ERT207 Analytical Chemistry

Oxidation-Reduction Titration

Pn Syazni Zainul KamalPPK Bioproses

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

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

• 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 :

oxidationFe2+ → Fe3+ + e- (2)

reductionCe4+ + e- → Ce3+ (3)

•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

• Electrodes – anode & cathode

oxidation reduction

2 types :1) galvanic cells - the chemical reaction occurs spontaneously to produce electrical energy. Eg battery2) electrolytic cell - electrical energy is used to force the non spontaneous chemical reaction.Eg electrolysis of water

Salt bridge allow redox reaction take placeAllow transfer of electron but prevent mixing of the 2 solutions

Galvanic cell Electrolytic cell

• 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

• 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

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, v

olt

s

• 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 = EºFe – 0.059 log [Ce3+] / [Ce4+]

n

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.

EXERCISE

• 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