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ANALYTICAL CHEMISTRY
Redox titration
1. REDOX REACTION
3
Oxidation - Reduction reactions (Redox rxns) involve the transfer of electrons from one species of the reactants to another. This results in an increase in oxidation number (O.N.) of a specific species and a complementary decrease in oxidation number of another species. Example:
Ce4+ + Fe2+ D Ce3+ + Fe3+
The O.N. of cerium was decreased while that of iron was increased. Cerium is reduced while iron is oxidized. A process that involves an increase in O.N. is an oxidation process and vice versa
4
Usually, a Redox reaction can be separated into two halves.
Ce4+ + e D Ce3+ Reduction
Fe2+ D Fe3+ + e Oxidation
Electrons appear in each half reaction while they do not show up in the overall equations.
5
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.
6
An example of a non-Redox reaction can be written where no change in O.N. occurs,
Na2CO3 + 2 HCl = 2 NaCl + CO2 + H2O
+1 +4 -2 +1 -1 +1 -1 +4 -2 +1 -2
There is no change in O.N. of any species involved in the reaction, which indicates that this is not a Redox reaction.
Balancing Redox Equations
• Simple reactions easy to balance• Complex reactions
– Usually encountered in laboratory– Can include CrO4
2-, Cr2O72-, MnO4
-, NO3- and
SO42-
– Follow balancing guidelines
Balancing Redox Reactions
• Guidelines– Step 1: Write the unbalanced equation for the reaction in ionic
form.– Step 2: Separate the equation into two half-reactions– Step 3: Balance each half-reaction for number and type of atoms
and charges. (Look at medium)– Step 4: Add the two half-reactions together and balance the final
equation by inspection. The electrons on both sides must cancel. (Be sure they are equal)
– Step 5: Verify that the equation contains the same type and numbers of atoms and the same charges on both sides of the equation.
Balancing Redox Equations
1. Write the unbalanced equation for the reaction ion ionic form.
The oxidation of Fe2+ to Fe3+ by Cr2O72- in acid solution?
Fe2+ + Cr2O72- Fe3+ + Cr3+
2. Separate the equation into two half-reactions.
Oxidation:
Cr2O72- Cr3+
+6 +3
Reduction:
Fe2+ Fe3++2 +3
3. Balance the atoms other than O and H in each half-reaction.
Cr2O72- 2Cr3+
Balancing Redox Reactions4. For reactions in acid, add H2O to balance O atoms and H+ to
balance H atoms.Cr2O7
2- 2Cr3+ + 7H2O14H+ + Cr2O7
2- 2Cr3+ + 7H2O
5. Add electrons to one side of each half-reaction to balance the charges on the half-reaction.
Fe2+ Fe3+ + 1e-
6e- + 14H+ + Cr2O72- 2Cr3+ + 7H2O
6. If necessary, equalize the number of electrons in the two half-reactions by multiplying the half-reactions by appropriate coefficients.
6Fe2+ 6Fe3+ + 6e-
6e- + 14H+ + Cr2O72- 2Cr3+ + 7H2O
Balancing Redox Reactions7. Add the two half-reactions together and balance the final
equation by inspection. The number of electrons on both sides must cancel.
6e- + 14H+ + Cr2O72- 2Cr3+ + 7H2O
6Fe2+ 6Fe3+ + 6e-Oxidation:
Reduction:
14H+ + Cr2O72- + 6Fe2+ 6Fe3+ + 2Cr3+ + 7H2O
8. Verify that the number of atoms and the charges are balanced.
14x1 – 2 + 6x2 = 24 = 6x3 + 2x3
9. For reactions in basic solutions, add OH- to both sides of the equation for every H+ that appears in the final equation.
17B Calculating Redox Equilibrium Constants
Ex 17-6: Calculate the equilibrium constant for the reaction
2Fe3+ + 3I- 2Fe2+ + I3-
Sol: 2Fe3+ + 2e- 2Fe2+ E0 = 0.771 V
I3- + 2e- 3I- E0 = 0.536 V
23
22
/FeFe0
/FeFe ]Fe[
]Fe[log
2
0592.0EE 23
23
][I
]I[log
2
0592.0EE
3
3
/II0
/II 33
]I[
]I[log
]Fe[
]Fe[log
0.0592
)E-2(E
]I[
]I[log
2
0592.0E
]Fe[
]Fe[log
2
0592.0E
EE
3
3
23
22/II
0/FeFe
0
3
3
/II0
23
22
/FeFe0
/II/FeFe
--3
23
323
323
3233
22
]I[]Fe[
]I[][Felog
0592.0
)EE(2
]I[]Fe[
]I[][Felog
/II0
/FeFe0
3233
223
23
7eq
/II0
/FeFe0
eq
107.894.7loganti
94.70592.0
)536.0771.0(2
0592.0
)EE(2log 3
23
K
K
Example 17-7
Calculate the equilibrium constant for the reaction
2MnO4- + 3Mn2+ + 2H2O 5MnO2(s) + 4H+
Sol: 2MnO4- + 8H+ +6e- 2MnO2(s) + 4H2O E0 = +1.695 V
3MnO2(s) + 12H+ + 6e- 3Mn2+ + 6H2O E0 = +1.23 V
EMnO4-/MnO2
= EMnO2/Mn2+
12
32
824 ]H[
]Mn[log
6
0592.023.1
]H[]MnO[
1log
6
0592.0695.1
83224
12
]H[]Mn[]MnO[
]H[log
0592.0
)23.1695.1(6
eq3224
4
log]Mn[][MnO
]H[log1.47 K
47eq 1011.47loganti K
32
12
824 ]Mn[
]H[log
]H[]MnO[
1log
0592.0
)23.1695.1(6
Calculating potentials of electrochemical cells
Ecell = Eright – Eleft
Ex. Cu Cu2+(0.0200M) Ag+(0.0200M) Ag
Cu (s) + 2Ag+ = Cu2+(aq) + 2Ag(s)
Ag+ + e = Ag(s) Eo = + 0.799 V
Cu2+ + 2e = Cu(s) Eo = + 0.337 V
2Ag+ + 2e = 2Ag(s) Eo = + 0.799 V E = 0.799 – (0.05916/1) log (1/0.0200) = 0.6984 V
Cu2+ + 2e = Cu(s) Eo = + 0.337 V E = 0.337 – (0.05916/2) log (1/0.0200) = 0.2867V
Cu (s) + 2Ag+ = Cu2+(aq) + 2Ag(s)
Eocell = + 0.799 V – (+ 0.337 V) Ecell = Eright – Eleft = 0.6984 – 0.2864
= + 0.462 V = + 0.412 V
G = – nFE = – 2 ×96485 C × 0.412 V = 79,503 J (18.99 kcal)
EX Calculate the potential of the following cell and indicate the reaction that would occur spontaneously if the cell were short-circuited. Pt U4+(0.200M), UO2
2+(0.0150M), H+(0.300M) Fe2+(0.0100M), Fe3+ (0.0250M) Pt
The two half-reactions are Fe3+ + e_ ↔ Fe2+ Eo = + 0.771V UO2
2+ + 4H+ + 2e- ↔ U4+ + 2H2O Eo = + 0.334V
The electrode potential for the right-hand electrode is Eright = 0.771 – 0.05916 log [Fe 2+]/[Fe 3+] = 0.771 – 0.05916 log 0.0100/0.0250 = 0.771 – (– 0.0236) = 0.7946 V
The electrode potential for the left-hand electrode is ELeft = 0.334 – (0.0592 / 2) log [U4+ ] / [UO2
2+][ H+]4
= 0.334 – (0.0592 / 2) log 0.200 / (0.0150) (0.0300) 4
= 0.334 – 0.2136 = 0.1204 V
And E cell = E right – E left = 0.7946 – 0.2136 = 0.674V
The positive sign means that the spontaneous reaction is the oxidation of U4+ on the left and the reduction of Fe 3+ on the right, or U4+ +2 Fe 3+ + 2H2O ↔ UO2
2+ +Fe 2+ + 4H+
EXCalculate the cell potential for Ag AgCl (sat’d), HCl (0.0200M) H2(0.800atm ), Pt
Cell without liquid junction
Note that this cell does not require two compartments (nor a salt bridge) because molecular H2 has little tendency to react directly with the low concentration of Ag+ in the electrolyte solution. This is an example of a cell without liquid junction.
The two half- reactions and their corresponding standard electrode potentials are
2H+ + 2e - ↔ H2 EoH+/H2 = 0.000V
AgCl(s) + e - ↔ Ag (s) + Cl– EoAgCl/Ag = 0.222V
The two electrode potentials are Eright = 0.000 – (0.0592/2)log pH2 / [H+]2
= – (0.0592/2) log 0.800 / (0.0200) 3 = – 0.0977 V Eleft = 0.222 – 0.0592log [Cl-] = 0.222 – 0.0592 log 0.0200 = 0.3226V
The cell potential is thus E cell = E right – E left = -0.0977 – 0.3226 = – 0.420V
The negative sign indicates that the cell reaction as considered 2H+ + 2 Ag(s) ↔ H2 + 2AgCl(s)
is nonspontaneous. To get this reaction to occur, we would have to apply an external voltage and construct an electrolytic cell.
Ex
Calculate the potential for the following cell using (a) concentration and (b) activities:
Zn ZnSO4(5.00 ×10–4 M), PbSO4 (sat’d) Pb
(a) [SO42–] = CZnSO4 = 5.00 ×10–4
PbSO4(s) + 2e Pb (s) + SO42– Eo
= – 0.350 V
Zn2+ + 2e Zn (s) Eo = – 0.763 V
Eright = Eo – (0.05916 / 2) log [SO4
2–] = – 0.350 – (0.05916 / 2) log (5.00 ×10–4)
= – 0.252 V
Eleft = Eo – (0.05916 / 2) log (1 / [Zn2+]
= – 0.763 – (0.05916 / 2) log {1 / (5.00 ×10–4)} = – 0.860 V
Ecell = Eright – Eleft = – 0.252 – (– 0.860) = 0.608 V
(b) Ionic strength for 5.00 ×10–4 M ZnSO4 :
= (1/2) {(5.00 ×10–4) ×(+2)2 + (5.00 ×10–4) ×(– 2)2 } = 2.00 ×10–3
Debye-Huckel Eq. : – log = (0.51 Z2 ) / ( 1+ 3.3 ) SO42– = 0.4 nm , Zn2+ = 0.4 nm
SO42– = 0.820, Zn2+ = 0.825
Activity = [C]
Eright = Eo – (0.05916 / 2) log { SO42– [SO4
2–] }
= – 0.350 – (0.05916 / 2) log (0.820 × 5.00 ×10–4) = – 0.250 V
Eleft = Eo – (0.05916 / 2) log {1 / ( Zn2+ [Zn2+])}
= – 0.763 – (0.05916 / 2) log {1 / (0.825 × 5.00 ×10–4)} = – 0.863 V
Ecell = Eright – Eleft = – 0.250 – (– 0.863) = 0.613 V
Redox systems in the respiratory chain, P=phosphate ion.(From P.Karlson, Introduction to modern Biochemistry.)
Calculation Redox Equilibrium Constant
Cu Cu2+(x M) Ag+(y M) Ag
Cu (s) + 2Ag+ = Cu2+(aq) + 2Ag(s)
Keq = [Cu2+] / [Ag+]2
2Ag+ + 2e = 2Ag(s) Eo = + 0.799 V
Cu2+ + 2e = Cu(s) Eo = + 0.337 V
Ecell = Eright – Eleft = EAg – ECu = 0 or Eright = Eleft = EAg = ECu
EoAg – (0.05916/2) log (1/[Ag+]2) = Eo
Cu – (0.05916/2) log (1/[Cu2+])
EoAg – Eo
Cu = (0.05916/2) log (1/[Ag+]2) – (0.05916/2) log (1/[Cu2+])
EoAg – Eo
Cu = (0.05916/2) log (1/[Ag+]2) + (0.05916/2) log ([Cu2+]/1)
2 (EoAg – Eo
Cu ) / 0.05916 = log ([Cu2+]/[Ag+]2 ) = log Keq
ln Keq = –Go/RT= – nFEocell / RT
ln Keq = – nEocell / 0.05916 = – n (Eo
right – Eoleft ) / 0.05916 <At 25oC >
REDOX TITRATION
Redox titrations• The methods of acid-base volumetric analysis
can be applied to redox titrations• A redox titration involves a controlled reaction
between a solution containing an oxidising agent and another solution containing a reducing agent
• The equivalence point occurs when the oxidising agent and reducing agent have reacted chemically equivalent amounts
• The end point is found using suitable indicators
Common Primary Standards• In redox titrations two common primary
standards used are:– Ammonium iron(II) sulfate (NH4)2Fe(SO4)2
• Active species Fe2+(aq)• Fe2+(aq) Fe3+ + e-
– Oxalic acid H2C2O4·2H2O• Active species H2C2O4(aq)
• H2C2O4(aq) 2CO2(g) + 2H+ + 2e-
Copyright © 2008 Pearson Prentice Hall, Inc.
Chapter 4/28
Redox Titrations
If the unknown concentration is the potassium permanganate solution, MnO4
1-, it can be slowly added to a known amount of oxalic acid, H2C2O4, until a faint purple color persists.
Titration: A procedure for determining the concentration of a solution by allowing a carefully measured volume to react with a solution of another substance (the standard solution) whose concentration is known.
5H2C2O4(aq) + 2MnO41-(aq) + 6H1+(aq)
10CO2(g) + 2Mn2+(aq) + 8H2O(l)
Copyright © 2008 Pearson Prentice Hall, Inc.
Chapter 4/32
Redox Titrations
5H2C2O4(aq) + 2MnO41-(aq) + 6H1+(aq)
10CO2(g) + 2Mn2+(aq) + 8H2O(l)
A solution is prepared with 0.2585 g of oxalic acid, H2C2O4. 22.35 mL of an unknown solution of potassium permanganate are needed to titrate the solution. What is the concentration of the potassium permanganate solution?
Moles ofH2C2O4
Mass ofH2C2O4
Moles ofKMnO4
Molarity ofKMnO4
Mole Ratio Molarity of KMnO4
Molar Mass of H2C2O4
Copyright © 2008 Pearson Prentice Hall, Inc.
Chapter 4/33
Redox Titrations
5H2C2O4(aq) + 2MnO41-(aq) + 6H1+(aq)
10CO2(g) + 2Mn2+(aq) + 8H2O(l)
Moles of H2C2O4 available:
90.04 g
1 mol= 0.002871 mol H2C2O4
0.2585 g H2C2O4
Moles of KMnO4 reacted:
5 mol H2C2O4
2 mol KMnO4
= 0.001148 mol KMnO40.002871 mol H2C2O4
x
x
Copyright © 2008 Pearson Prentice Hall, Inc.
Chapter 4/34
1 L
1000 mL
Redox Titrations
5H2C2O4(aq) + 2MnO41-(aq) + 6H1+(aq)
10CO2(g) + 2Mn2+(aq) + 8H2O(l)
Concentration of KMnO4 solution:
= 0.05136 M KMnO422.35 mLx
0.001148 mol KMnO4
18B Reducing Agents
• Thiosulfate: stable to air oxidation• Iron(II): E0 = 0.771V
– for titration of cerium(IV), chromium(VI), vanadium(V)
– indicator:ferroin or diphenylamine sulfonate.
18C Oxidizing Agents
Potassium permanganate (KMnO4)– E0=1.51
In neutral solution: MnO4-MnO2
In acid solution: MnO4-Mn2+
– Autocatalytic decomposition:– Standardization: Na2C2O4
5H2C2O4+2MnO4-+6H+
10CO2+2Mn2++8H2O
Cerium (IV): Ce4+ / H2SO4: E0 = 1.44V;
Ce4+ / HClO4: E0 = 1.70V
1) the rate of oxidation of chloride ion is slow
2) is stable in H2SO4
3) (NH4)2Ce(NO3)6 can be obtained as a primary standard.
indicator: Ferroin
• Potassium dichromate: K2Cr2O7
a slightly weaker oxidizing agent than KMnO4 primary standard
Cr2O72- Cr3+
E0 = 1.33~1.00V in 1M HCl
Dr. S. M. Condren
Dr. S. M. Condren
Dr. S. M. Condren
Dr. S. M. Condren
Dr. S. M. Condren
Dr. S. M. Condren
Dr. S. M. Condren
REDOX TITRATION CURVE
16-1 The shape of a redox titration curve
• A redox titration is based on an oxidation-reduction reaction between analyte and titrant.
• Consider the titration of iron(II) with standard cerium(IV), monitored potentiometrically
with Pt and calomel electrodes.
The potentials show above is in 1 M HClO4
solution. Note that equilibria 16-2 and 16-3are both established at the Pt electrode.
• There are three distinct regions in the titration of iron(II) with standard cerium(IV), monitored potentiometrically with Pt and
calomel electrodes.
1. Before the equivalence point, where the potential at Pt is dominated by the analyte redox pair.
2. At the equivalence point, where the potential at the indicator electrode is the average of their conditional potential.
3. After the equivalence point, where the potential was determined by the titratant redox pair.
Before the equivalence point: using analyte’s concentration to calculate E+
At the equivalence point: needs both redox pairs to calculate (why?)
E of the cell
After the equivalence point:
Summary• The greater the difference in reduction potential between analyze and
titrant, the sharper will be the end point. • The voltage at any point in this titration depends only on the ratio of
reactants; it will be independent of dilution.
• Prior to the equivalence point, the half-reaction involving analyze is used to find the voltage because the concentrations of both the oxidized and the reduced forms of analyte are known.
• After the equivalence point, the half-reaction involving titrant is employed. At the equivalence point, both half-reactions are used simultaneously to find the voltage.
There are two special points during the above titration process: (1) when V = ½ Ve, [Fe3+] = [Fe2+] and E+ = E°(Fe3+ | Fe2+) ; (2) when V = 2
Ve, [Ce4+] = [Ce3+] and E+ = E°(Ce4+ | Ce3+) = 1.70 V.
• The curve is not symmetric about the equivalence point. This is because the stoichiometry of reactants is 2:1, not 1:1.
• Still, the curve is so steep near the equivalence point that negligible error is introduced if the center of the steep part is taken as the end point.
For the titration of Tl+ (Thallium) by IO3- in 1.00 M HCl solution
17C Constructing Redox Titration Curves
Example 17-8
Obtain an expression for the equivalence-point potential in the titration of 0.0500 M U4+ with 0.1000 M Ce4+. Assume
that both solutions are 1.0 M in H2SO4.
U4+ + 2Ce4+ + 2H2O UO22+ + 2Ce3+ + 4H+
Sol: UO22+ + 4H+ + 2e- → U4+ + 2H2O E0 = 0.334 V
Ce4+ + e- Ce3+ E0' = 1.44 V42
2
4
/UUO0
eq]][HUO[
]U[log
2
0592.0EE 42
2
]Ce[
]Ce[log
1
0592.0EE
4
3
/CeCe0
eq34
422
4
/UUO0
eq]H][UO[
]U[log0592.0E2E2 42
2
4422
34
/CeCe0
/UUO0
eq]H][Ce][UO[
]Ce][[Ulog0592.0EE2E3 3442
2
2
]Ce[]U[
44
2
]Ce[]UO[
32
2
443
34/CeCe
0/UUO
0
eq ]H][[Ce]Ce[2
]Ce][Ce[2log
3
0592.0
3
EE2E
34422
4
/CeCe0
/UUO0
]H[
1log
3
0592.0
3
EE2 34422
Electrode Potential versus SHE in Titrations with 0.100 M Ce4+
Potential, V vs. SHEReagent Volume,
mL50.00 mL of
0.0500 M Fe2+50.00 mL of
0.02500 M U4+
5.00 0.64 0.316
15.00 0.69 0.339
20.00 0.72 0.352
24.00 0.76 0.375
24.90 0.82 0.405
25.00 1.06 ←Equivalence →Point
0.703
25.10 1.30 1.30
26.00 1.36 1.36
30.00 1.40 1.40
Table 17-1
Note: H2SO4 concentration is such that [H+] = 1.0 M throughout.
Titration curve : potentiometric titration of Fe2+ with Ce4+.
Titration reaction: Ce4+ + Fe2+ Ce3+ + Fe3+
Cell: HgHg2Cl2Cl– || Ce4+ , Ce3+ , Fe3+, Fe2+ | Pt
Apparatus for potentiometric titration of Fe2+ with Ce4+.
Reaction at the SCE reference electrode:
2Hg(l) + 2 Cl– Hg2Cl2(s) + 2 e Eo = 0.241V
Reaction at the Pt indicator electrode:
Fe3+ + e Fe2+ Eo = 0.767V (in 1M HClO4 )
Ce4+ + e Ce3+ Eo = 1.70V (in 1M HClO4 )
Cell reaction: 2Fe3+ + 2Hg(l) + 2 Cl– 2 Fe2+ + Hg2Cl2(s)
2Ce4+ + 2Hg(l) + 2 Cl– 2 Ce3+ + Hg2Cl2(s)
ECe = EFe = Esolution
Ecell = Ecathode – Eanode = Esolution – ESCE
Eocell = (0.767)– (0.241) = 0.526 V
The equilibium constant for this titration reaction
K = 10 nEo/0.05916
log K = n Eocell / 0.05916 = (1) (0.526) / (0.05916) =
K = 1.7 1017
Theoretical titration curve for titration of 100ml of 0.050M Fe2+ with 0.100M Ce4+ in 1M HClO4.
½Ve Ve
Initial Fe2+
[Ce4+] = [Ce3+][Fe2+] = [Fe3+] [Ce3+] = [Fe3+]
[Ce4+]
Region Major constituents Comment
Before the titration begins Fe2+ No calculation possible
Before the equivalence point Fe2+ , Fe3+, Ce3+ Use the Nernst equation
for the analyte half reaction
At the equivalence point Fe3+, Ce3+ Use e.p equation
After the equivalence point Fe3+, Ce3+, Ce4+ Use the Nernst equation for
the titrant half reaction
1) Before the equivalence point
(amount Fe2+ remaining) = (amount Fe2+ initial) – (amount Ce4+ added amount Fe2+ used)
(amount Fe3+ produced) = (amount Fe2+ used)
[Fe2+] = amount Fe2+ (mmol) / vol(ml), [Fe3+] = amount Fe3+ (mmol) / vol(ml)
Ecell = Ecathode – Eanode = Esolution – ESCE
= (0.767 – 0.05916 log{[Fe2+]/[Fe3+]}) – (0.241)
= 0.526 – 0.05916 log{[Fe2+]/[Fe3+]}
When V = (1/2) Ve, [Fe2+] = [Fe3+], Ecell = 0.526 .
2) At the equivalence point
[Ce3+] = [Fe3+]
[Ce4+] = [Fe2+]
Eeq = EoCe4+ – 0.05916 log[Ce3+]/ [Ce4+]
Eeq = EoFe3+ – 0.05916 log[Fe2+]/ [Fe3+]
2Eeq = EoCe4+ + Eo
Fe3+– 0.05916 log[Ce3+]/ [Ce4+] – 0.05916 log[Fe2+]/ [Fe3+]
= EoCe4+ + Eo
Fe3+– 0.05916 log[Ce3+] [Fe2+] / [Ce4+] [Fe3+]
= EoCe4+ + Eo
Fe3+
= 1.70 + 0.767
= 2.467
Eeq = (EoCe4+ + Eo
Fe3+ ) / 2 = 1.23 V
E(cell) = Eright – Ecalomel = 1.23 – 0.241 = 0.99V
3) After the equivalence point
[Fe2+] = 0
E(cell) = E+ – Ecalomel
= ( 1.70 – 0.05916 log [Ce3+]/[Ce4+]) – 0.241
amount Ce4+ remaining = amount Ce4+ added – amount Ce4+ used
= amount Ce4+ added – (amount Fe2+ initial × reacting ratio )
Titration curves for 0.1000M Ce4+ titration , Curve A: Titration of 50.00mL of 0.05000 M Fe2+. Curve B: Titration of 50.00 ml of 0.02500 M U4+.
General approach to redox titration
( titration with an oxidizing agent)
T + A T– + A+
titrant analyte
T + e T–
E = EoT – 0.05916 log [T– ] / [T]
[T– ] / [T] = 10 ( EoT – E) / 0.05916
[T– ] = [T]
A+ + e A
E = EoA – 0.05916 log [A ] / [A+ ]
[A ] / [A+ ] = 10 ( EoA – E) / 0.05916
[A ] = [A+ ]
Mass balance
[T– ] + [T] = Ttotal
[T– ] + (1/ ) [T– ] = Ttotal
[T– ] = ( Ttotal) / (1+ )
[A+ ] = Atotal / (1+ )
[T– ] = [A+ ] 1:1 mole ratio
( Ttotal) / (1+ ) = Atotal / (1+ )
Fraction of titration
= Ttotal / Atotal ( = 1 at eq point)
= (1+) / { ( 1+ )}
Titration of a mixtureThe titration of two species will exhibit two breaks if the standard potentials of the redox couples are sufficiently different.
Example :
Titration reactions
First : IO3– + 2Sn2+ + 2Cl– + 6H+ ICl2
– + 2Sn4+ + 3H2O
Second : IO3– + 2Tl+ + 2Cl– + 6H+ ICl2
– + 2Tl3+ + 3H2O
Half reactions
IO3– + 2Cl– + 6H+ + 4e ICl2
– + 3H2O Eo = 1.24 V
= [ICl2– ] /[IO3
– ] = 10 [4(1.24 – E) / 0.05916] – 2 pCl – 6 pH
Sn4+ + 2e Sn2+ Eo = 0.139 V
1 = [Sn2+]/[Sn4+] = 10 2 (0.139 – E) / 0.05916
Tl3+ + 2e Tl+ Eo = 0.77 V
2 = [Tl+]/[Tl3+] = 10 2 (0.77 – E) / 0.05916
Theoretical curve for titration of 100 ml of 0.01M Tl+ with 0.01M IO3
– in 1M HCl
Theoretical curves for 0.01M Tl+ plus 0.01M Sn2+ titrated with 0.01M IO3
–. The initial volume of analyte is 100ml and all solutions contain 1M HCl.
plot for the Fe2+/Ce4+ system.
Totration curve calculated using the inverse master equation approach.
f = Fe3+ /Ce3+
Factors affecting the shape of titration curves :
1) concentration
2) completeness of reactionV(ml)
V(ml)
E(V)
E(V)
concentrated
diluted
Higher Eo titrant
Lower Eo titrant
Effect of titrant electrode potential on reaction completeness.
Dr. S. M. Condren
Dr. S. M. Condren
Dr. S. M. Condren
Redox Titration Curve
Derivation of a titration curve
Fe+2 + Ce+4 <=> Ce+3 + Fe+3
EXAMPLE: Derive the titration curve for 50.00 mL of 0.0500 M Fe+2 with 0.1000 M Ce+4 in a medium that is 1.0 M in H2SO4.
Dr. S. M. Condren
EXAMPLE: Derive the titration curve for 50.00 mL of 0.0500 M Fe+2 with 0.1000 M Ce+4 in a medium that is 1.0 M in H2SO4.
Fe+2 + Ce+4 <=> Ce+3 + Fe+3
E = Eo - (0.0592/n) log([red]/[ox])
At 0.00 mL of Ce+4 added, inital point
no Ce+4 present; minimal, unknown [Fe+3]; thus, insufficient information to calculate E
Dr. S. M. Condren
EXAMPLE: Derive the titration curve for 50.00 mL of 0.0500 M Fe+2 with 0.1000 M Ce+4 in a medium that is 1.0 M in H2SO4.
Fe+2 + Ce+4 <=> Ce+3 + Fe+3
At 15.00 mL of Ce+4 added, VFeMFe > VCeMCe
= 2.308 x 10-2 M
VCe MCe (15.00 mL)(0.1000 M)[Fe+3] = --------------- = ----------------------------
VFe + VCe (50.00 + 15.00)mL
Dr. S. M. Condren
EXAMPLE: Derive the titration curve for 50.00 mL of 0.0500 M Fe+2 with 0.1000 M Ce+4 in a medium that is 1.0 M in H2SO4.
Fe+2 + Ce+4 <=> Ce+3 + Fe+3
At 15.00 mL of Ce+4 added, VFeMFe > VCeMCe
[Fe+3] = 2.308 x 10-2 M VFe MFe - VCe MCe
[Fe+2] = ------------------------ VFe + VCe
Dr. S. M. Condren
EXAMPLE: Derive the titration curve for 50.00 mL of 0.0500 M Fe+2 with 0.1000 M Ce+4 in a medium that is 1.0 M in H2SO4.
Fe+2 + Ce+4 <=> Ce+3 + Fe+3
At 15.00 mL of Ce+4 added, VFeMFe > VCeMCe
[Fe+3] = 2.308 x 10-2 M (50.00 mL)(0.0500 M) - (15.00 mL)(0.1000 M)
[Fe+2] = --------------------------------------------------------(50.00 + 15.00)mL
= 1.54 x 10-2 M
Dr. S. M. Condren
EXAMPLE: Derive the titration curve for 50.00 mL of 0.0500 M Fe+2 with 0.1000 M Ce+4 in a medium that is 1.0 M in H2SO4.Fe+2 + Ce+4 <=> Ce+3 + Fe+3
At 15.00 mL of Ce+4 added, VFeMFe > VCeMCe
[Fe+3] = 2.308 x 10-2 M[Fe+2] = 1.54 x 10-2 Mfor Fe+2 -> Fe+3 Eo = 0.69 v
0.0592 [Fe+2]Ecell = Ecell
o - ----------- log ----------- n [Fe+3]
Dr. S. M. Condren
EXAMPLE: Derive the titration curve for 50.00 mL of 0.0500 M Fe+2 with 0.1000 M Ce+4 in a medium that is 1.0 M in H2SO4.Fe+2 + Ce+4 <=> Ce+3 + Fe+3
At 15.00 mL of Ce+4 added, VFeMFe > VCeMCe
[Fe+3] = 2.308 x 10-2 M[Fe+2] = 1.54 x 10-2 Mfor Fe+2 -> Fe+3 Eo = 0.69 v
0.0592 1.54 x 10-2
Ecell = 0.68 v - ---------- log ----------------- v = 0.69 v 1 2.31 x 10-2
Dr. S. M. Condren
EXAMPLE: Derive the titration curve for 50.00 mL of 0.0500 M Fe+2 with 0.1000 M Ce+4 in a medium that is 1.0 M in H2SO4.
Fe+2 + Ce+4 <=> Ce+3 + Fe+3
At 25.00 mL of Ce+4 added,
VFeMFe = VCeMCe, equivalence point
nFe EFeo + nCe ECe
o 1(1.44) + 1(0.68)
Ecell = ----------------------- = ------------------------ v nFe + nCe 1 + 1
= 1.06 v
Dr. S. M. Condren
EXAMPLE: Derive the titration curve for 50.00 mL of 0.0500 M Fe+2 with 0.1000 M Ce+4 in a medium that is 1.0 M in H2SO4.Fe+2 + Ce+4 <=> Ce+3 + Fe+3
At 26.00 mL of Ce+4 added, VCeMCe > VFeMFe
VFe MFe (50.00 mL)(0.0500 M)
[Ce+3] = ---------------- = -------------------------------VFe + VCe (50.00 + 26.00)mL
= 3.29 x 10-2 M
Dr. S. M. Condren
EXAMPLE: Derive the titration curve for 50.00 mL of 0.0500 M Fe+2 with 0.1000 M Ce+4 in a medium that is 1.0 M in H2SO4.Fe+2 + Ce+4 <=> Ce+3 + Fe+3
At 26.00 mL of Ce+4 added, VCeMCe > VFeMFe
[Ce+3] = 3.29 x 10-2 M
VCe MCe - VFe MFe[Ce+4] = ---------------------------
VFe + VCe
Dr. S. M. Condren
EXAMPLE: Derive the titration curve for 50.00 mL of 0.0500 M Fe+2 with 0.1000 M Ce+4 in a medium that is 1.0 M in H2SO4.Fe+2 + Ce+4 <=> Ce+3 + Fe+3
At 26.00 mL of Ce+4 added, VCeMCe > VFeMFe
[Ce+3] = 3.29 x 10-2 M
(26.00 mL)(0.1000 M) - (50.00 mL)(0.0500 M)[Ce+4] = -------------------------------------------------------
(50.00 + 26.00)mL= 1.32 x 10-3 M
Dr. S. M. Condren
EXAMPLE: Derive the titration curve for 50.00 mL of 0.0500 M Fe+2 with 0.1000 M Ce+4 in a medium that is 1.0 M in H2SO4.
Fe+2 + Ce+4 <=> Ce+3 + Fe+3
At 26.00 mL of Ce+4 added, VCeMCe > VFeMFe
[Ce+3] = 3.29 x 10-2 M [Ce+4] = 1.32 x 10-3 M
for Ce+4 -> Ce+3 Eo = 1.44 v
0.0592 [Ce+3]Ecell = Ecell
o - ----------- log ----------n [Ce+4]
Dr. S. M. Condren
EXAMPLE: Derive the titration curve for 50.00 mL of 0.0500 M Fe+2 with 0.1000 M Ce+4 in a medium that is 1.0 M in H2SO4.
Fe+2 + Ce+4 <=> Ce+3 + Fe+3
At 26.00 mL of Ce+4 added, VCeMCe > VFeMFe
[Ce+3] = 3.29 x 10-2 M [Ce+4] = 1.32 x 10-3 M
for Ce+4 -> Ce+3 Eo = 1.44 v
0.0592 3.29 x 10-2
Ecell = 1.44 v - ------------ log --------------- v = 1.39 v 1 1.32 x 10-3
3. TITRATION END POINT
Detection of the end point
1- Self indication: If the titrant is highly colored, this color may be used to detect the end point. For example; potassium permanganate is deep purple, the production of its reduction Mn2+ is nearly colorless, being a very faint pink.
2- Starch indicator: this indicator is used for titrations involving iodine. Starch forms a complex color with I2, That is a very dark blue color. The color is sensitive to very small amounts of iodine. In titration of reducing agents with iodine, the solution remains colorless up to the equivalent point. A fraction of a drop of excess titrant turns the solution a definitive blue.
3- Redox indicators: most types of redox titrations are detected using redox indicators. These are highly colored dyes that are weak reducing or oxidizing agents that can be oxidized or reduced, the colors of the oxidized and reduced forms are deferent. The oxidation state of the indicator and its color will depend on the potential of the solution. A half-reaction and Nernst equation can be written for the indicator
Ox indicator + ne-↔ Redindicator
So the ratio [Redin]/[Oxin], and therefore the color will change as the potential of the solution changes. The redox indicator reaction must be rapid and eversible, if the reaction is slow or is irreversible, the color change will be gradual and a sharp end point will be not detected.
16-2 Finding the end point
• A redox indicator is a compound that changes color when it goes from its oxidized to its reduced
state.
or
For ferroin, with E° = 1.147 V
we expect the color change to
occur in the approximate range
1.088 V to 1.206 V with respect SHE
A redox titration is feasible if the difference between analyte and titrantis > 0.2 V.If the difference in the formal potential is > 0.4 V, then a redox indicator usually gives a satisfactory end point.
• Starch is the indicator of choice for those procedures involving iodine because it forms an intense blue complex with iodine. Starch is not a redox indicator; it responds specifically to the presence of I2, not to a change in redox potential.
• The active fraction of starch is amylose, a polymer of the sugar α-d-glucose.
• In the presence of starch, iodine forms I6 chains inside the amylose helix and the color turns dark blue
Starch-Iodine Complex
17D Oxidation/Reduction Indicators
• Self-indication:If the titrant is highly colored, this color may be
used to detect the end point.
Ex : MnO4- Mn2+
purple faint pink
• Starch indicator:
This indicator is used for titrations involving iodine
Starch + I2 dark-blue color complex
• Redox Indicators:These are highly colored dyes that are weak
reducing or oxidizing agents that can be oxidized or reduced
Oxind + ne- Redind
ind
ind0indind Ox
Redlog
n
0.059EE
10
1
]In[
]In[
ox
red 10
]In[
]In[
ox
red
A potential equal to 2×(0.059/n)V is required for a sharp color change
n = 1 0.12Vn = 2 0.060V
The redox indicator reaction must be rapid and reversible.
Table 17.2Ex:
(1) Ferroin: [tris(1,10-phenanthroline)ion(II) sulfate]
for titrations with cerium(IV)(2) Starch/Iodine soln.
Redox indicator
In(oxidized) + ne In(reduced)
E = Eo – (0.05916 / n) log [In(reduced)] / [In(oxidized)]
[In(reduced)] = [In(oxidized)] E = Eo ± (0.05916 / n)
[In(reduced)] / [In(oxidized)] (10/1) [In(reduced)]
[In(reduced)] / [In(oxidized)] (1/10) [In(oxidized)]
Indicator transition range = transition range – E(calomel) vs SCE vs SHE
Color changes for general redox indicators depend only on the potential of the system.The range of potentials over which a color change occurs (the transition potential) is often pH dependent.
Iron complexes of 1,10-phenanthrolines (orthophenanthrolines)
Starch-Iodine complex
Starch solution(05~ 1%) is not redox indicator.
The active fraction of starch is amylose, a polymer of the sugar -D-glucose ( 1,4 bond).
The polymer exists as a coiled helix into which small molecules can fit.
In the presence of starch and I–, iodine molecules form long chains of I5– ions
that occupy the center of the amylose helix.
••••[I I I I I]– ••••[I I I I I]– ••••
Visible absorption by the I5– chain bound within the helix gives rise to the
characteristic starch-iodine color.
Structure of the repeating unit of the sugar amylose.
Schematic structure of the starch-iodine complex. The amylose chain forms a helix around I6 unit.
View down the starch helix, showing iodine, inside the helix.
Potentiometric end points for redox titrations:
reference electrode || analyte solution | Pt
Reference electrode : SCE
The end point can be determined from a plot of the measured potential as afunction of titrant volume.
E(vs SCE)
Titrant volume
4. ADJUSTMENT OF OXIDATION STATE
16-3 Adjustment of analyte oxidation state
• Sometimes one needs to adjust the oxidation state of analyte before it can be titrated.
• Pre-adjustment must be quantitative and one must eliminate excess pre-adjustment reagent so that it won’t interference the subsequent titration.
• Pre-oxidation: powerful oxidants that can be easily removed after preoxidation include peroxydisulfate, silver(II) oxide, sodium bismuthate.
Definition:• Disproportionation: a reactant oxidizes and reduces itself, such as H2O2 in
boiling water. • Pre-reduction: Process of reducing an analyte to a lower oxidation state
prior to performing a titration with an oxidizing agent.• Amalgam: a solution of anything in mercury.
ADJUSTMENT OF ANALYTE OXIDATION STATES
Before titration we adjust the oxidation state of the analyte, eg Mn2+ can be pre-oxidised to MnO4
- and then titrated with Fe2+
Pre-adjustment must be quantitative and all excess reagent must be destroyed.
Pre-oxidation :
Persulphate S2O82- is a powerful oxidant that requires Ag+ as a
catalyst: S2 O82- + Ag+ SO4
2- + SO4- + Ag2+
Two powerful oxidants
Excess reagent is destroyed after by boiling the solution after oxidation is complete
2S2O82- + 2 H2O boiling 4SO4
2- + O2 + 4H+
H2O2 is a good oxidant in basic solution and reductant in acidic solution. The excess spontaneously disproportionate in boiling water.
H2O2 H2O + O2
PRE-REDUCTION
Stannous & chromous chloride, SO2 , H2S are used to pre-reduce analytes to a lower oxidation state.
An important pre-reduction technique uses a packed column to pre-reduce analyte to a lower oxidation state (analyte is drawn by suction).
Jones reductor, which contains Zn coated with Zn amalgam. Zn is a powerful reducing agent (Eo = -0.764V) making the Jones reductor unselective, other species eg, Cr3+ are reduced and may interfere with the titration analysis.
Preadjustment of analyte oxidation stateIt is necessary to adjust the oxidation state of the analyte to one that can be titrated
with an auxiliary oxidizing or reducing agent.
Ex. Preadjustment by auxiliary reagent
Fe(II), Fe(III) Fe(II)4–
Titration
Ce4+
Preoxidation : Peroxydisulfate ( (NH4)2S2O8 )
2– )
Sodium bismuthate ( NaBiO3)
Hydrogen peroxide (H2O2)
Prereduction : Stannous chloride ( SnCl2 )
Chromous chloride
Jones reductor (zinc coated with zinc amalgam)
Walden reductor ( solid Ag and 1M HCl)
Jones reductor :
2Zn (s) + Hg2+ Zn2+ + Zn(Hg) (s)
Redox Titrations Common Redox Reagents
2.) Adjustment of Analyte Oxidation State Before many compounds can be determined by Redox Titrations, must be
converted into a known oxidation state- This step in the procedure is known as prereduction or preoxidation
Reagents for prereduction or preoxidation must:- Totally convert analyte into desired form- Be easy to remove from the reaction mixture- Avoid interfering in the titration
Examples:- Preoxidation:
a) Peroxydisulfate or persulfate (S2O82-) with Ag+ catalyst
Powerful oxidants
Oxidizes Mn2+, Ce3+, Cr3+, VO2+
excess S2O82- and Ag+ removed by boiling the solution
Redox Titrations Common Redox Reagents
2.) Adjustment of Analyte Oxidation State Examples:
- Preoxidation:b) Silver(II) oxide (AgO) in concentrated mineral acids also yields Ag2+
excess removed by boiling
c) Hydrogen peroxide (H2O2) is a good oxidant to use in basic solutionsOxidizes Co2+, Fe2+, Mn2+
Reduces Cr2O72-, MnO4-
excess removed by boiling
- Prereduction:a) Stannous chloride (SnCl2) in hot HCl
Reduce Fe3+ to Fe2+
excess removed by adding HgCl2b) Jones reductor (Zn + Zn amalgam – anything in mercury)
5. PERMANGANAT
16-4 Oxidation with potassium permanganate
• KMnO4 is a strong oxidant with an intense violet color. In strongly acidic solutions (pH < 1), it is reduced to Mn2+.
• In neutral or alkaline solution, it is reduced to brown solid MnO2.
• In strongly alkaline solution ( 2 M NaOH), green manganate ion (MnO4
2-) is produced.
Dr. S. M. Condren
Redox Titrations Common Redox Reagents
3.) Common Titrants for Oxidation Reactions Potassium Permanganate (KMnO4)
- Strong oxidant- Own indicator
Titration of VO2+ with KMnO4
Before Near AfterEquivalence point
Eo = 1.507 VViolet colorless
pH ≤ 1
Eo = 1.692 V
pH neutral or alkaline
Violet brown
pH strolngly alkaline
Eo = 0.56 VViolet green
Using KMnO4 as Oxidising Agents
• One of the most common oxidising agent used in redox titrations is KMnO4.
• KMnO4. reacts differently in both acid and bases.
In Acidic Solution–In acid solutions, the reduction of KMnO4 can be represented by the following equation:
MnO4- + 8H+ + 5e- Mn2+ + 4H2O
Sulfuric acid is the most suitable acid, as it does not react with permanganate in dilute solutionsWith hydrochloric acid, there is a likelihood of the reaction
2MnO4- +10Cl- + 16H+ +2e- Mn2+ + 4H2O + 5Cl2
Using KMnO4. as Oxidising Agents
In Basic Solution– In basic solutions, the reduction of KMnO4
will be reduced to form manganese dioxide and can be represented by the following equation:
MnO4- + 8H20 + 2e- MnO2 + 4OH-
Permanganate TitrationIn using Potassium permanganate, the following conditions must be used
–Stored at low temperatures and away from light
–Experiments performed at 60oC–Sulfuric acid used in the titration
Permanganate Titration
• The reason for refrigeration
• Refrigeration is used to prevent the degradation of potassium permanganate by light.
The reason why the experiment was performed at 60◦C
• The reaction does not proceed if the temperature is low. A temperature greater than 60 ◦C is necessary for this experiment. The oxidative reaction by potassium permanganate in sulfuric acid is sometimes performed even at 100 ◦C. The reaction why sulfuric acid is used in this experiment
• If hydrochloric acid is used, Cl- is oxidized by potassium permanganate. Since nitric acid itself is oxidizing agent, its use is also inappropriate.
Common oxidising agents• Problems using potassium permanganate
– The solid is rarely pure– The solutions are relatively unstable due to partial
decomposition of the permanganate ion
– The production of MnO2 has a catalytic effect, increasing the rate of decomposition
– Dust and organic matter in the air also cause oxidation to MnO2
– Solutions of permanganate are stored in dark bottles away from sunlight to prevent decomposition
)l(2)s(2)aq()aq(4 OH2MnOe3H4MnO
Standardisation of potassium permanganate
• Potassium permanganate cannot be used as a primary standard because it is unstable
• Ammonium iron sulfate, sodium oxalate and oxalic acid are used as primary standards to standardise potassium permanganate
• The permanganate goes in the buret and the primary standard in the conical flask
Titration Setup for volumetric analysis using permanganate
Permanganate goes in the buret
Indicator is unnecessary
Primary standard solution goes in the
conical flask
End Point• Acidified potassium permanganate solution,
which is a deep purple in colour, is a commonly used secondary standard. It has the advantage of being self indicating as its products are almost colourless
Common titrations using potassium permanganate solution as the oxidising agent
Reducing agent Redox equation Endpoint, (KMnO4 in
burette) Oxalic acid solution H2C2O4
2MnO4- + 6H+ + 5 H2C2O4 10CO2 + 2Mn2+ +
8H20 Colourless to a faint pink
Ammonium iron(II) sulphate (NH4)2Fe(SO4)2
MnO4- + 8H+ + 5Fe2+ 5Fe3+ + 2Mn2+ + 4H20
This strong electrolyte to dissociate producing NH4
+, Fe2+ and SO42-. Of these ions only Fe2+ is
involved in the reaction
Colourless to faint pink
Hydrogen peroxide H2O2
2MnO4- + 6H+ + 5H2O2 5O2 + 2Mn2+ + 8H20
Colourless to faint pink
Problem : Redox Titration- I
Volume (L) of KMnO4 Solution
Moles of KMnO4
M (mol/L)
Molar ratioin redox reaction.
Chemical Formulas
Moles of CaC2O4
Problem: Calcium Oxalate was precipitated from 1.00 mL blood by the addition of Sodium Oxalate so the Ca2+ conc. in the blood could be determined. This precipitate wasdissolved in a sulfuric acid solution,which then required 2.05 mL of 4.88 x 10-4 M KMnO4 to reach the endpoint via the rxn. a) Calculate the moles of Ca2+.b) Calculate the Ca2+ conc. in blood.Plan: a) Calculate the moles ofCa2+ in the H2SO4 solution (and blood sample).b) Convert the Ca2+ conc.into units of mg Ca2+/ 100 mL blood.
Moles of Ca+2
a)
b)
c)
Equation:2 KMnO4 (aq) + 5 CaC2O4 (aq) + 8 H2SO4 (aq) 2 MnSO4 (aq) + K2SO4 (aq) + 5 CaSO4 (aq) + 10 CO2 (g) + 8 H2O(l)
Problem :Redox Titration - Calculation - II
a) Moles of KMnO4
b) Moles of CaC2O4
c) Moles of Ca+2
Problem : Redox Titration - III
Moles of Ca2+/ 1 mL of blood
Moles of Ca2+/ 100 mL blood
Mass (g) of Ca2+/ 100 mL blood
Mass (mg) of Ca2+ / 100 mL blood
multiply by 100 a) Calc of mol Ca2+ per 100 mL
M (g/mol) b) Calc of mass of Ca2+ per 100 mL
1g = 1000mg c) convert g to mg!
Standardisation of permanganate
• Using oxalic acid – H2C2O4
• The half reactions are:
)l(22
)aq()aq(-4(aq) OH4Mne5H8MnO
} x 5
Overall:
e2H2CO2OCH )aq()g(2)aq(422
} x 2
)g(2)l(22
)aq()aq()aq(4)aq(422 CO10OH8Mn2H6MnO2OCH5
Example: Standardisation of permanganate with oxalic acid
Example:
• A 20 mL solution of 0.0512 M oxalic acid was used to standardise an unknown solution of potassium permanganate. An average of 21.24 mL of permanganate was required.
• Calculate the concentration of the potassium permanganate solution
Apply solution map
Example: A 20 mL solution of 0.0512 M oxalic acid was used to standardise an unknown solution of potassium permanganate. An average of 21.24 mL of permanganate was required.Calculate the concentration of potassium permanganate solution
Given: 20 mL of 0.0512 M H2C2O4, 21.24 mL K2MnO4
Find: Conc. K2MnO4 (moles/L)
SM: L H2C2O4→mol. H2C2O4→mol. K2MnO4→conc. K2MnO4
44
422
4422422 MnO mol 10096.4
OCH mol 5
MnO mol 2
L 1
OCH mol 0512.0
mL 1000
1LOCH mL 0.20
L 02124.0mL 1000
L 1mL 21.24 MnO liters -
4
M 0193.0L 02124.0
MnO moles10096.4
solution liters
solute moles molarity 4
-4
)g(2)l(22
)aq(
)aq()aq(4422
CO10OH8Mn2
H6MnO2OCH5
Question
• 25 mL of 0.0627 M acidified oxalic acid solution required 31.2 mL of potassium permanganate solution for complete oxidation.
• Calculate the concentration of permanganate solution
DICHROMATE
Volumetric analysis using potassium dichromate
• Potassium dichromate can be used as an oxidising agent
• It has the advantages that it forms stable solutions and can be used as a primary standard
• The half reaction is:
)l(23
)aq()aq(2
)aq(72 OH7Cr2e6H14OCr
Oxidation with potassium dichromate
Cr2O72– + 14H+ + 6e = 2Cr3+ + 7H2O Eo = 1.36 V
K2Cr2O7 is a primary standard.
Indicator : diphenylamine sulphonic acid
Ex. Redox titration ( hydroquinone vs dichromate standard solution )
HO OH O O + 2H+ + 2e Eo= 0.700
Cr2O72– + 14H+ + 6e 2 Cr3+ + 7 H2O Eo=
1.33
3
3 HO
OH + Cr2O72– + 8H+ 3 O O + 2 Cr3+ + 7
H2O
Eo= Eocathode – Eo
anode = 1.33 – 0.700 = 0.63 V
K = 10 nEo/0.05916 = 10 6(0.63) / 0.05916 = 10 64
redox indicator : diphenylamine
colorless to violet
Very large : quantitative : complete reaction
CERIUM (CE4+)
16-5 Oxidation with Ce4+
• Reduction of Ce4+ to Ce3+ proceeds cleanly in acidic solutions.
• The aquo ion, Ce(H20)n4+, probably does not exist in any
of these solutions, because Ce(IV) binds anions (ClO4-,
SO42- NO3
-, Cl-) very strongly.
Oxidation with Ce4+
Ce4+ + e = Ce3+ 1.7 V in 1 N HClO4
yellow colorless 1.61 V in 1N HNO3
1.47 V in 1N HCl
1.44 V in 1M HSO4
Indicator : ferroin, diphenylamine
Preparation and standardization:
Ammonium hexanitratocerate, (NH4)2Ce(NO3)6, (primary standard grade)
Ce(HSO4)4, (NH4)4Ce(SO4)4·2H2O
Standardized with Sodium oxalate.
Applications of cerimetry
(1) Menadione (2-methylnaphthoquinon: vitamin K3)
O
O
CH3
OH
OH
CH3
2 Ce(SO4)2
HCl, Zn
Reduction
(2) Iron
2FeSO4 + 2 (NH4)4Ce(SO4)4 = Fe2(SO4)3 + Ce2(SO4)3 + 4 (NH4)2SO4
IODINE
Titration Involving Iodine: Iodimetry And Iodometry
Redox titrations are among the most important types of analysis performed in many areas of application. Example in clinical laboratories are rare, since most analyses are for traces, but these titrations are still extremely useful for standardizing reagents.
A- Iodimetry
iodine is a moderately strong oxidizing agent and can be used to titrate reducing agents.
Titration with I2 are called Iodimetric methods. These titrations are usually performed in neutral or mildly alkaline (pH 8) to weakly acid solutions. If the pH is too alkaline, I2 will disproportionate o hypoiodate and iodide:
I2 +2OH = IO- + I- +H2O
There are three reasons for keeping he solution from becoming strongly acidic:
1-the starch used for the end point detection tends to hydrolyze or decompose in strong acid which will affect the EP.
2-the reducing power of several reducing agents is increased in neutral solution.
3-the I- produced in the reaction tends to be oxidized by dissolved oxygen in acid solution:
4I- +O2 + 4H+ → 2I2 +2H2O
Iodine has a low solubility in water but the complex,I3-, is
very soluble. So iodine solutions are prepared by dissolving I2 in a concentrated solution of potassium iodide:
I2 + I- → I3-
I3- IS Therefore the actual species used in the reaction.
IodometryIodide ion is a weak reducing agents and will reduce strong
oxidizing agents. It is not used, however, as a titrant mainly because of lack of convenient visual indicator system, as well as other factors such as speed of the reaction.
When an excess of iodide is added to a solution of an oxidizing agent I2 is produced in an amount equivalent to the oxidizing agent present.
This I2 can therefore be titrated with a reducing agent and the result will be the same as if the oxidizing agent were titrated directly. The titrating agent used is SODUIM THIOSULFATE.
Analysis of an oxidizing agent in this way is called an Iodometric method. Consider, for example, the determination of dichromate:
Cr2O72- + 6I- (excess) + 14H+ → 2Cr 3+ +3I2 + 7H2O
I2 + 2S2O3 2- → 2I- + S4O6
2-
Why not titrate the oxidizing agents directly with the thiosulfate? Because strong oxidizing agents oxidize thiosulfate to oxidation states higher than that of terathionate, but the reaction is generally not stochiometric.
The end point for iodometric titrations is detected with starch. The disappearance of the blue starch-I2 color indicates the end of the titration. The starch is not added at the beginning of the titration when iodine concentration is high. instead, it is added just before the EP when the dilute iodine color becomes pale yellow.
There are tow reasons for such timing:1- the iodine-starch complex is only slowly dissociated, and
a diffuse end point would result if a large amount of the iodine were adsorbed on the starch.
2- most iodometric titrations are performed in strongly acid medium and the starch has a tendency to hydrolyze in acid solution.
The titration should be performed rapidly to minimize air oxidation of the iodide. Stirring should be efficient in order to prevent local excesses of thiosulfate, because it is decomposed in acid solution.
In iodometric methods, a large excess of iodide is added to promote the reaction. The unreacted iodide dose not interfere, but it may be air-oxidized if the titration is not performed immediately.
Sodium thiosulfate solution is standardized iodometrically against a pure oxidizing agent such as K2Cr2O7, KIO3, KBrO3, or metallic copper (dissolved to give Cu2+). With potassium dichromate, the deep green color of the resulting chromic ion makes it a little more difficult to determine the iodine-starch end point.
Common Titrant for Oxidation ReactionsIodine (Solution of I2 + I-)
I3- is actual species used in titrations with iodine
Either starch of Sodium Thiosulfate (Na2S2O3) are used as indicator
K = 7 x 102
Before endpoint
Before endpoint
At endpoint
I3- + StarchI3
- I3- + S2O3
2-
16-7 Methods Involving Iodine
• Iodimetry: a reducing analyte is titrated directly with iodine (to produce I−).
• iodometry, an oxidizing analyte is added to excess I− to produce iodine, which is then titrated with standard thiosulfate solution.
• Iodine only dissolves slightly in water. Its solubility is enhanced by interacting with I-
• A typical 0.05 M solution of I2 for titrations is prepared by dissolving 0.12 mol of KI plus 0.05 mol of I2 in 1 L of water. When we speak of using iodine as a titrant, we almost always mean that we are using a solution of I2 plus excess I−.
Preparation and Standardization of Solutions
• Acidic solutions of I3- are unstable because the excess I−
is slowly oxidized by air:
• In neutral solutions, oxidation is insignificant in the
absence of heat, light, and metal ions. At pH 11, ≳triiodide disproportionates to hypoiodous acid (HOI), iodate, and iodide.
• An excellent way to prepare standard I3- is to add a
weighed quantity of potassium iodate to a small excess of KI. Then add excess strong acid (giving pH ≈ 1) to produce I2 by quantitative reverse disproportionation:
Volumetric analysis using iodine
• Iodine (I2) is used as a weak oxidising agent
• The half equation is:
• Starch is used as an indicator. Excess iodine causes the formation of a blue starch-iodine complex
)aq()aq(2 I2e2I
Starch
Analysis of wine
• The sulfur dioxide content of wine can be determined by oxidation with iodine
• The iodine only reacts with the sulfur dioxide in wine and not with the alcohols (as would permanganate or dichromate)
• The oxidation reaction for SO2 is: e2H4SOOH2SO )aq(
2)aq(4)l(2)g(2
16-7 Methods Involving Iodine
• Iodimetry: a reducing analyte is titrated directly with iodine (to produce I−).
• iodometry, an oxidizing analyte is added to excess I− to produce iodine, which is then titrated with standard thiosulfate solution.
• Iodine only dissolves slightly in water. Its solubility is enhanced by interacting with I-
• A typical 0.05 M solution of I2 for titrations is prepared by dissolving 0.12 mol of KI plus 0.05 mol of I2 in 1 L of water. When we speak of using iodine as a titrant, we almost always mean that we are using a solution of I2 plus excess I−.
Preparation and Standardization of Solutions
• Acidic solutions of I3- are unstable because the excess I−
is slowly oxidized by air:
• In neutral solutions, oxidation is insignificant in the
absence of heat, light, and metal ions. At pH 11, ≳triiodide disproportionates to hypoiodous acid (HOI), iodate, and iodide.
• An excellent way to prepare standard I3- is to add a
weighed quantity of potassium iodate to a small excess of KI. Then add excess strong acid (giving pH ≈ 1) to produce I2 by quantitative reverse disproportionation:
Bromatimetry
KBrO3 BrO3– + 5Br– + 6H+ 3Br2 + H2O
2I– + Br2 I2 + 2Br–
I2 + 2 S2O32– 2I– + 2S4O6
2–
Substitution reactions BrO3– + 5Br– + 6H+ 3Br2 + H2O
2I– + Br2 I2 + 2Br– I2 + 2 S2O3
2– 2I– + S4O62–
pH 4-9
Al3+ + 3HOC9H6N Al(OC9H6N)3 (s) + 3H+
hot 4M HCl
Al(OC9H6N)3 (s) 3HOC9H6N + Al3+
3HOC9H6N + 6 Br2 3HOC9H4NBr2 + 6HBr
1 mol Al3+ 3 mol HOC9H6N 6 mol Br2 2 mol KBrO3
Addition reactions
Determining water with the Karl Fisher Reagent
The Karl Fisher reaction :
I2 + SO2 + 2H2O 2HI + H2SO4
For the determination of small amount of water, Karl Fischer(1935) proposed a reagent prepared as an anhydrous methanolic solution containing iodine, sulfur dioxide and anhydrous pyridine in the mole ratio 1:3:10. The reaction with water involves the following reactions :
C5H5N•I2 + C5H5N•SO2 + C5H5N + H2O 2 C5H5N•HI + C5H5N•SO3
C5H5N+•SO3– + CH3OH C5H5N(H)SO4CH3
Pyridinium sulfite can also consume water.
C5H5N+•SO3– + H2O C5H5NH+SO4H–
It is always advisable to use fresh reagent because of the presence of various side reactions involving iodine. The reagent is stored in a desiccant-protected container.
The end point can be detected either by visual( at the end point, the color changes from dark brown to yellow) or electrometric, or photometric (absorbance at 700nm) titration methods. The detection of water by the coulometric technique with Karl Fischer reagent is popular.
Pyridine free Karl Fisher reagent
In recent years, pyridine, and its objectionable odor, have been replaced in the Karl Fisher reagent by other amines, particularly imidazole.
(1) Solvolysis 2ROH + SO2 RSO3– + ROH2
+
(2) Buffering B + RSO3– + ROH2
+ BH+SO3R– + ROH
(3) Redox B•I2 + BH+SO3R– + B + H2O BH+SO4R– + 2 BH+I–
Applications
• Combustion• Bleaching• Batteries and fuel cells• Metallurgy• Corrosion• Respiration
Copyright © 2008 Pearson Prentice Hall, Inc.
Chapter 1/176
Determination of Amount of Vitamin C in a Sample by Redox Titration
Experiment 3
Experiment 3• Goal:
– To detemine the amount of vitamin C (L-ascorbic acid) in a sample
• Method:– Use AA/DCP redox titration reaction – Perform titrations with known concentrations– Perform titrations with sample of unknown vit
C concentration
Vitamin C Oxidation Half-Reaction
Oxidation: Loss of e-
C
C C
O
C
O
C
H2C OHH
OH
HO
HO
H
Ascorbic Acidreduced,
vitamin form
C
C C
O
C
O
C
H2C OHH
OH
O
O
H
Ascorbic Acidoxidized,
excreted form
+ 2e- + 2H+
Reduced L-Ascorbic
Acid(vitamin)
Oxidized L-Dehydroascorbic
Acid(excreted)
DCP Reduction Half-Reaction
N
Cl
Cl
O
OH
Oxidized DCP(Dark Blue in
BasePink in acid)
NH
Cl
Cl
HO
OH
Reduced DCP(Colorless in acid)
+ 2e- + 2H+
Reduced DCP Colorless
Oxidized DCP Dark blue in base Red-pink in acid
Reduction: Gain of e-
DCP = Dichloroindophenol
Redox Reaction
e2H2AAAA oxred
redox DCPe2H2DCP
redoxoxred DCP AA DCP AA
Oxidation:
Reduction:Overall:
Redox Reaction
redoxoxred DCP AA DCP AA colorless pink colorless colorless
DCP: titrant and indicator
Solution: colorless until all AAred is gone
Then: excess DCPox appears pink
Amount Profiles
mL DCP added
mole
s
AAred
DCPr
ed
AAox
AAred = 0
Titration
DCP titrant
Vit C analyte
At as close to endpoint as possible:
Pink should fade after 30 seconds
Part 1•Standardization: Find mg AA/mL DCP
Use 10.00 mL AA solution VAA=10.00 mL• of known conc. in ~30 mL buffer CAA
Titrate AA with DCP to endpoint VDCP
Repeat 4 times (1* trial, 3 good)
Find mg AA consumed per mL DCP mgAA/mL DCP
Part 1•Standardization mg AA/mL DCP example
– 0.0500 g AA weighed– Dissolved in 100.0 mL– 10.00 mL aliquot used– 20.12 mL DCP required
DCP mL
AA mg 0.246
AA g 1
mgAA1000
DCP mL 20.12
mLAA 0.001
mL 100.0
gAA0500.0
weighed out
total volume
mL AA used
mL DCP req’d
Part 2
•Determine AA content in sampleUse Kool-Aid providedEstimate mg AA in sample mgAAest
Want to use ~10-20 mL DCP Calculate required dilution dilution factor• and aliquot volume Valiqout
Titrate 4 times (1* trial; 3 consistent) VDCP
Aliquot dilution OK
Part 2•Estimated AA content desired example
•Want to use ~ 10 - 20 mL DCP…•How many mg AA desired in sample aliquot?
AA mg 92.4DCP mL
AA mg .2460DCP mL 20
from Part 1
Part 2•AA content in Countrytime® lemonade example
– Serving: 18 g– Vit C: 10% RDA – RDA: 60 mg
CTL g 8.14AA mg 6
CTL g 18AA mg .924
AA in CTLAmount CTL to use
CTL g 18
AA mg 6so
AA desired
Part 2•Vit C content in fruit example•1. 5.00 g fruit → 20.00 mL acid & prepped•2. 5.00 mL liquid → diluted to 30.0 mL•3. 13.13 mL DCP req’d
•Find mg AA/gram fruit:
fruit g
AA mg 2.58
fruit g 5.00
nsol'AA mL 0.002
DCP mL
AA mg .2460
nsol'AA mL 5.00
DCL mL 13.13
mL DCP req’d
sample volume
from Part 1 sol’n to fruit conversion
PROBLEMS
Problems
• Would indigo tetrasulfonate be a suitable redox indicator for the titration of Fe(CN)6
4- with Tl3+ in 1 M HCl?
• Solution:
Standard potentials: indigo tetrasulfonate is 0.36 V;
Fe(CN)63-/ Fe(CN)6
4- is 0.356 V;
Tl3+/Tl+, 0.77 V.
The end-point potential will be between 0.356 and 0.77 V. Indigo tetrasulfonate changes color near 0.36 V. Therefore it will not be a useful indicator for this titration.
• Given the following data select the best indicator for the titration
of iron(II) with thallium(III), using a S.C.E. reference electrode.
(a) methylene blue(b) diphenylamine sulfonic acid(c) diphenylamine
• Which of the following titrations would give a titration curve symmetric around the equivalence point?
(a) Na2S2O3 titrated with I2.(b) ascorbic acid titrated with I2.(c)As4O6 titrated with I2.
Problems
• Would indigo tetrasulfonate be a suitable redox indicator for the titration of Fe(CN)6
4- with Tl3+ in 1 M HCl?
• Solution:
Standard potentials: indigo tetrasulfonate is 0.36 V;
Fe(CN)63-/ Fe(CN)6
4- is 0.356 V;
Tl3+/Tl+, 0.77 V.
The end-point potential will be between 0.356 and 0.77 V. Indigo tetrasulfonate changes color near 0.36 V. Therefore it will not be a useful indicator for this titration.
H18 – C4
• 4.22, 4.23, 4.29, 4.85, 4.87, 4.91*, 4.107*, 4.113*
Gustav Robert Kirchhoff(1824-1877) was a German physicist who made many important contributions to physics and chemistry. In addition to his work in spectroscopy, he is known for Kirchhoff’s laws of current and voltage in electrical circuits. These laws can be summarized by the following equations: I = 0 and E = 0. These equations state that the sum of the currents into any circuit point (node) is zero and the sum of the potential differences around any circuit loop is zero.