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Redox reactins half-reactions: Reduction 2Fe 3+ + 2e -  2Fe 2+ oxidation Sn 2+  Sn 4+ + 2e - 2Fe 3+ + Sn 2+  2Fe 2+ + Sn 4+

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Presentation on theme: "Redox reactins half-reactions: Reduction 2Fe 3+ + 2e -  2Fe 2+ oxidation Sn 2+  Sn 4+ + 2e - 2Fe 3+ + Sn 2+  2Fe 2+ + Sn 4+"— Presentation transcript:

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2 Redox reactins half-reactions: Reduction 2Fe e -  2Fe 2+ oxidation Sn 2+  Sn e - 2Fe 3+ + Sn 2+  2Fe 2+ + Sn 4+

3 Redox reactins occurring in 1) solution 2) electrochemical cell. 2Fe 3+ + Sn 2+  2Fe 2+ + Sn 4+

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

5 Standard Reduction Potentials Reduction Half-Reaction E  (V) F 2 (g) + 2e -  2F - (aq) 2.87 Au 3+ (aq) + 3e -  Au(s) 1.50 Cl 2 (g) + 2 e -  2Cl - (aq) 1.36 Cr 2 O 7 2- (aq) + 14H + (aq) + 6e -  2Cr 3+ (aq) + 7H 2 O 1.33 O 2 (g) + 4H + + 4e -  2H 2 O(l) 1.23 Ag + (aq) + e -  Ag(s) 0.80 Fe 3+ (aq) + e -  Fe 2+ (aq) 0.77 Cu 2+ (aq) + 2e -  Cu(s) 0.34 Sn 4+ (aq) + 2e -  Sn 2+ (aq) H + (aq) + 2e -  H 2 (g) 0.00 Sn 2+ (aq) + 2e -  Sn(s) Ni 2+ (aq) + 2e -  Ni(s) Fe 2+ (aq) + 2e -  Fe(s) Zn 2+ (aq) + 2e -  Zn(s) Al 3+ (aq) + 3e -  Al(s) Mg 2+ (aq) + 2e -  Mg(s) Li + (aq) + e -  Li(s) Ox. agent strength increases Red. agent strength increases

6 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 H 2 O 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.

7 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 H 2 O 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.

8 Balancing of redox reactions H 2 O 2 (aq) + Cr 2 O 7 -2 (aq )  Cr 3+ (aq) + O 2 (g) Redox reaction ====================================== 1)write 2 half reactions Half Rxn (oxid): Cr 2 O 7 -2 (aq)  Cr 3+ Half Rxn (red): H 2 O 2 (aq)  O 2 2)Atom balance Cr 2 O 7 -2 (aq)  2Cr 3+ H 2 O 2 (aq)  O 2

9 Balancing of redox reactions 3)Oxygen balance Half Rxn (oxid): Cr 2 O 7 -2 (aq)  2Cr H 2 O Half Rxn (red): H 2 O 2 (aq)  O 2 4)Hydrogen balance Half Rxn (oxid): 14H + + Cr 2 O 7 -2 (aq)  2Cr H 2 O Half Rxn (red): H 2 O 2 (aq)  O 2 + 2H + 5)Electron balance 6e H + + Cr 2 O 7 -2 (aq)  2Cr H 2 O H 2 O 2 (aq)  O 2 + 2H + + 2e -

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

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

12 کاهنده کمکی آنالیت باید در یک حالت اکسایش باشد ( Fe 3+ یا (Fe 2+ Zn-Al-Cd-Pb-Ni - ملغمه روی 2Zn(s)+Hg 2+ → Zn 2+ + Zn(Hg) (s) ملغمه روی باعث کاهش +Hنمی شود

13 اکسنده کمکی سدیم بیسموتات NaBio 3 (s) + 4H + + 2e- → BiO+ +Na + + 2H 2 O اضافی ← صاف کردن آمونیوم پراکسی دی سولفات 2e-+S 2 O 8 2- → 2SO 4 2- E o = 2.0) اضافی ← جوشاندن هیدروژن پراکسید H 2 O 2 +2H + +2e- → 2H 2 O Eo=1.78 اضافی ← جوشاندن

14 اکسنده های استاندارد: MnO 4 - → Mn 2+ E o =1.51 Ce 4+ → Ce 3+ E o =1.44 Cr2O7 2- → Cr 3+ E o =1.30 I - E o =0.54 → I 2 Eo پایین ید می تواند عوامل کاهنده قوی را در حضور عوامل کاهنده ضعیف اندازه گیری نماید.

15 شناساگرها شناساگر معروف: فروئین چسب نشاسته در منگانومتری شناساگر لازم نیست

16 تهيه محلول پرمنگنات با محاسبه پرمنگنات توزين و... ← گرم ← صاف ← استاندارد 2MnO H 2 C 2 O 4 +6H + → 2Mn CO2(g) +8H2O واکنش کند است Mn 2+ کاتالیز می کند (در ابتدا واکنش خیلی کند است –بعد سریع می شود)

17 پایداری نقطه پایانی منگنانومتری: 2MnO Mn H 2 O → 5MnO 2 (S) + 4H + K=1*10 47 در محیط باقی نمی ماند ولی سرعت واکنش کم است حدود 30 ثانیه رنگ باقی می ماند

18 پایداری محلولهای آبی پرمنگنات: 4MnO H 2 O→ 4MnO 2 (S) + 3O OH - K تقریبا بالا است ولی سرعت پایین است و محلولها معمولا پایدار می باشد نور-گرما-اسید-باز-Mn 2+ -MnO 2 واکنش را کاتالیز می کنند.

19 ید- تیوسولفات روش مستقیم (یدیمتری) →2I - I 2 +S روش غیرمستقیم ( یدومتری ) I - +S oxi →I 2

20 ید- تیوسولفات S 4 O I 2 +2S 2 O 3 2- →2I تیوسولفات←تتراتیونات n=2 - I 2 →2I S 4 O 6 2- n=1 S 2 O 3 2- → ید براي سنجش تيوسولفات منحصر به فرد است زيرا سایر اکسنده های قوی تتراتیونات راهم اکسید می کنند

21 تهیه محلول آبکی ید: [ I 2 ]=0.001 M حلالیت در آب I 2 +I - ⇄ I 3 - K=700 ید فرار است 4I - + O 2 (g)+ 4H + →2I 2 +2H 2 O اسید- گرما و نور واکنش را کاتالیز می کند.

22 تهيه محلول آبكي يد!!! I 2 +I - ↔ I 3 - IO 3 -+5I - →3I 2 2Cu 2+ +4I - →2CuI+I 2

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

24 پايداري تيوسولفات S 2 O H + →HSO 3 - +S (S) pHمحلول - غلظت محلول -- Cu 2+ نور خورشید و میکروارگانیزم ها واکنش را تشدید می کنند.

25 24 Electrolysis of Copper Concentration Cells 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.

26 25 Cu│Cu 2+ (1.0M) ‖ Cu 2+ (0.1 M)│Cu Anod cathod E=E /2Log(0.1/1.0) = Concentration Cells Cu+Cu 2+ (1.0 M)  Cu 2+ (0.1M)+Cu

27 26 The pH Meter Prentice-Hall © 2002General Chemistry: Chapter 21 Slide 26 of 52 2 H + (1 M) → 2 H + (x M) Pt | H 2 (1 atm)|H + (x M) ||H + (1.0 M) |H 2 (1 atm) | Pt(s) 2 H + (1 M) + 2 e - → H 2 (g, 1 atm) H 2 (g, 1 atm) → 2 H + (x M) + 2 e - H 2 (g, 1 atm) +2 H + (1 M) → 2 H + (x M) + H 2 (g, 1 atm)

28 27 Slide 27 of 52 E cell = E cell ° - log n V x2x E cell = 0 - log V x2x2 1 E cell = V log x E cell = (0.059 V) pH 2 H + (1 M) → 2 H + (x M) E cell = E cell ° - log Q n V pH = E cell /(0.059)

29 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 1.It is difficult to be used and to keep H 2 ­ gas at one atmosphere during all determinations. 2.It needs periodical replating of Pt. Sheet with Pt. Black Standard Hydrogen Electrode

30 Calomel electrode E voltKCl 0.241Saturated M M

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

32 Ag/AgCl Disadvantage of silver-silver chloride electrode 1.It is more difficult to prepare than SCE. 2.AgCI in the electrode has large solubility in saturated KCl Advantage of Ag-AgCI electrodes over SCE. 1.It has better thermal stability. 2.Less toxicity and environmental problems with consequent cleanup and disposal difficulties.

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34 Indicator electrode E cell =E indicator -E reference 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.

35 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: E 25 = Eo Ag/AgCl log [Cl - ]

36 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

37 3. indicator electrodes for neutralization reaction Glass Membrane Electrode

38 Measurement of pH pH meters use electrochemical reactions. Ion selective probes: respond to the presence of a specific ion. pH probes are sensitive to H 3 O +. Specific reactions: Hg 2 Cl 2 (s) + 2e - 2Hg(l) + 2Cl - (aq) E° 1/2 = 0.27 V Hg 2 Cl 2 (s) + H 2 (g) 2Hg(l) + 2H + (aq) + 2Cl - (aq) H 2 (g)2H + (aq) + 2e - E° 1/2 = 0.0 V

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

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

41 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

42 Glass Membrane Electrode Advantages of glass electrode: It can be used in presence of oxidizing, reducing, complexing Disadvantage: 1.Delicate, it can’t be used in presence of dehydrating agent e.g. conc. H2SO4, ethyl alcohol…. 2.Interference from Na+ occurs above pH 12 i.e Na+ excghange together with H+ above pH 12 and higher results are obtained. 3.It takes certain time to come to equilibrium due to resistance of glass to electricity.

43 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+

44 Redox titration Cu (S) +2Ag +  Cu Ag (S) Cu(S)  Cu e - oxi. E o = Ag + + e -  Ag (s) Red. E o = Ag + + Cu (S)  Ag (s) + Cu 2+ Redox E o =0.462 Cu│Cu 2+ (xM) ││ Ag + (yM) │ Ag k eq= [Cu2+]/ [ Ag+] 2

45 Redox titration EAg=E o Ag /2*Log1/[Ag + ] 2 ECu=E o Cu /2*Log1/[Cu 2+ ] E Cell=0 → EAg+ =ECu2+ درتعادل

46 EAg + =ECu 2+ E o Ag /2*Log1/[Ag + ] 2 =E o Cu /2*Log1/[Cu 2+ ] E o Ag + - E o Cu 2+ = 0.059/2*Log1/[Ag+ ] /2*Log1 /[Cu 2+ ] = 0.059/2*Log1/[Ag+ ] /2*Log [Cu 2+ ] = 0.059/2*Log [Cu 2+ ] /[Ag+ ] 2 E o Ag + - E o Cu 2+ = E o Redox E o Redox = 0.059/2*Log [Cu 2+ ] /[Ag+ ] 2 2(E o Redox)/0.059= Log Cu 2+ /[ Ag + ] 2 =LogK eq 2( )/0.059=15.6 K eq =4.1*10 15

47 محاسبه ثابت تعادل Mno Fe 2+ +8H +  Mn 2+ +5Fe 3+ +4H 2 0 Mno e - +8H + → Mn H 2 0 E=1.51 n=5 5Fe 2 + → 5Fe 3+ +5e E = n=1 LogK eq =5( )/0.059=62.5

48 ترسيم منحني تيتراسيون 100 ml Fe M WITH Mno M Mno Fe 3+  Mn 2+ +5Fe 2+ Fe 3+  Fe 2+ E= /5*Log[Fe2+] 5 /[Fe3+] 5 E=nE 0 OX + mE 0 Red /(m+n) E= E= /5 *Log [Mn 2+ ]/[Mno 4 - ][H + ] 8 ml تيترانت E سلول 0???


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