1. Redox reactins half-reactions: Reduction 2Fe3+ + 2e-  2Fe2+
2Fe3+ + Sn2+  2Fe2+ + Sn4+
half-reactions:
Reduction
2Fe3+ + 2e-  2Fe2+
oxidation
Sn2+  Sn4+ + 2e-

2. Redox reactins occurring in 1) solution 2) electrochemical cell.
2Fe3+ + Sn2+  2Fe2+ + Sn4+

3. Electrochemical Reactions
1)chemical  electric:
primary cell (Galvanic cell)
2)electric  chemical:
electrolytic cell

4. Standard Reduction Potentials
Reduction Half-Reaction
E(V)
F2(g) + 2e-  2F-(aq)
2.87
Au3+(aq) + 3e-  Au(s)
1.50
Cl2(g) + 2 e-  2Cl-(aq)
1.36
Cr2O72-(aq) + 14H+(aq) + 6e-  2Cr3+(aq) + 7H2O
1.33
O2(g) + 4H+ + 4e-  2H2O(l)
1.23
Ag+(aq) + e-  Ag(s)
0.80
Fe3+(aq) + e-  Fe2+(aq)
0.77
Cu2+(aq) + 2e-  Cu(s)
0.34
Sn4+(aq) + 2e-  Sn2+(aq)
0.15
2H+(aq) + 2e-  H2(g)
0.00
Sn2+(aq) + 2e-  Sn(s)
-0.14
Ni2+(aq) + 2e-  Ni(s)
-0.23
Fe2+(aq) + 2e-  Fe(s)
-0.44
Zn2+(aq) + 2e-  Zn(s)
-0.76
Al3+(aq) + 3e-  Al(s)
-1.66
Mg2+(aq) + 2e-  Mg(s)
-2.37
Li+(aq) + e-  Li(s)
-3.04
Ox. agent strength increases
Red. agent strength increases 4

5. Balancing of redox reactions. Under Acidic conditions
1. Identify oxidized and reduced species
Write the half reaction for each.
2. Balance the half rxn separately except H & O’s.
Balance: Oxygen by H2O
Balance: Hydrogen by H+
Balance: Charge by e -
3. Multiply each half reaction by a coefficient.
There should be the same # of e- in both half-rxn.
4. Add the half-rxn together, the e - should cancel.

6. Balancing of redox reactions. Under Basic conditions
1. Identify oxidized and reduced species
Write the half reaction for each.
2. Balance the half rxn separately except H & O’s.
Balance: Oxygen by H2O
Balance: Hydrogen by OH-
Balance: Charge by e -
3. Multiply each half reaction by a coefficient.
There should be the same # of e- in both half-rxn.
4. Add the half-rxn together, the e - should cancel.

7. Balancing of redox reactions
H2O2 (aq) + Cr2O7-2(aq )  Cr 3+ (aq) + O2 (g) Redox reaction
======================================
1)write 2 half reactions
Half Rxn (oxid):
Cr2O7-2 (aq)  Cr3+
Half Rxn (red):
H2O2 (aq)  O2
2)Atom balance
Cr2O7-2 (aq)  2Cr3+

8. Balancing of redox reactions
3)Oxygen balance
Half Rxn (oxid): Cr2O7-2 (aq)  2Cr H2O
Half Rxn (red): H2O2 (aq)  O2
4)Hydrogen balance
Half Rxn (oxid): 14H+ + Cr2O7-2 (aq)  2Cr H2O
Half Rxn (red): H2O2 (aq)  O H+
5)Electron balance
6e H+ + Cr2O7-2 (aq)  2Cr H2O
H2O2 (aq)  O H e-

9. Balancing of redox reactions
6) Equalize of produced and consumed electrons
6e H+ + Cr2O7-2 (aq)  2Cr H2O
( H2O2 (aq)  O H e- ) x 3
7)Multiply each half reaction
8 H H2O2 + Cr2O72-  2Cr O H2O

10. تیتراسیونهای Redox - کالریمتری - واکنشهای معمولی (با استفاده از معرف)
- کالریمتری - واکنشهای معمولی (با استفاده از معرف)
- پتانسیومتری - واکنشهای الکتروشیمی(با استفاده از پتانسیومتر)

11. آنالیت باید در یک حالت اکسایش باشد ( Fe3+یا (Fe2+
کاهنده کمکی
آنالیت باید در یک حالت اکسایش باشد ( Fe3+یا (Fe2+
Zn-Al-Cd-Pb-Ni
- ملغمه روی
2Zn(s)+Hg2+→ Zn2++ Zn(Hg) (s)
ملغمه روی باعث کاهش +Hنمی شود

12. اکسنده کمکی سدیم بیسموتات NaBio3 (s) + 4H++ 2e- → BiO+ +Na++ 2H2O
اضافی ← صاف کردن
آمونیوم پراکسی دی سولفات
2e-+S2O82-→ SO E o = 2.0)
اضافی ← جوشاندن
هیدروژن پراکسید
H2O2+2H++2e- → 2H2O Eo=1.78
اضافی ← جوشاندن

13. اکسنده های استاندارد: MnO4-→ Mn2+ Eo=1.51 Ce4+ → Ce3+ Eo=1.44
Cr2O72- → Cr Eo=1.30
I Eo= → I2
Eo پایین ید می تواند عوامل کاهنده قوی را در حضور عوامل کاهنده ضعیف اندازه گیری نماید.

14. شناساگرها شناساگر معروف: فروئین چسب نشاسته
در منگانومتری شناساگر لازم نیست

15. تهيه محلول پرمنگنات با محاسبه پرمنگنات توزين و ... ← گرم ← صاف ←
با محاسبه پرمنگنات توزين و ... ←
گرم ←
صاف ←
استاندارد
2MnO4-+5H2C2O4+6H+ → 2Mn CO2(g) +8H2O
واکنش کند است
Mn2+ کاتالیز می کند
(در ابتدا واکنش خیلی کند است –بعد سریع می شود)

16. پایداری نقطه پایانی منگنانومتری:
2MnO4-+3Mn2++ 2 H2O → 5MnO2(S) + 4H+
K=1*1047
در محیط باقی نمی ماند
ولی سرعت واکنش کم است
حدود 30 ثانیه رنگ باقی می ماند

17. پایداری محلولهای آبی پرمنگنات:
4MnO4-+2H2O→ 4MnO2(S) + 3O2+ 4 OH-
K تقریبا بالا است
ولی سرعت پایین است
و محلولها معمولا پایدار می باشد
نور-گرما-اسید-باز-Mn2+-MnO2 واکنش را کاتالیز می کنند.

18. ید- تیوسولفات روش مستقیم (یدیمتری) →2I- I2+S روش غیرمستقیم ( یدومتری )
I-+Soxi→I2

19. ید- تیوسولفات S4O62-+ - I2+2S2O32-→2I تیوسولفات←تتراتیونات
n= I2 →2I
S4 O n= S2O32-→
ید براي سنجش تيوسولفات منحصر به فرد است زيرا
سایر اکسنده های قوی تتراتیونات راهم اکسید می کنند

20. تهیه محلول آبکی ید: [ I2]=0.001 M حلالیت در آب I2+I-⇄ I3- K=700
ید فرار است
4I-+ O2(g)+ 4H+→2I2+2H2O
اسید- گرما و نور واکنش را کاتالیز می کند.

21. تهيه محلول آبكي يد!!!
I2+I-↔ I3-
IO3-+5I-→3I2
2Cu2++4I-→2CuI+I2

22. يد در محيط قليائي I2+OH-→IO-+I-+H+ 3IO-→Io3-+2I-
تیتراسیون در محیط اسیدی ضعیف یا خنثی انجام می شود.

23. پايداري تيوسولفات S2O32-+H+→HSO3-+S(S)
pH محلول - غلظت محلول -- Cu2+ نور خورشید و میکروارگانیزم ها واکنش را تشدید می کنند.

24. Electrolysis of Copper
Concentration Cells
Electrolysis of Copper
A concentration cell based on the Cu/Cu2+ half-reaction. A, Even though the half-reactions involve the same components, the cell operates because the half-cell concentrations are different. B, The cell operates spontaneously until the half-cell concentrations are equal. Note the change in electrodes (exaggerated here for clarity) and the equal color of solutions.

25. Concentration Cells Cu+Cu2+ (1.0 M)Cu2+ (0.1M)+Cu
Cu│Cu2+ (1.0M)‖ Cu2+ (0.1 M)│Cu Anod cathod
E=E /2Log(0.1/1.0) =
Cu+Cu2+ (1.0 M)Cu2+ (0.1M)+Cu

26. General Chemistry: Chapter 21
Chemistry 140 Fall 2002
The pH Meter
Pt | H2 (1 atm)|H+(x M) ||H+(1.0 M) |H2(1 atm) | Pt(s)
H2(g, 1 atm) → 2 H+(x M) + 2 e-
2 H+(1 M) + 2 e- → H2(g, 1 atm)
H2(g, 1 atm) +2 H+(1 M) → 2 H+(x M) + H2(g, 1 atm)
2 H+(1 M) → 2 H+(x M)
General Chemistry: Chapter 21
Prentice-Hall © 2002
Slide 26 of 52

27. pH = Ecell /(0.059) Ecell = Ecell° - log Q n 0.059 V
2 H+(1 M) → 2 H+(x M)
Ecell = Ecell° log n
0.059 V x2 12
Ecell = log 2
0.059 V x2 1
Ecell = V log x
Ecell = (0.059 V) pH
pH = Ecell /(0.059)
Slide 27 of 52

28. Standard Hydrogen Electrode
It’s a primary reference electrode. Its potential is considered to be zero.
Electrode reaction:
half cell: pt, H2 / H+ (1N) 
Eo = zero
d-Limitation
It is difficult to be used and to keep H2­ gas at one atmosphere during all determinations.
It needs periodical replating of Pt. Sheet with Pt. Black

29. Calomel electrode
E volt
KCl
0.241
Saturated
0.280 1M
0.334
0.1 M

30. Ag/AgCl, electrode reaction Ag+ + e = Ago half cell
Ag/AgCl, saturated KCl ||
or 1 N KCl ||
or 0.1 N KCl ||
design
The Nernest equation for the electrode:

31. Ag/AgCl Disadvantage of silver-silver chloride electrode
It is more difficult to prepare than SCE.
AgCI in the electrode has large solubility in saturated KCl
Advantage of Ag-AgCI electrodes over SCE.
It has better thermal stability.
Less toxicity and environmental problems with consequent cleanup and disposal difficulties.

33. Indicator electrode Ecell=Eindicator-Ereference It must be:
(a) give a rapid response and
(b) its response must be reproducible.
Metallic electrodes: where the redox reaction takes place at the electrode surface.
Membrane (specific or ion selective) electrodes: where charge exchange takes place at a specific surfaces and as a result a potential is developed.

34. Electrodes for precipitemetry and complexometry
a- First-order electrodes for cations:
e.g. in determination of Ag+ a rode or wire of silver metal is the indicator electrode, it is potential is:
It is used for determination of Ag+ with Cl-, Br- and CN-. Copper, lead, cadmium, and mercury
b) Second order electrodes for anions
A metal electrode is also indirectly responsive to anions that form slightly soluble precipitates or stable complexes with its cation. The electrode reaction is AgCl + e = Ag+ + Cl-, and the electrode potential is given by:
E25 = EoAg/AgCl log [Cl-]

35. 2. Indicators electrodes for redox reaction:
Electrodes formed from platinum or gold
inert and the potential it developed depends upon the potential of oxidation-reduction systems of the solution in which it is immersed
for example the potential of a platinum electrode in a solution containing Ce(III) and Ce(IV)ions is given by

36. 3. indicator electrodes for neutralization reaction
Glass Membrane Electrode

37. Measurement of pH pH meters use electrochemical reactions.
Ion selective probes: respond to the presence of a specific ion. pH probes are sensitive to H3O+.
Specific reactions:
Hg2Cl2(s) + 2e- 2Hg(l) + 2Cl-(aq) E°1/2 = 0.27 V
E°1/2 = 0.0 V
H2(g) 2H+(aq) + 2e-
Hg2Cl2(s) + H2(g) 2Hg(l) + 2H+(aq) + 2Cl-(aq)

38. Measurement of pH (cont.)
Hg2Cl2(s) + H2(g) 2Hg(l) + 2H+(aq) + 2Cl-(aq)
• What if we let [H+] vary?
Ecell = E°cell - (0.0591/2)log(Q)
Ecell = E°cell - (0.0591/2)(2log[H+] + 2log[Cl-])
constant
Ecell = E°cell - (0.0591)(log[H+] + log[Cl-])
saturate

39. Measurement of pH (cont.)
Ecell = E°cell - (0.0591)log[H+] + constant
• Ecell is directly proportional to log [H+]
electrode

40. Glass Membrane Electrode
E = K (pH1 - pH2)
K= constant known by the asymmetry potential.
PH1 = pH of the internal solution 1.
PH2 = pH of the external solution 2.
The final equation is:
E = K pH

41. Glass Membrane Electrode
Advantages of glass electrode:
It can be used in presence of oxidizing, reducing, complexing
Disadvantage:
Delicate, it can’t be used in presence of dehydrating agent e.g. conc. H2SO4, ethyl alcohol….
Interference from Na+ occurs above pH 12 i.e Na+ excghange together with H+  above pH 12 and higher results are obtained.
It takes certain time to come to equilibrium due to resistance of glass to electricity.

42. Application of potentiometric titration in
a) Neutralization reactions: glass / calomel electrode for determination of pH
b) Precipitation reactions: Membrane electrodes for the determination of the halogens using silver nitrate reagent
c) Complex formation titration: metal and membrane electrodes for determination of many cations (mixture of Bi3+, Cd2+ and Ca2+ using EDTA)
d) Redox titration: platinum electrode For example for reaction of Fe3+/ Fe2+ with Ce4+/Ce3+

43. Redox titration Cu(S) +2Ag+  Cu2+ + 2Ag (S)
Cu(S)  Cu2+ + 2e oxi. Eo=-0.337
Ag+ + e-  Ag(s) Red. Eo =0.799
Ag+ + Cu(S)  Ag(s) + Cu2+ Redox Eo =0.462
Cu│Cu2+ (xM) ││ Ag+(yM) │ Ag
k eq= [Cu2+]/ [ Ag+]2

44. Redox titration EAg=Eo Ag+-0.059/2*Log1/[Ag +]2
ECu=Eo Cu /2*Log1/[Cu 2+]
E Cell=0 → EAg+ =ECu2+ درتعادل

45. Eo Ag+-0.059/2*Log1/[Ag +]2 =Eo Cu2+-0.059/2*Log1/[Cu 2+]
EAg+ =ECu2+
Eo Ag /2*Log1/[Ag +]2 =Eo Cu /2*Log1/[Cu 2+]
Eo Ag+- Eo Cu2+=
0.059/2*Log1/[Ag+ ] /2*Log1 /[Cu2+] =
0.059/2*Log1/[Ag+ ] /2*Log [Cu2+] = 0.059/2*Log [Cu2+] /[Ag+ ]2
Eo Ag+- Eo Cu2+= Eo Redox
Eo Redox = 0.059/2*Log [Cu2+] /[Ag+ ]2
2(Eo Redox)/0.059= Log Cu2+ /[ Ag+]2 =LogKeq
2( )/0.059=15.6
Keq=4.1*1015

46. محاسبه ثابت تعادل LogKeq=5(1.51-0.771)/0.059=62.5
Mno4-+5Fe2++8H+  Mn2++5Fe3++4H20
Mno4-+5e-+8H+→ Mn2++ 4H20 E= n=5
5Fe2 +→ 5Fe3+ +5e E= n=1
LogKeq=5( )/0.059=62.5

47. ترسيم منحني تيتراسيون 100 ml Fe2+ 0.5 M WITH Mno4- 0.5 M
Mno4-+5Fe3+  Mn2++5Fe2+
Fe3+  Fe2+
E= /5*Log[Fe2+]5/[Fe3+]5
E=nE0 OX+ mE0 Red/(m+n)
E=E= /5 *Log [Mn2+]/[Mno4-][H+]8
ml تيترانت
E سلول
??? 10 20 30