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ERT207 Analytical Chemistry Oxidation-Reduction Titration Pn Syazni Zainul Kamal PPK Bioproses

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CO4: ABILITY TO DIFFERENTIATE AND CALCULATE CONCENTRATION OF ANALYTES OF VARIOUS TITRIMETRIC METHODS (ACID- BASE, COMPLEXATION, REDOX, PRECIPITATION)

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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

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Oxidation Reduction Reaction Called redox reaction – occur between a reducing and an oxidizing agent Ox 1 + Red 2 ⇌ Red 1 + Ox 2 Ox 1 is reduced (gain e-) to Red 1 and Red 2 is oxidized (donate e-) to Ox 2 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)

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Ce 4+ + Fe 2+ → Ce 3+ + Fe 3+ (1) Redox reaction can be separated into two half reaction : oxidation Fe 2+ → Fe 3+ + e- (2) reduction Ce 4+ + e- → Ce 3+ (3)

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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 bride to avoid reaction between the reacting species. Electrochemical Cells

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Electrodes – anode & cathode oxidationreductio n 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

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

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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)

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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 Ce 4+ + Fe 2+ → Ce 3+ + Fe 3+ When half reaction written in reduction form : Ce 4+ + e - Ce 3+ E°= 1.610 V Fe 3+ + e - Fe 2+ E° = 0.771 V

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

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If a solution containing Fe 2+ is mixed with another solution containing Ce 4+, 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 Fe 2+ is being oxidised at the platinum wire (the anode): Fe 2+ → Fe 3+ + e - The electron thus produced will flow through the wire to the other beaker where the Ce 4+ is reduced (at the cathode). Ce 4+ + e - → Ce 3+

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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

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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)

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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

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Redox Titrations

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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.

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RedOx Titration Curve

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Consider titration of 100 ml 0.1 M Fe 2+ with 0.1 M Ce 4+ Before titration started – only have a solution of Fe 2+. So cannot calculated the potential. Titration proceed – a known amount of Fe 2+ is converted to Fe 3+. So we know the ratio of [Fe 2+ ]/[Fe 3+ ]. The potential can be calculated from Nernst equation of this couple : E = Eº Fe – 0.059 log [Fe 2+ ] / [Fe 3+ ] n Potential is near the E° value of this couple

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At equivalence point – E Fe = E Ce 2E = E° Fe + E° Ce – 0.059 log [Fe 2+ ][Ce 3+ ] n [Fe 3+ ][Ce 4+ ] Beyond equivalence point – excess Ce 4+ E = Eº Fe – 0.059 log [Ce 3+ ] / [Ce 4+ ] n

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50.0 ml of 0.05 M Fe 2+ is titrated with 0.10 M Ce 4+ 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 Ce 4+ is added. Use 0.68 V as the formal potential of the Fe 2+ - Fe 3+ system in sulphuric acid and 1.44 V for the Ce 3+ - Ce 4+ system. EXERCISE

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Ce 4+ + Fe 2+ → Ce 3+ + Fe 3+ When half reaction written in reduction form : Ce 4+ + e - Ce 3+ E°= 1.44 V Fe 3+ + e - Fe 2+ E° = 0.68 V aOx + ne ⇌ bRed The Nernst equation is: E = E° - 0.059 log [Red] b n [Ox] a

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a) Addition of 0.0ml Ce 4+ No addition of titrant, therefore the solution does not give any potential. Hence E is not known

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b) Addition of 5.0 ml Ce 4+ Initial mmol Fe 2+ = 50.0 ml x 0.05 M = 2.50 mmol mmol Ce 4+ added = 5.0 ml x 0.10 M = 0.50 mmol = mmol Fe 3+ formed mmol Fe 2+ left = 2.00 mmol E = Eº Fe – 0.059 log [Fe 2+ ] n [Fe 3+ ] E = 0.68 – 0.059 log 2.00 1 0.50 = 0.64 V

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c) Addition of 25.0 ml Ce 4+ Initial mmol Fe 2+ = 50.0 ml x 0.05 M = 2.50 mmol mmol Ce 4+ added= 25.0 ml x 0.10 M = 2.50 mmol = mmol Fe 3+ formed Equivalence point reached. E Fe = E Ce, so E = E° Ce -0.059 log [Ce 3+ ] n [Ce 4+ ] E = E° Fe - 0.059 log [Fe 2+ ] n [Fe 3+ ]

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Adding the two expression, At equivalence the concentration of Fe 3+ = Ce 3+ and Fe 2+ = Ce 4+ 2E = Eº Fe + E° Ce – 0.059 log [Fe 2+ ] [Ce 3+ ] n [Fe 3+ ] [Ce 4+ ] E = 0.68 + 1.44 - 0 2 = 1.06 V

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d) Addition of 30.0 ml Ce 4+ concentration of Fe 2+ is very small and we can neglect the value and for convenience, we will utilise the Ce 4+ electrode potential to calculate the solution potential. Initial mmol Fe 2+ = 50.0 ml x 0.05 M = 2.50 mmol = mmol Ce 3+ formed mmol Ce 4+ added =30.0 ml x 0.10 M = 3.00 mmol mmol Ce 4+ excess = 0.50 mmol E = Eº Ce – 0.059 log [Ce 3+ ] n [Ce 4+ ] E = 1.44 – 0.059 log 2.50 1 0.50 = 1.4 V

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