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

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

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.

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)

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

Balancing simple redox reactions 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 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

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

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-

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.

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

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

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)

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- Ce3+ E°= 1.610 V Fe3+ + e- Fe2+ E° = 0.771 V

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

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). Ce4+ + e- → Ce3+

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

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

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

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° - 0.059 log [Red]b n [Ox]a

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)

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

Redox Titrations

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.

RedOx Titration Curve

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 – 0.059 log [Fe2+] / [Fe3+] n Potential is near the E° value of this couple

At equivalence point – EFe = ECe 2E = E°Fe+ E°Ce – 0.059 log[Fe2+][Ce3+] n [Fe3+][Ce4+] Beyond equivalence point – excess Ce4+ E = EºCe – 0.059 log [Ce3+] / [Ce4+] n

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.

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- Ce3+ E°= 1.44 V Fe3+ + e- Fe2+ E° = 0.68 V aOx + ne ⇌ bRed The Nernst equation is: E = E° - 0.059 log [Red]b n [Ox]a

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

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 – 0.059 log [Fe2+] n [Fe3+] E = 0.68 – 0.059 log 2.00 1 0.50 = 0.64 V

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 -0.059 log [Ce3+] n [Ce4+] E = E°Fe - 0.059 log [Fe2+] n [Fe3+]

Adding the two expression, At equivalence the concentration of Fe3+ = Ce3+ and Fe2+ = Ce4+ 2E = EºFe + E°Ce– 0.059 log [Fe2+] [Ce3+] n [Fe3+] [Ce4+] E = 0.68 + 1.44 - 0 2 = 1.06 V

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 – 0.059 log [Ce3+] n [Ce4+] E = 1.44 – 0.059 log 2.50 1 0.50 = 1.4 V