REDOX Equilibrium (I) C.P. Huang University of Delaware

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1. 2. 3. 4. 5. 6. 7.

Definition Oxidation number Balancing redox reactions Electromotive force, EMF Electrochemical cells Potentiometric titration Eh-pH diagram

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1.0 Definition • • • • • •

Electron transfer reaction Electron: reducing agent Oxidation: electron donating Reduction: electron accepting Oxidants: electron acceptors Reductants: electron donors

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2.0 Oxidation number NO3(N2O5)

V

H2S

-II

Cl-

-I

HCO3

IV

NO2(N2O3)

III

S8

0

Cl2

0

HCOOH

II

N2

0

SO32-

IV

ClO-

I

C6H12O6

0

NH3

-III

SO42-

VI

ClO2-

III

CH3OH

-II

S2O32-

II

ClO2

IV

CH4

-IV

S4O62-

2.5

ClO3-

V

C6H5COOH -2/7

S2O62-

V

ClO4-

VIII

4

5

Stumm & Morgan

3.0 Balancing Redox reactions • Mn2+ to MnO4- by PbO2 • S2O32- to S4O62- by O2

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Mn2+ to MnO4- by PbO2 2

 4



Mn  4 H 2O  MnO  8H  5e PbO

4 Mn

2

 4 MnO 2

2

 4 H   4 e  Pb

2

 2 H 2O

 16 H 2 O  5 PbO 2  20 H  4

 32 H



 5 Pb

2



 20 e

 10 H 2 O  20 e  4

2

4 Mn  5PbO2  6 H 2O  4 MnO  5Pb  12 H 7



S2O32- to S4O62- by O2 2 S 2O

2 3

 S 4O

2 6

 2e



O  4 H  4 e  2 H 2O 2

4 S 2O 32   O 2  4 H   4 e  2 S 4O 62   2 H 2O  4 e

4 S 2O 32   O 2  4 H   2 S 4O 62   2 H 2O

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4.0 Electromotive force, EMF OX + ne = Red Go = -nFEo K = [Red]/([Ox][e]n) log K = log{[Red]/[Ox]} - n log [e] 2.303 RT log K = 2.303RT log {[Red]/[Ox]} - 2.303RT n log[e] = RTln{[Red]/[Ox]} – G G =Go + RT ln{[Red]/[Ox]} G = 2.303 n RT log[e] = -nFE E = -G/nF = -(2.303 RT/F) log [e] = - (RT/F) ln[e] -Go

E = -(2.303RT/F) log (e) E = -0.05969 log(e) = 0.05915 pe

O2

H2O

1.229 v

Fe3+

Fe2+

0.772 v

Cu2+

Cu

0.3402 v

AgCl

Ag

0.2223 v

H+

H2

0.000 v

Pb2+

Pb

Cd2+

Cd

-0.4026 v

H2O Na+

H2

-0.82777 v

Na

-2.7109 v

K

-2.924v

K+

-0.1263 v

[e] = 10 (-E/0.05969) =101-6.906E

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5.0 Electrochemical cells Ox + ne = Red

Reduction potential

Eox ,red  Eoox ,red 

RT Re d ln Ox  nF

Oxidation potential

-

+ o Ered,ox  Ered ,ox 

RT [Ox ] ln nF [Re d]

Ered,ox  Eox,red 10

Electrochemical Cells I e

e

Anode (-) Cathode (+) Galvanic Cell Galvanic cell: energy generation

Cathode (-)

Anode (+)

Electrolytic Cell Electrolytic cells: energy dissipation

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Standard Cell E  EH ,H  EAgCl,Ag 2

H2 + 2AgCl = 2Ag + 2H+ +2Cl-

AgCl

Ag 0.2223 v 2AgCl + 2e = 2Ag +2Cl-

RT [H ]2 EH ,H  E ln  2 2F PH2 RT [Cl ]2 o ln E AgCl,Ag  E AgCl,Ag  F 1 o H2 ,H

E  EH

2 ,H



E

RT [H ] RT [ Cl ]   EoAgCl,Ag  ln ln 2F PH2 F 1

E

E

o H2 ,H

H+ H2 0.0000 v H2 = 2H+ + 2e

(-) H2 | H+,Cl- || AgCl | Ag (+) (-) anode

(+) cathode

 EAgCl,Ag

o H2 ,H

o AgCl,Ag

RT [H ][ Cl ]2  ln 2F PH2

RT [H ]2 [Cl ]2 E  0.2222  ln 2F PH2

Eo  EHo ,H  EoAgCl,Ag 2

 0  0.2222  0.2222v 13

Standard Cell 2H+ +Zn = Zn2+ + H2

E  EH ,H  E 2

Zn,Zn2





PH2 RT RT [ Zn2 ] o  ln  2  E Zn,Zn2   ln 2F [H ] 2F 1

 EHo  ,H

2

H+

H2

0v



 EHo  ,H  EoZn,Zn2  2





2 RT PH2 [ Zn ]  ln 2F [H ]2 

2 RT PH2 [ Zn ]  (0  0.76)  ln 2F [H ]2

Zn2+

Zn

-0.76 v

 0.76 



PH2 [ Zn2 ]

RT ln 2F [H ]2

(-) Zn|Zn2+ (a=1M)||H+(a=1M)|H2(g),Pt (+) 14

Standard Cell (-)Cd +2Ag+ = Cd2+ +2Ag(+)

E  E Ag ,Ag  E  E Ag ,Ag  E

Ag+

Ag 0.8 v

 EoAg ,Ag 

Cd,Cd2





Cd2 ,Cd

RT 1 RT 1 o   ln E ln  Cd2 ,Cd 2F [ Ag ]2 2F [Cd2  ] 

RT [Cd2 ] o o  E Ag ,Ag  E 2   ln Cd ,Cd 2F [ Ag ]2 

Cd2+

Cd -0.4 v

RT [Cd2 ]  0 .8  (  0 .4 )  ln 2F [ Ag ]2 

RT [Cd2 ]  1 .2  ln 2F [ Ag ]2

(-) Cd|Cd2+(a=1M)||Ag+(a=1M)|Ag(+) (-) Cd|Cd2+||Ag+|Ag(+) 15

Standard Cell 2H+ +2Ag + 2Cl- = H2+2AgCl

H+

H2

0v

E  EH ,H  E Ag,AgCl 2

PH2 RT 1 RT E  ln ln  2  EoAg,AgCl  2F [H ] 2F [Cl ]2 PH2 RT  (0  (0.222))  ln 2F [H ]2 [Cl ]2 PH2 0.05915  0.222  ln  2  2 2 [H ] [Cl ] o H ,H2

AgCl

Ag

-0.2222v

(-) Ag|Ag+(a=1M)||H+(a=1M)|H2(g),Pt(+) 16

Concentration Cell (-) Zn(Hg)(C1)|Zn2+(C3)|Zn(Hg)(C2) (+)

C1>C2

Anode (-) Zn(Hg)(C1) = Zn2+(C3) + 2e

Zn-Amalgam

Zn2+(C3) + 2e = Zn(Hg) (C2) (+) Cathode 0.05915 C2 0.05915 [ Zn2 ] E ln  ln 2 [ Zn2 ] 2 C1 

0.05915 C2 ln 2 C1

(+) Zn(Hg)(C1)|Zn2+(C3)|Zn(Hg)(C2)(-)

C1 C2; E > 0;  C1 P2; E > 0

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(-) (Pt) Cl2(P1= 0.5 atm| Cl-|| Cl2(P2= 0.2 atm(Pt)(+)

   

PCl2 0.05915 E log 2 PCl2

2

P1 >P2; E > 0

1

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Battery PbSO4+2e= Pb+SO42-;

o EPbSO  0.3563 4 ,Pb

o PbO2 + SO42- + 4H+ + 2e = PbSO4 + 2H2O; EPbO2 ,PbSO4  1.685(v)

PbO2 + Pb + 2SO42- + 4H+ = 2PbSO4 + 2H2O; Eo o o Eo = E PbO 2 ,PbSO+4 EPbSO =2.041 4 ,Pb

PbO2

PbSO4 1.685 v

o Ecell  EPbO  2 ,PbSO 4

 2.041 

0.05915 1 0.05915 1 o  EPb log log ,PbSO4   4 2 [SO4 ][H ] [SO42 ] 2 2

0.05915 1 log [SO 24 ]2 [H  ]4 2

[H2SO4] = 0.5 M PbSO4

Pb

-O.356 v

[H+] = 1M [SO42-] = 0.5 M

Ecell = 2.022 v

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Example What is the cell voltage of a new battery which H2SO4 concentration is 0.5 M? What is the cell voltage after discharge for one hour at 2.5 Amp current? [H2SO4] = 0.5 M [H+] = 1M

[SO42-] = 0.5 M

Ecell = 2.022 v

Q = it Q = 2.5x60x60 = 9000 C 9000/96500 =0.093 eq/L = 0.0465M

W = (98/2)x(0.093) = 4.56 g of H2SO4

[H2SO4] = 0.5-0.0465 = 0.4535 M o  Ecell  EPbO 2 ,PbSO 4

 2.041 

1 0.05915 1 0.05915 o   log E log Pb ,PbSO 4 2 [ SO24 ] 2 [ SO24 ][H ]4

0.05915 1 log 2 2 2 [ SO 4 ] [H ]4

E = 2.041+ 0.05915 (log(0.4535)+ 2*(log(2*0.4535)) =2.041- 0.4281 = 1.16128 v 20

Hydrogen fuel cells • Proton Exchange Membrane (PEM) • • • • • • • • • •

Direct Methanol (DMFC) Phosphoric Acid (PAFC) Solid Oxide (SOFC) Molten Carbonate (MCFC) Alkaline (AFC) Zinc-Air (ZAFC) Regenerative Aluminum Air Formic Acid Microbial

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Proton exchange membrane (PEM)

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Direct methanol fuel cell (DMFC) (anode) CH3OH + H2O = CO2 + 6H+ + 6e2H2= 4H+ + 4e(cathode) 1.5O2 + 6H+ + 6e- = 3H2O O2 + 4H+ + 4e- = 2H2O CH3OH + 1.5 O2 = CO2 + 2H2O 2H2 + O2 = 2H2O Platinum (Pt) : anode in the presence of methanol, CO, formed as a reaction intermediate, irreversibly absorbs to the Pt surface, rapidly lowering its activity. Pt/Ru bifunctional catalysts. Ru-OH + Pt-CO = Ru + Pt + CO2 + H+ + e23

Phosphoric acid fuel cell (PAFC) The PAFC uses an electrolyte that is phosphoric acid (H3PO4) that can approach 100% concentration. The ionic conductivity of phosphoric acid is low at low temperatures, so PAFCs are operated at the upper end of the range 150ºC–220ºC. The charge carrier in this type of fuel cell is the hydrogen ion (H+, proton). This is similar to the PEFC where the hydrogen introduced at the anode is split into its protons and electrons. The protons migrate through the electrolyte and combine with the oxygen, usually from air, at the cathode to form water. The electrons are routed through an external circuit where they can perform useful work. This set of reactions in the fuel cell produces electricity and by-product heat.

Anode Reaction:

2 H2 = 4 H+ + 4 e-

Cathode Reaction:

O2(g) + 4 H+ + 4 e- = 2 H2O

Overall Cell Reaction:

2 H2 + O2 = 2 H2O

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Solid oxide fuel cell (SOFC) The Solid Oxide Fuel Cell (SOFC) is currently the highest-temperature fuel cell in development and can be operated over a wide temperature range from 600ºC– 1000ºC allowing a number of fuels to be used. To operate at such high temperatures, the electrolyte is a thin, solid ceramic material (solid oxide) that is conductive to oxygen ions (O2-). The charge carrier in the SOFC is the oxygen ion (O2-). At the cathode, the oxygen molecules from the air are split into oxygen ions with the addition of four electrons. The oxygen ions are conducted through the electrolyte and combine with hydrogen at the anode, releasing four electrons. The electrons travel an external circuit providing electric power and producing by-product heat.

Anode Reaction: 2 H2 + 2 O2- = 2 H2O + 4 eCathode Reaction: O2 + 4 e- = 2 O2Overall Cell Reaction: 2 H2 + O2 = 2 H2O 25

Molten carbonate fuel cell (MCFC) The electrolyte typically consists of a combination of alkali (Na and K) carbonates retained in a ceramic matrix of LiAlO2. The anode is made from Ni while the cathode is made from nickel oxide.

anode: H2 + CO32- = H2O + CO2 + 2ecathode: 0.5 O2 + CO2 + 2e- = CO32overall: H2 + 0.5 O2 + CO2 (cathode) = H2O + CO2 (anode)

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Alkaline fuel cell (AFC) Alkaline fuel cells use an electrolyte that is an aqueous (water-based) solution of potassium hydroxide (KOH) retained in a porous stabilized matrix.

Anode Reaction: Cathode Reaction: Overall Net Reaction:

2 H2 + 4 OH- = 4 H2O + 4 eO2 + 2 H2O + 4 e- = 4 OH2 H2 + O2 = 2 H2O 27

Regenerative fuel cell (RFC)

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Microbial electrolysis cell

Anode: C12H22O11 + 13H2O = 12CO2 + 48H+ + 48eCathode: O2 + 4H+ + 4e = 2H2O Overall C12H22O11 + 12 O2 = 12CO2 + 11 H2O

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Microbial electrolysis cell

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Zinc-carbon battery Anode (-)

Zn(s) → Zn2+(aq) + 2 e− ; E° = −1.04 volts Cathode (+):

the overall reaction:

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Li -battery LiCoO 2  Li1 x CoO 2  xLi  xe

xLi  xe  6C  Li x C6 Li  e  LiCoO 2  Li2O  CoO LiCoO 2  Li  CoO 2

anode cathode overall Over charged at 5.2 V

Liquid electrolytes: LiPF6, LiBF4 or LiClO4 in ethylene carbonate, dimethyl carbonate, and diethyl carbonate.

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Lithium battery (anode) Average potential difference (V)

Specific capacity (mA·h/g)

Specific energy (kW·h/kg)

LiCoO2

3.7

140

0.518

LiMn2O4

4.0

100

0.400

LiNiO2

3.5

180

0.630

LiFePO4

3.3

150

0.495

Li2FePO4F

3.6

115

0.414

LiCo1/3Ni1/3Mn1/3O2

3.6

160

0.576

Li(LiaNixMnyCoz)O2

4.2

220

0.920

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Li battery (cathode) Electrode material

Average potential difference (V)

Specific capacity Specific energy (mA-h/g) (kW-h/g)

Graphite (LiC6)

0.1-0.2

372

0.0372-0.0744

Hard Carbon (LiC6)

?

?

?

Titanate (Li4Ti5O12)

1-2

160

0.16-0.32

Si (Li4.4Si)

0.5-1

4212

2.106-4.212

Ge (Li4.4Ge)

0.7-1.2

1624

1.137-1.949

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Li-battery

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6.0 Potentiometric Titration • Can titration take place? • Can one predict the end point?

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Titrate Fe2+ with Ce4+ E



Fe 3 ,Fe 2



E

[Fe 2  ]  0.05915 log [Fe 3  ]

o Fe 3  ,Fe 2 



Ce4+

Ce3+

1.61 v

E



Ce 4 ,Ce 3



E

o 

Ce 4 ,Ce 3



 0.05915 log

[Ce 3 ] 

[Ce 4 ]

Before end point: [Fe2+] = C(1-f) [Fe3+] = Cf Fe3+

Fe2+

0.76 v



[Fe 2 ] 

[Fe 3 ]

E  EFe3  ,Fe2  o  EFe 3 ,Fe 2 



1 f f

[Fe 2  ] E  0.05915 log [Fe 3  ] 1 f  0.05915 log f o Fe 3  ,Fe 2 

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Titrate Fe2+ with Ce4+ E ep

At f = 0.5

[Fe2+] = [Fe3+]

o E  EFe3  ,Fe2   EFe  0.05915 log 3 ,Fe 2 

0 .5 0 .5

o  EFe  0.76(v) 3 ,Fe 2 

At f = 1

o 0.05915 log K  EFe  EoCe 4  ,Ce3  2 ,Fe 3 

[Fe2+] =[Ce4+] [Fe 2  ] [Ce 4  ]  [Fe 3  ] [Ce 3  ] K

3

E ep

[Fe 2 ] E  0.05915 log [Fe 3 ] 0.05915 o  EFe log K 3 2  ,Fe 2 [Ce 3 ] o  ECe 4  ,Ce3   0.05915 log [Ce 4  ] 0.05915  EoCe 4  ,Ce3   log K 2 o Fe 3  ,Fe 2 

[Fe3+] =[Ce3+]

E ep  E

o Fe 3  ,Fe 2 



o  EoCe 4  ,Ce3  EFe 2 ,Fe 3 

2 o o o  ECe 2(EFe )  EFe 3 2 4 ,Fe 2  ,Fe 3  ,Ce 3 

3

[Fe ][Ce ] [Fe 2  ][Ce 4  ]

2

[Fe 3  ] [Ce 3  ]   K [Fe 2  ] [Ce 4  ]



Eep=(0.76+1.61)/2 = 1.19 v

o o EFe 3 2  E ,Fe Ce 4  ,Ce 3 

2 38

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Titrate Fe2+ with MnO4MnO4- + 5e +8H+ = Mn2+ + 4H2O; Eo(MnO4-;Mn2+) = 1.52 Fe3+ + e = Fe2+; Eo(Fe3+,Fe2+)=0.76 v At f