negative the Eo, the better reducing agent is the reduced form (e.g., Zn).

22. 22
However, SHE is troublesome and more convenient
reference electrodes are used. Saturated calomel
electrode (SCE) and Ag/AgCl electrodes are most
common.

23. 23
Commercial saturated calomel electrode
The S.C.E. is a common reference electrode.
The cell half-reaction is: Hg2Cl2 + 2e- = 2Hg + 2Cl-.
E = 0.242 V for saturated KCl.

24. The potential of the following half cell reaction
at standard state was measured versus SCE
giving a value of 0.529V. Find Eo of the half
cell reaction:
Fe3+ + e D Fe2+

25. 25
Schematic representation of Zn2+/Zn(s) electrode potential
relative to different reference electrodes (Eo
SHE = 0.000V)
Zn2+ + 2e D Zn (s) Eo = - 0.762 V

26. The potential of the following half cell reaction
at standard state was measured versus SCE
giving a value of - 0.103V. Find Eo of the half
cell reaction:
Sn4+ + 2e D Sn2+

27. 27 Cell for potentiometric measurements
A complete cell consists of an indicating electrode that responds to the analyte
and a reference electrode of fixed potential.
The potential difference between the two is measured.

28. 28
Calculating the Cell Potential
The process of calculating the cell potential is
simple and involves calculation of the potential
of each electrode separately, then the overall
cell potential can be determined from the simple
equation:
Ecell = Ecathode - Eanode
Sometimes, this relation is written as:
Ecell = Eright – Eleft
This is true since the convention is to place the
cathode to the right of the cell while the anode is
placed to the left of the cell

29. 29
Cell Representation
Zn(s) | ZnCl2 (0.200 M) II CuSO4 (0.100 M) I Cu(s)

30. 30
This cell is read as follows: a zinc electrode is
immersed in a 0.200 M ZnCl2 solution (this is
the first half-cell, anode), 0.100 M CuSO4
solution in which a Cu(s) electrode is
immersed (this is the second half-cell,
cathode).
If the Ecell is a positive value, the reaction is
spontaneous
and if the value is negative, the reaction is
nonspontaneous in the direction written and
will be spontaneous in the reverse direction.

31. 31
Effect of Concentration on Electrode Potential
The IUPAC convention for writing half-cell
reactions is to represent the process as a
reduction. The more positive half-cell
reaction is the oxidizing agent, and the less
positive half-cell reaction is the reducing
agent. The relationship between the
concentration and the electrode potential for
a half-cell reaction is represented by Nernst
equation where for the half-cell reaction we
have:
aA + bB + ne D cC + dD Eo = x V

32. 32
The Nernst equation can be written as:
E = Eo – (RT/nF) ln {[C]c[D]d/[A]a[B]b}
Where, R is the molar gas constant (R = 8.314 J
mol-1 K-1), T is the absolute temperature in
Kelvin ( T = oC + 273) and F is the Faraday
constant ( F = 96485 Coulomb.eq-1) and n is
the number of electrons per mole.
One may write after substitution:
E = Eo – (0.0592/n) log {[C]c[D]d/[A]a[B]b}

33. 33
Example
Calculate the electrode potential for the half-
cell below if the solution contains 0.100 M
Cu2+.
The half-cell reaction is:
Cu2+ + 2e D Cu (s) Eo = 0.342 V
Solution
E = Eo – (0.0592/n) log [Cu(s)]/[Cu2+]
E = 0.342 – (0.0592/2) log 1/0.100
E = 0.312 V

34. 34
Example
Calculate the electrode potential of a half-cell
containing 0.100 M KMnO4 and 0.0500 M
MnCl2 at pH 1.000.
Solution
MnO4
- + 8 H+ + 5 e D Mn2+ + 4 H2O Eo = 1.51 V
E = Eo – (0.0592/n) log [Mn2+]/[MnO4
-][H+]8
E = 1.51 – (0.0592/5) log 0.0500/0.100*(0.100)8
E = 1.42 V

35. 35
Redox Indicators
Redox visual indicators are of two types. The
first type is called specific while the other
type is called nonspecific.
Nonspecific redox indicators are very weak
oxidants or reductants which have different
colors of the oxidized and reduced forms.
When all the analyte is consumed in a redox
reaction, the first drops of excess titrant will
react with the indicator, thus changing its
color.

36. 36
An example of a specific indicator is starch
where it forms a blue complex with iodine
but not with iodide. When iodine is
consumed, the blue complex disappears and
the solution turns almost white indicating the
end point.
Another specific indicator is the permanganate
ion where it is also called a self indicator.
The purple color of the permanganate ion is
converted to the colorless Mn2+.

37. 37
The redox indicator equilibrium can be
represented by the equation below:
InOX + n e = InRED Eo = x V
E = EIn
o – (0.0592/nIn) log [InRED]/[InOX]
The color of the oxidized form can be clearly
distinguished when 10 [InRED] = [InOX] and the
color of the reduced form can be clearly
distinguished when [InRED] = 10 [InOX].

38. 38
Substituting into the electrode potential
equation we get:
DEColor Change = EIn
o + (0.0592/nIn)
Therefore, the color change of the indicator
occurs around Eo
In and, in fact, very close to
it. For a real titration, Eo
In should be as close
as possible to the electrode potential at the
equivalence point, Eep, and should thus be at
the potential range of the sharp break of the
titration curve

39. 39
Examples of some redox indicators

40. 40
Redox Titrations
The idea of redox titrations is that the potential of a
redox reaction will change as the concentration
changes. Therefore, consider the following
redox reaction,
Fe2+ + Ce4+ D Fe3+ + Ce3+
If we titrate Fe2+ with Ce4+, the concentration of Fe2+ will
decrease upon addition of Ce4+ and thus the potential
of an electrode immersed in a Fe2+ solution will start
to change. This means that the potential of the cell will
be a function of Fe2+ concentration. Thus, if the
volume of Ce4+ added is plotted against the electrode
potential, a titration curve will be obtained.

41. 41
Titration Curves
Oxidation-reduction titrations and
titration curves can be followed by
measuring the potential of an indicator
electrode (e.g. platinum) as compared
to a reference electrode (has a fixed
constant potential, like Saturated
Calomel Electrode, SCE).
The potential of the cell measured is:
Ecell = Esolution – Ereference electrode

42. 42

43. 43
Ecell = Esolution – Ereference electrode
This means that:
Esolution = Ecell + Ereference electrode
Assume the process of titration of Fe2+
with Ce4+, we have
Fe2+ + Ce4+ D Fe3+ + Ce3+
Fe3+ + e D Fe2+ Eo = 0.771 V
Ce4+ + e D Ce3+ Eo = 1.70 V
We have three cases to consider:

44. 44
1. Before equivalence point we have
Fe2+/Fe3+ couple and we can use the
Nernst equation to calculate the
electrode potential where:
EFe = EFe
o – (0.0592/nFe) log [Fe2+]/[Fe3+]
Therefore, we calculate the concentration
of Fe2+ and Fe3+ and insert the values
in the Nernst equation to find the
resulting potential of the electrode .

45. 45
We can use the other half reaction to
calculate the potential of the cell but it
is easier to use the Fe2+/Fe3+ couple
since the concentration of Ce4+ is
difficult to calculate at this stage. The
Nernst equation for such potential
calculation is:
ECe = ECe
o – (0.0592/nCe) log [Ce3+]/[Ce4+]

46. 46
2. At the equivalence point, we may combine
the two Nernst equations for the two half
reactions by adding them up. We get:
nFe EFe + nCe ECe = nFe EFe
o + nCe ECe
o – (0.0592)
{log [Fe2+]/[Fe3+] + log [Ce3+]/[Ce4+]}
nFe EFe + nCe ECe = nFe EFe
o + nCe ECe
o – (0.0592)
{log [Fe2+] [Ce3+]/[Fe3+] [Ce4+]}
But [Fe3+] = [Ce3+] and
[Fe2+] = [Ce4+] at equivalence point.

47. 47
Therefore, the equation simplifies to:
nFe EFe + nCe ECe = nFe EFe
o + nCe ECe
o
Also, at equivalence point we have
EFe = ECe = Eeq pt, therefore:
(nFe + nCe) Eeq pt = nFe EFe
o + nCe ECe
o
The final equation can be written as:
Eeq pt = (nFe EFe
o + nCe ECe
o)/ (nFe + nCe)

48. 48
3. After equivalence point, it is easier to
calculate the cell potential from the
cerium half-cell using the equation:
ECe = ECe
o – (0.0592/nCe) log [Ce3+]/[Ce4+]
It is easier to calculate the Ce3+ and Ce4+
concentrations and it will be difficult to
calculate the Fe2+ concentration since it
is totally consumed, except for an
equilibrium concentration.

49. 49
Example
Find the potential of a 50 mL solution of a 0.10
M Fe2+ after addition of 0, 25, 40, 50, and 75
mL of 0.1 M Ce4+.
Fe3+ + e D Fe2+ Eo = 0.771 V
Ce4+ + e D Ce3+ Eo = 1.70 V
Solution
1. After addition of 0 mL Ce4+
The solution contains Fe2+ only and attempts to
apply Nernst equation will not be possible
since assumably no Fe3+ will be present.
EFe = EFe
o – (0.0592/nFe) log [Fe2+]/[Fe3+]

50. 50
EFe = EFe
o – (0.0592/nFe) log [Fe2+]/[Fe3+]
EFe = EFe
o – (0.0592/nFe)log 0.1/0
This means that as Fe3+ approaches zero, the
electrode potential approaches negative infinity.
This, in fact, is not practically true since it
suggests that a Fe2+ solution will have
unbelievably high reducing power. The way out
of this problem is to always assume the
presence of very small Fe3+ concentration in
equilibrium with Fe2+, even in very pure Fe2+
solutions. The conclusion with regards to
calculation of electrode potential after addition
of 0 mL Ce4+ is that we can not calculate it using
direct Nernst equation.

51. 51
2. After addition of 25 mL Ce4+
Initial mmol Fe2+ = 0.10 x 50 = 5.0
mmol Ce4+ added = 0.10 x 25 = 2.5
mmol Fe2+ left = 5.0 – 2.5 = 2.5
[Fe2+] = 2.5/75 M
mmol Fe3+ formed = 2.5
[Fe3+] = 2.5/75 M
Application of Nernst equation gives:
EFe = EFe
o – (0.0592/nFe)log {(2.5/75)/(2.5/75)}
EFe =0.771 V
Therefore, the standard electrode potential for Fe2+/Fe3+
couple can be calculated at half the way to the
equivalence point.

52. 52
3. After addition of 40 mL Ce4+
Initial mmol Fe2+ = 0.10 x 50 = 5.0
mmol Ce4+ added = 0.10 x 40 = 4.0
mmol Fe2+ left = 5.0 – 4.0 = 1.0
[Fe2+] = 1.0/90 M
mmol Fe3+ formed = 4
[Fe3+] = 4.0/90 M
Application of Nernst equation gives:
EFe = EFe
o – (0.0592/nFe) log {(1.0/90)/(4.0/90)}
EFe = 0.771 – 0.0592 log 1.0/4.0
EFe = 0.807 V

53. 53
4. After addition of 50 mL Ce4+
Initial mmol Fe2+ = 0.10 x 50 = 5.0
mmol Ce4+ added = 0.10 x 50 = 5.0
mmol Fe2+ left = 5.0 – 5.0 = ??
This is the equivalence point
mmol Fe3+ formed = 5.0
[Fe3+] =5.0/100 = 0.05 M, [Ce3+] = 0.05 M
At equivalence point, we have from the above
discussion:
Eeq pt = (nFe EFe
o + nCe ECe
o)/ (nFe + nCe)
Eeq pt = (1 * 0.771 + 1 * 1.70)/(1 + 1)
Eeq pt =1.24 V

54. 54
3. After addition of 75 mL Ce4+
Initial mmol Fe2+ = 0.10 x 50 = 5.0
mmol Ce4+ added = 0.10 x 75 = 7.5
mmol Ce4+ excess = 7.5 – 5.0 = 2.5
[Ce4+] = 2.5/125 M
mmol Ce3+ formed = 5.0
[Ce3+] = 5.0/125 M
Application of Nernst equation gives:
ECe = ECe
o – (0.0592/nCe) log [Ce3+]/[Ce4+]
ECe =1.70 – (0.0592/1)log {(5.0/125)/(2.5/125)}
ECe = 1.68 V

55. 55
Example
Find the electrode potential of a 50 mL of
a 0.10 M solution of Fe2+ after addition
of 0, 5, 10, and 20 mL 0f 0.10 KMnO4, at
pH 1.
Solution
MnO4
- + 8H+ + 5 Fe2+ D Mn2+ + 5 Fe3+ + 4 H2O
Fe3+ + e D Fe2+ Eo = 0.771 V
MnO4
- + 8H+ + 5e D Mn2+ + 4 H2O Eo = 1.51 V

56. 56
1. After addition of 0 mL MnO4
-
The solution contains Fe2+ only and attempts to apply
Nernst equation will not be possible since
assumably no Fe3+ will be present.
EFe = EFe
o – (0.0592/nFe) log [Fe2+]/[Fe3+]
EFe = EFe
o – (0.0592/nFe)log 0.1/0
This means that as Fe3+ approaches zero, the electrode
potential approaches negative infinity. This, in fact,
is not practically true since it suggests that a Fe2+
solution will have unbelievably high reducing power.
The conclusion with regards to calculation of
electrode potential after addition of 0 mL MnO4
- is
that we can not calculate it using direct Nernst
equation.

57. 57
2. After addition of 5 mL MnO4
-
mmol Fe2+ left = initial mmol Fe2+ - mmol Fe2+ reacted
mmol Fe2+ reacted = 5 mmol MnO4
-
mmol Fe2+ left = initial mmol Fe2+ - 5 mmol MnO4
-
mmol Fe2+ left = 0.1*50 – 5 *(0.1*5)
mmol Fe2+ left = 5.0 – 2.5 = 2.5 , [Fe2+] = 2.5/55 M
mmol Fe3+ formed = 2.5, [Fe3+] = 2.5/55 M
Application of Nernst equation gives:
E = EFe
o – (0.0592/nFe)log {(2.5/55)/(2.5/55)}
E =0.771 V
Therefore, the standard electrode potential for Fe2+/Fe3+
couple can be calculated at half the way to the
equivalence point.

58. 58
3. After addition of 10 mL MnO4
-
mmol Fe2+ left = initial mmol Fe2+ - mmol Fe2+
reacted
mmol Fe2+ left = initial mmol Fe2+ - 5 mmol MnO4
-
mmol Fe2+ left = 5.0 – 5 * (0.1*10) = ??
This is the equivalence point
At the equivalence point we have:
Electrode potential for the iron oxidation =
electrode potential for permanganate reduction
EMnO4
-/Mn
2+ = Eo
MnO4
-/Mn
2+– (0.0592/n MnO4
-/Mn
2+) log
[Mn2+]/[MnO4
-][H+]8
EFe = EFe
o – (0.0592/nFe) log [Fe2+]/[Fe3+]

59. 59
Eeq = Eo
MnO4
-/Mn
2+– (0.0592/n MnO4
-/Mn
2+) log [Mn2+]/[MnO4
-][H+]8
Eeq = EFe
o – (0.0592/nFe) log [Fe2+]/[Fe3+]
Addition of the two equations gives:
(5Eep+ Eep) = (5Eo
Mn + EFe
o) – 0.0592 log {
[Mn2+][Fe2+]/[Fe3+][MnO4
-][H+]8}
Substitute for the following:
[Fe2+] = 5 [MnO4
-]
[Fe3+] = 5[Mn2+]
6Eep = (5Eo
Mn + EFe
o) – 0.0592 log { [Mn2+] * 5[MnO4
-]/5
[Mn2+] [MnO4
-][H+]8}
(5 + 1) Eep = (5 * 1.51 + 1 * 0.771) – 0.0592 log 1/(0.10)8
Eep = 1.31 V

60. 60
4. After addition of 20 mL MnO4
-
mmol MnO4
- excess = mmol MnO4
- added - mmol MnO4
-
reacted
mmol MnO4
- reacted = 1/5 mmol Fe2+
mmol MnO4
- excess = mmol MnO4
- added - 1/5 mmol Fe2+
mmol MnO4
- excess = 0.1*20 – 1/5* (0.1*50) = 1
[MnO4
-]= 1/70 M
mmol Mn2+ formed = 1
[Mn2+] = 1/70 M
E = Eo
MnO4
-/Mn
2+– (0.0592/n MnO4
-/Mn
2+) log [Mn2+]/[MnO4
-
][H+]8
E = 1.51 – (0.0592/5) log (1/70)/{(1/70)(0.1)8}
E = 1.42 V