1. ERT207 Analytical Chemistry Oxidation-Reduction Titration
Dr Akmal Hadi Ma’ Radzi
PPK Bioproses

2. Types of titrimetric methods
Classified into four groups based on type
of reaction involve:
1. Acid-base titrations
2. Complexometric titrations
3. Redox titrations
4. Precipitation titrations

3. Oxidation Reduction Reaction
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.

4. Oxidation Reduction Reaction
Called redox reaction – occur between a reducing and an oxidizing agent
Ox1 + Red2 ⇌ Red1 + Ox2
Ox1 is reduced (gain e-) to Red1 and Red2 is oxidized (donate e-) to Ox2
Redox reaction involve electron transfer
Oxidizing agent gain electron and reduced
(electronegatif element O,F,Cl)
Reducing agent donate electron and oxidized
(electropositif elements Li,Na,Mg)

5. Ce4+ + Fe2+ → Ce3+ + Fe (1)
Redox reaction can be separated into two half reaction :
oxidation
Fe2+ → Fe3+ + e (2)
reduction
Ce e- → Ce (3)

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

7. 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 atoms:
Balancing
oxygens:
MnO4-(aq) Mn+2(aq) + 4H2O

8. Balancing complex redox reactions
Balancing hydrogens:
Reaction happening in an acidic medium
MnO4-(aq)+8H Mn+2(aq) + 4H2O
Oxidation
numbers: Mn = +7,
O = 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

9. 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) Fe+3(aq) + 5e-
MnO4-(aq)+8H++ 5e Mn+2(aq) + 4H2O
5Fe2++MnO4-(aq)+8H++ 5e-
5Fe3+ +Mn+2(aq) + 4H2O + 5e-

10. Electrochemical Cells
Electro. Cells – cell that have a pair of electrodes or conductors immersed in electrolytes.
Electrodes connected to outside conductor, current will be produced.
Oxidation will take place at the surface of one electrode (anode) and reduction at another surface of electrode (cathode)
Solution of electrolytes is separated by salt bridge to avoid reaction between the reacting species.

11. Electrodes – anode & cathode
oxidation
reduction
2 types :
1) galvanic cells - the chemical reaction occurs spontaneously to produce electrical energy. Eg battery
2) electrolytic cell - electrical energy is used to force the non spontaneous chemical reaction.
Eg electrolysis of water

12. Electrolytic cell
Galvanic cell
Salt bridge allow redox reaction take place
Allow transfer of electron but prevent mixing of the 2 solutions

13. Each electrode has the electrical potential which is determine by the potential of the ions tendency to donate or accept electron
Also know as electrode potential
No method to measure electrode potential separately
But potential difference between the two electrode can be measured (using voltmeter placed between two electrode)

14. Electrode potential
The electrode potentials are given +ve or –ve signs.
All half reactions are written in reduction form ; therefore the potentials are the standard reduction potential
Example Ce4+ + Fe2+ → Ce3+ + Fe3+
When half reaction written in reduction form :
Ce4+ + e Ce E°= V
Fe3+ + e Fe E° = V

15. As E° becomes more +ve, the tendency for reduction is greater
As E° becomes more –ve, the tendency for oxidation is greater.

17. 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).
Ce e- → Ce3+

19. Nernst equation
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

20. 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

21. 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

22. Nernst equation
This equation shows a relationship between potential (E) and concentration (activity) of species
Generally for half-cell reduction reaction:
aOx + ne ⇌ bRed
The Nernst equation is:
E = E° log [Red]b
n [Ox]a

23. E = reduction potential for a specific concentration
E° = standard reduction potential for half-cell
n = number of electron involved in the half-reaction (equivalents per mole)

24. Redox titration
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.
the change in potential during redox titration is determine by immersing 2 electrodes in test solution during titration
Reference electrode & indicator electrode

25. Redox Titrations

26. Reference electrode – give constant potential electrode towards the changes in a mixed solution. (SHE, NHE, calomel)
Indicator electrode – shows change of potential quantitatively towards changes in mixed solution
The Diff between the two electrode potentials (indicator&reference) change with respect to the vol of titrant added
The titration curve – plotting the potential (V) against titrant volume.

27. RedOx Titration Curve

28. Consider titration of 100 ml 0.1 M Fe2+ with 0.1 M Ce4+
Before titration started – only have a solution of Fe2+. So cannot calculated the potential.
Titration proceed – a known amount of Fe2+ is converted to Fe3+. So we know the ratio of [Fe2+]/[Fe3+]. The potential can be calculated from Nernst equation of this couple :
E = EºFe – log [Fe2+] / [Fe3+] n
Potential is near the E° value of this couple

29. At equivalence point – EFe = ECe
2E = E°Fe+ E°Ce – log[Fe2+][Ce3+]
n [Fe3+][Ce4+]
Beyond equivalence point – excess Ce4+
E = EºCe – log [Ce3+] / [Ce4+] n

30. EXERCISE
50.0 ml of 0.05 M Fe2+ is titrated with 0.10 M Ce4+ in a sulphuric acid media at all times. Calculate the potential of the inert electrode in the solution when 0.0, 5.0, 20.0, 25.0, 30.0 ml 0.10 M Ce4+ is added. Use 0.68 V as the formal potential of the Fe2+ - Fe3+ system in sulphuric acid and 1.44 V for the Ce3+ - Ce4+ system.

31. The Nernst equation is: E = E° - 0.059 log [Red]b n [Ox]a
Ce4+ + Fe2+ → Ce3+ + Fe3+
When half reaction written in reduction form :
Ce4+ + e Ce E°= 1.44 V
Fe3+ + e Fe E° = 0.68 V
aOx + ne ⇌ bRed
The Nernst equation is:
E = E° log [Red]b
n [Ox]a

32. a) Addition of 0.0ml Ce4+
No addition of titrant, therefore the solution does not give any potential. Hence E is not known

33. b) Addition of 5.0 ml Ce4+
Initial mmol Fe2+ = 50.0 ml x 0.05 M = 2.50 mmol
mmol Ce4+ added = 5.0 ml x 0.10 M = 0.50 mmol = mmol Fe3+ formed
mmol Fe2+ left = 2.00 mmol
E = EºFe – log [Fe2+]
n [Fe3+]
E = 0.68 – log 2.00
= 0.64 V

34. c) Addition of 25.0 ml Ce4+
Initial mmol Fe2+ = 50.0 ml x 0.05 M = 2.50 mmol
mmol Ce4+ added= 25.0 ml x 0.10 M = 2.50 mmol = mmol Fe3+ formed
Equivalence point reached. EFe = ECe, so
E = E°Ce log [Ce3+]
n [Ce4+]
E = E°Fe log [Fe2+]
n [Fe3+]

35. Adding the two expression, At equivalence the concentration of Fe3+ = Ce3+ and Fe2+ = Ce4+
2E = EºFe + E°Ce– log [Fe2+] [Ce3+] n [Fe3+] [Ce4+]
E = 2
= 1.06 V

36. d) Addition of 30.0 ml Ce4+
concentration of Fe2+ is very small and we can neglect the value and for convenience, we will utilise the Ce4+ electrode potential to calculate the solution potential.
Initial mmol Fe2+ = 50.0 ml x 0.05 M = 2.50 mmol
= mmol Ce3+ formed
mmol Ce4+ added =30.0 ml x 0.10 M = 3.00 mmol
mmol Ce4+ excess = 0.50 mmol
E = EºCe – log [Ce3+]
n [Ce4+]
E = 1.44 – log 2.50
= 1.4 V