Redox reactins

half-reactions:

Reduction

2Fe3+ + 2e- 2Fe2+

oxidation

Sn2+ Sn4+ + 2e-

2Fe3+ + Sn2+ 2Fe2+ +

Sn4+

Redox reactinsoccurring in

1) solution

2) electrochemical cell.

2Fe3+ + Sn2+ 2Fe2+ +

Sn4+

Electrochemical Reactions

1)chemical electric:

primary cell (Galvanic cell)

2)electric chemical:

electrolytic cell

Standard Reduction PotentialsReduction 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.

age

nt s

tren

gth

incr

ease

sR

ed. agent strength increases

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 H2OBalance: 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.

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 H2OBalance: 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.

Balancing of redox reactionsH2O2 (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 balanceCr2O7

-2 (aq) 2Cr3+

H2O2 (aq) O2

Balancing of redox reactions

3)Oxygen balanceHalf Rxn (oxid): Cr2O7

-2 (aq) 2Cr3+ + 7 H2O

Half Rxn (red): H2O2 (aq) O2

4)Hydrogen balanceHalf Rxn (oxid): 14H+ + Cr2O7

-2 (aq) 2Cr3+ + 7 H2O

Half Rxn (red): H2O2 (aq) O2 + 2H+

5)Electron balance6e- + 14H+ + Cr2O7

-2 (aq) 2Cr3+ + 7 H2O

H2O2 (aq) O2 + 2H+ + 2e-

Balancing of redox reactions

6) Equalize of produced and consumed electrons

6e- + 14H+ + Cr2O7-2

(aq) 2Cr3+ + 7 H2O

( H2O2 (aq) O2 + 2H+ + 2e- ) x 3

7)Multiply each half reaction8 H+ + 3H2O2 + Cr2O7

2- 2Cr+3 + 3O2 + 7H2O

Redoxتیتراسیونهای

با- ( کالریمتری - معمولی واکنشهای( معرف از استفاده

( پتانسیومتری- با- الکتروشیمی واکنشهای( پتانسیومتر از استفاده

کمکی کاهنده

باشد ( اکسایش حالت یک در باید یا +Fe3آنالیت)Fe2+

Zn-Al-Cd-Pb-Ni

روی - ملغمه

2Zn(s)+Hg2+→ Zn2++ Zn(Hg) (s) کاهش + باعث روی شود Hملغمه نمی

کمکی اکسنده

بیسموتات • سدیم

•NaBio3 (s) + 4H++ 2e- → BiO+ +Na++ 2H2O کردن ← • صاف اضافیسولفات • دی پراکسی آمونیوم•2e-+S2O8

2-→ 2SO42- E o = 2.0 (

جوشاندن ← • اضافیپراکسید • هیدروژن•H2O2+2H++2e- → 2H2O Eo=1.78

جوشاندن ← • اضافی

: استاندارد های اکسنده

• MnO4-→ Mn2+ Eo=1.51

•Ce4+ → Ce3+ Eo=1.44

•Cr2O72- → Cr3+ Eo=1.30

• I- Eo=0.54 → I2

•Eo حضور در را قوی کاهنده عوامل تواند می ید پایین. نماید گیری اندازه ضعیف کاهنده عوامل

شناساگرها

• : فروئین معروف شناساگرنشاسته • چسبشناساگر • منگانومتری در

نیست الزم

پرمنگنات تهيه محلول

و ... ← • توزين پرمنگنات محاسبه باگرم ←•صاف ←•استاندارد ••2MnO4

-+5H2C2O4+6H+ → 2Mn2++10 CO2(g) +8H2O

است • کند واکنش•Mn2+ کند می کاتالیزسریع( – • بعد است کند خیلی واکنش ابتدا در

( شود می

پایانی نقطه پایداری منگنانومتری:

•2MnO4-+3Mn2++ 2 H2O → 5MnO2(S) + 4H+

• K=1*1047

ماند • نمی باقی محیط دراست • کم واکنش سرعت ولیماند 30حدود • می باقی رنگ ثانیه

آبی محلولهای پایداریپرمنگنات:

•4MnO4-+2H2O→ 4MnO2(S) + 3O2+ 4 OH-

•K است باال تقریبااست • پایین سرعت ولیباشد • می پایدار معموال محلولها و

•- - - باز- اسید گرما را Mn2+-MnO2نور واکنش. کنند می کاتالیز

تیوسولفات- ید

•( یدیمتری ( مستقیم روش →2I- I2+S

یدومتری ( ) • غیرمستقیم روش

• I-+Soxi→I2

•

تیوسولفات- ید

• S4O62-+ - I2+2S2O3

2-→2I

تتراتیونات← • تیوسولفات

•n=2 - I2 →2I

•S4 O 62- n=1 S2O3

2-→

فرد • به منحصر تيوسولفات سنجش براي یدزيرا است

راهم • تتراتیونات قوی های اکسنده سایرکنند می اکسید

: ید آبکی محلول تهیه

• [I2=]0.001 M آب در حاللیت

•I2+I-⇄ I3- K=700

است • فرار ید

•4I-+ O2(g)+ 4H+→2I2+2H2O

می- • کاتالیز را واکنش نور و گرما اسیدکند.

•

!!! يد آبكي محلول تهيه

•I2+I-↔ I3-

•IO3-+5I-→3I2

•2Cu2++4I-→2CuI+I2

قليائي محيط در يد

•I2+OH-→IO-+I-+H+

•3IO-→Io3-+2I-

یا • ضعیف اسیدی محیط در تیتراسیون. شود می انجام خنثی

تيوسولفات پايداري

•S2O32-+H+→HSO3

-+S(S)

•pH - - محلول غلظت نور +Cu2- محلولرا واکنش ها میکروارگانیزم و خورشید

. کنند می تشدید

24

Electrolysis of Copper

•Concentration Cells

A concentration cell based on the A concentration cell based on the Cu/Cu2+ half-reaction.Cu/Cu2+ half-reaction. AA, Even , Even

though the half-reactions involve the though the half-reactions involve the same components, the cell operates same components, the cell operates

because the half-cell concentrations are because the half-cell concentrations are different. different. BB, The cell operates , The cell operates

spontaneously until the half-cell spontaneously until the half-cell concentrations are equal. Note the concentrations are equal. Note the

change in electrodes (exaggerated here change in electrodes (exaggerated here for clarity) and the equal color of for clarity) and the equal color of

solutionssolutions..aliasadipour@yahoo.com

aliasadipour@yahoo.com25

Cu│Cu2+ (1.0M)‖ Cu2+ (0.1 M)│Cu Anod cathod

E=E0-0.059/2Log(0.1/1.0) =+0.0296

Concentration Cells

Cu+Cu2+ (1.0 M)Cu2+ (0.1M)+Cu

26

The pH Meter

Prentice-Hall © 2002General Chemistry: Chapter 21 Slide 26 of 52

2 H+(1 M) → 2 H+(x M)

Pt | H2 (1 atm)|H+(x M) ||H+(1.0 M) |H2(1 atm) | Pt(s)

2 H+(1 M) + 2 e- → H2(g, 1 atm)

H2(g, 1 atm) → 2 H+(x M) + 2 e -

H2(g, 1 atm) +2 H+(1 M) → 2 H+(x M) + H2(g, 1 atm)

27 Slide 27 of 52

Ecell = Ecell° - logn

0.059 V x2

12

Ecell = 0 - log2

0.059 V x2

1

Ecell = - 0.059 V log x

Ecell = (0.059 V) pH

2 H+(1 M) → 2 H+(x M)Ecell = Ecell° - log Qn

0.059 V

aliasadipour@yahoo.com

pH = Ecell /(0.059)

It’s a primary reference electrode. Its potential is considered to be zero.

Electrode reaction: half cell: pt, H2 / H+ (1N) Eo = zero d-Limitation1. It is difficult to be used and to

keep H2 gas at one atmosphere during all determinations.

2. It needs periodical replating of Pt. Sheet with Pt. Black

Standard Hydrogen Electrode

Calomel electrode

22Hg/Hg25 ][Hg

1 log 0.059 - oE E2

KClE volt

Saturated0.241

1M0.280

0.1 M0.334

Ag/AgCl,

][Ag

1 log 0.059 - E E o

Ag/Ag

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:

Ag/AgCl

Disadvantage of silver-silver chloride electrode1. 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.

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.

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 - 0.059 log [Cl-]

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

3. indicator electrodes for neutralization reaction

Glass Membrane Electrode

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

Hg2Cl2(s) + H2(g) 2Hg(l) + 2H+(aq) + 2Cl-(aq)

H2(g) 2H+(aq) + 2e- E°1/2 = 0.0 V

Measurement of pH (cont.)Hg2Cl2(s) + H2(g) 2Hg(l) + 2H+(aq) + 2Cl-(aq)

• What if we let [H+] vary?

Q H 2Cl 2

Ecell = E°cell - (0.0591/2)log(Q)

Ecell = E°cell - (0.0591/2)(2log[H+] + 2log[Cl-])

Ecell = E°cell - (0.0591)(log[H+] + log[Cl-])saturate

constant

Measurement of pH (cont.)

Ecell = E°cell - (0.0591)log[H+] + constant

• Ecell is directly proportional to log [H+]

electrode

Glass Membrane Electrode

E = K + 0.059 (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 - 0.059 pH

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.

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+

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

Redox titration•EAg=Eo Ag+-0.059/2*Log1/]Ag +[2

•ECu=Eo Cu2+-0.059/2*Log1/]Cu 2+[

E Cell=0 → EAg+ =ECu2 +درتعادل

•EAg+ =ECu2+

•Eo Ag+-0.059/2*Log1/]Ag +[2 =Eo Cu2+-0.059/2*Log1/]Cu 2+[

•Eo Ag+- Eo Cu2+ =

•0.059/2*Log1/]Ag+ ]2 - 0.059/2*Log1 /]Cu2+[ =•0.059/2*Log1/]Ag+ ]2 + 0.059/2*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)0.799-0.337/(0.059=15.6

•Keq=4.1*1015

•

تعادل ثابت محاسبه

•Mno4-+5Fe2++8H+ Mn2++5Fe3++4H20

•Mno4

-+5e-+8H+→ Mn2++ 4H20 E=1.51 n=5

•5Fe2 +→ 5Fe3+ +5e E= -0.771 n=1

• LogKeq=5(1.51-0.771)/0.059=62.5

تيتراسيون منحني ترسيم•100 ml Fe2+ 0.5 M WITH Mno4

- 0.5 M•Mno4

-+5Fe3+ Mn2++5Fe2+

•Fe3+ Fe2+

•E=0.771-0.059/5*Log[Fe2+]5/[Fe3+]5

•E=nE0 OX+ mE0 Red/(m+n)

•E=E=1.51-0.059/5 *Log ]Mn2+[/]Mno4-[]H+[8

EE سلوسلولل

ml ml تيترانتيترانتت

??????00

1010

2020

3030