ib chemistry on gibbs free energy, equilibrium constant and cell potential

16
cell nFE G Relationship between Energetics and Equilibrium c K RT G ln S T H G Enthalpy change Entropy change Equilibrium constant Gibbs free energy change H G Relationship bet G, K c and E cell cell nFE G S T H G c K RT G ln c K Relationship between Energetics and Cell Potential G cell E Gibbs free energy change Cell potential F = Faraday constant (96 500 Cmol -1 ) n = number electron Relationship bet G, K c and E cell ΔG θ K c E θ /V Extent of rxn > 0 < 1 < 0 No Reaction Non spontaneous ΔG θ = 0 K c = 1 0 Equilibrium Mix reactant/product < 0 > 1 > 0 Reaction complete Spontaneous ΔG θ K c Eq mixture ΔG θ = + 200 9 x 10 -36 Reactants ΔG θ = + 10 2 x 1 -2 Mixture ΔG θ = 0 K c = 1 Equilibrium ΔG θ = - 10 5 x 10 1 Mixture ΔG θ = - 200 1 x 10 35 Products Relationship bet G and K c shift to left (reactant) shift to right (products) cell E G c K K nF RT E cell ln

Upload: lawrence-kok

Post on 09-Jan-2017

1.207 views

Category:

Education


4 download

TRANSCRIPT

Page 1: IB Chemistry on Gibbs Free Energy, Equilibrium constant and Cell Potential

cellnFEG

Relationship between Energetics and Equilibrium

cKRTG ln STHG

Enthalpy

change

Entropy

change

Equilibrium

constant

Gibbs free

energy change

HG

Relationship bet ∆G, Kc and E cell

cellnFEG STHG cKRTG ln

cK

Relationship between Energetics and Cell Potential

G cellE

Gibbs free

energy change

Cell potential

F = Faraday constant (96 500 Cmol-1)

n = number electron

Relationship bet ∆G, Kc and Ecell

ΔGθ Kc Eθ/V Extent of rxn

> 0 < 1 < 0 No Reaction Non spontaneous

ΔGθ = 0 Kc = 1 0 Equilibrium Mix reactant/product

< 0 > 1 > 0 Reaction complete Spontaneous

ΔGθ Kc Eq mixture

ΔGθ = + 200 9 x 10-36 Reactants

ΔGθ = + 10 2 x 1-2 Mixture

ΔGθ = 0 Kc = 1 Equilibrium

ΔGθ = - 10 5 x 101 Mixture

ΔGθ = - 200 1 x 1035 Products

Relationship bet ∆G and Kc

shift to left (reactant)

shift to right (products)

cellE

G

cKK

nF

RTE cell ln

Page 2: IB Chemistry on Gibbs Free Energy, Equilibrium constant and Cell Potential

Magnitude of Kc Extend of reaction

How far rxn shift to right or left?

Not how fast

cK

Position of equilibrium

cK

Temp dependent

Extend of rxn

Not how fast

Shift to left/ favour reactant

Shift to right/ favour product

cK

Relationship between Equilibrium and Energetics

cKRTG ln STHG

Enthalpy

change Entropy

change

Equilibrium

constant Gibbs free energy change

HG cK

G

Energetically Thermodynamically Favourable/feasible

ΔGθ ln K Kc Eq mixture

ΔGθ -ve < 0

Positive ( + )

Kc > 1 Product (Right)

ΔGθ +ve > 0

Negative ( - )

Kc < 1 Reactant (left)

ΔGθ = 0 0 Kc = 1 Equilibrium

Measure work available from system

Sign predict spontaneity of rxn

Negative (-ve) spontaneous

Positive (+ve) NOT spontaneous

veG veG

NOT favourable

Energetically favourable

Product formation NO product

cKRTG ln

Page 3: IB Chemistry on Gibbs Free Energy, Equilibrium constant and Cell Potential

Magnitude of Kc Extend of reaction

How far rxn shift to right or left?

Not how fast

cK

Position of equilibrium

cK

Temp dependent

Extend of rxn

Not how fast

Shift to left/ favour reactant

Shift to right/ favour product

cK

Relationship between Equilibrium and Energetics

cKRTG ln STHG

Enthalpy

change

Entropy

change

Equilibrium

constant

Gibbs free energy change

HG cK

ΔGθ ln K Kc Eq mixture

ΔGθ -ve < 0

Positive ( + )

Kc > 1 Product (Right)

ΔGθ +ve > 0

Negative ( - )

Kc < 1 Reactant (left)

ΔGθ = 0 0 Kc = 1 Equilibrium

cKRTG ln STHG

∆Hsys ∆Ssys ∆Gsys Description

- + ∆G = ∆H - T∆S

∆G = - ve Spontaneous, All Temp

+ - ∆G = ∆H - T∆S

∆G = + ve Non spontaneous, All Temp

+ + ∆G = ∆H - T∆S

∆G = - ve Spontaneous, High ↑ Temp

- - ∆G = ∆H - T∆S

∆G = - ve Spontaneous, Low ↓ Temp

Relationship bet ∆G and Kc

Page 4: IB Chemistry on Gibbs Free Energy, Equilibrium constant and Cell Potential

GEnergetically

Thermodynamically Favourable/feasible

Sign predict spontaneity of rxn

veG veG

NOT favourable

Energetically favourable

Product formation NO product

KRTG ln

Predict will rxn occur with ΔG and Kc

cK

Very SMALL Kc < 1

Shift to right/ favour product

Shift to left/ favour reactant

Very BIG Kc > 1

veG veG

KRTG ln

1cK 1cK

Negative (-ve) spontaneous

Positive (+ve) NOT spontaneous

Relationship bet ∆G and Kc

ΔGθ Kc Eq mixture

ΔGθ = + 200

9 x 10-36 Reactant

ΔGθ = + 10 2 x 1-2 Mixture

ΔGθ = 0 Kc = 1 Equilibrium

ΔGθ = - 10 5 x 101 Mixture

ΔGθ = - 200 1 x 1035 Products

shift to left (reactant)

shift to right (product)

G, Gibbs free energy

A

Mixture composition

B

100% A 100% B

∆G decreases ↓

30 % A 70 % B

Equilibrium mixture

∆G < 0

∆G = 0 (Equilibrium) ↓

Free energy minimum

∆G < 0

∆G < 0

∆G = 0

Free energy system is lowered on the way to equilibrium Rxn proceed to minimum free energy ∆G = 0

System seek lowest possible free energy Product have lower free energy than reactant

∆G < 0 product reactant

Page 5: IB Chemistry on Gibbs Free Energy, Equilibrium constant and Cell Potential

GEnergetically

Thermodynamically Favourable/feasible

Sign predict spontaneity of rxn

veG veG

NOT favourable

Energetically favourable

Product formation NO product

KRTG ln

cK

Very SMALL Kc < 1

Shift to right/ favour product

Shift to left/ favour reactant

Very BIG Kc > 1

veG veG

KRTG ln

1cK 1cK

Negative (-ve) spontaneous

Positive (+ve) NOT spontaneous

Relationship bet ∆G, Q and Kc

G, Gibbs free energy

A

B

100% A 100% B

∆G decreases ↓

30 % A 70 % B

Equilibrium mixture

∆G < 0

∆G = 0 (Equilibrium) ↓

Free energy minimum

∆G < 0

∆G < 0

∆G = 0

∆G < 0 product reactant

G, Gibbs free energy

reactant product ∆G < 0 A

B

∆G decreases ↓

100% A 100% B 30 % A 70 % B

∆G = 0

Q = K

∆G < 0

Q < K

∆G > 0

∆G < 0

Q > K

∆G > 0

A ↔ B A ↔ B

Equilibrium mixture

Predict will rxn occur with ΔG and Kc

Page 6: IB Chemistry on Gibbs Free Energy, Equilibrium constant and Cell Potential

Relationship bet ∆G and Kc

G, Gibbs free energy

A

B

100% A

100% B

∆G decreases ↓

30 % A 70 % B

Equilibrium mix close to product

∆G < 0

∆G = 0 (Equilibrium) ↓

Free energy minimum

∆G < 0

∆G < 0

∆G = 0

∆G < -10

Kc > 1

A ↔ B A ↔ B

G, Gibbs free energy

A

B

∆G decreases ↓

∆G < -100

100% A

100% B

∆G = 0 (Equilibrium) ↓

Free energy minimum

Kc > 1 Equilibrium mix close to product

10 % A 90 % B

∆G < 0

∆G < 0 ∆G = 0

∆G very –ve → Kc > 1 → (more product/close to completion) ∆G –ve → Kc > 1 → (more product > reactant)

A ↔ B

G, Gibbs free energy

100% A

100% B

A

B

∆G +ve → Kc < 1 → (more reactant > product)

∆G > +10

∆G = 0 (Equilibrium) ↓

Free energy minimum

Kc < 1

∆G increases ↑

70 % A 30 % B

Equilibrium mix close to reactant

∆G < 0

∆G = 0

A ↔ B

G, Gibbs free energy

∆G more +ve → Kc < 1 → (All reactant / no product at all)

A

∆G = 0 (Equilibrium) ↓

Free energy minimum

Kc < 1 100% A

100% B

Equilibrium mix close to reactant/ No reaction.

∆G > +100 B

90 % A 10 % B

∆G increases ↑

∆G = 0

∆G < 0

reactant

reactant

reactant

reactant

product product

product product

Page 7: IB Chemistry on Gibbs Free Energy, Equilibrium constant and Cell Potential

Relationship bet ∆G and Kc

shift to left (reactant)

shift to right (product)

G, Gibbs free energy

A

B

100% A

100% B

∆G decreases ↓

30 % A 70 % B

Equilibrium mixture

∆G < 0

∆G = 0 (Equilibrium) ↓

Free energy minimum

∆G < 0

∆G < 0

∆G = 0

Free energy system is lowered on the way to equilibrium Rxn proceed to minimum free energy ∆G = 0

System seek lowest possible free energy Product have lower free energy than reactant

∆G < -10

Kc > 1

A ↔ B A ↔ B

G, Gibbs free energy

A

B

∆G decreases ↓

∆G < -100

100% A

100% B

∆G = 0 (Equilibrium) ↓

Free energy minimum

Kc > 1 Equilibrium mixture

10 % A 90 % B

∆G < 0

∆G < 0 ∆G = 0

∆G very –ve → Kc > 1 → (All product/close to completion) ∆G –ve → Kc > 1 → (more product > reactant)

∆G

∆G = 0

∆G > 0

∆G < 0

No reaction/most reactants Kc <1

Complete rxn/Most products Kc > 1

Kc = 1 (Equilibrium) Reactants = Products

reactant reactant

ΔGθ Kc Eq mixture

ΔGθ = + 200 9 x 10-36 Reactant

ΔGθ = + 10 2 x 1-2 Mixture

ΔGθ = 0 Kc = 1 Equilibrium

ΔGθ = - 10 5 x 101 Mixture

ΔGθ = - 200 1 x 1035 Products

Page 8: IB Chemistry on Gibbs Free Energy, Equilibrium constant and Cell Potential

298314.8

)212000(ln

RT

GK c

Zn ↔ Zn2+ + 2e Eθ = +0.76 Cu2+ + 2e ↔ Cu Eθ = +0.34 Zn + Cu2+ → Zn 2+ + Cu Eθ = +1.10V

Zn half cell (-ve) Oxidation

Cu half cell (+ve) Reduction

Anode Cathode

Zn(s) | Zn2+(aq) || Cu2+

(aq) | Cu (s)

Cell diagram

Anode Cathode

Half Cell Half Cell

(Oxidation) (Reduction)

Salt Bridge Flow

electrons

Zn/Cu Voltaic Cell

-e -e

Zn/Cu half cells

Eθcell = Eθ

(cathode) – Eθ (anode)

Eθcell = +0.34 – (-0.76) = +1.10V

Zn 2+ + 2e ↔ Zn (anode) Eθ = -0.76V Cu2+ + 2e ↔ Cu (cathode) Eθ = +0.34V

Std electrode potential as std reduction potential

Find Eθcell (use reduction potential) Find Eθ

cell (use formula)

Eθcell = Eθ

(cathode) – Eθ(anode)

Zn 2+ + 2e ↔ Zn Eθ = -0.76V Cu2+ + 2e ↔ Cu Eθ = +0.34V

Oxidized sp ↔ Reduced sp Eθ/V

Li+ + e- ↔ Li -3.04 K+ + e- ↔ K -2.93 Ca2+ + 2e- ↔ Ca -2.87 Na+ + e- ↔ Na -2.71 Mg 2+ + 2e- ↔ Mg -2.37 Al3+ + 3e- ↔ AI -1.66 Mn2+ + 2e- ↔ Mn -1.19 H2O + e- ↔ 1/2H2 + OH- -0.83

Zn2+ + 2e- ↔ Zn - 0.76

Fe2+ + 2e- ↔ Fe -0.45 Ni2+ + 2e- ↔ Ni -0.26 Sn2+ + 2e- ↔ Sn -0.14 Pb2+ + 2e- ↔ Pb -0.13 H+ + e- ↔ 1/2H2 0.00 Cu2+ + e- ↔ Cu+ +0.15 SO4

2- + 4H+ + 2e- ↔ H2SO3 +0.17

Cu2+ + 2e- ↔ Cu + 0.34 1/2O2 + H2O +2e- ↔ 2OH- +0.40

+

+1.10 V

Eθ Zn/Cu = 1.10V

Cu2+

-

-

-

-

Zn Cu

+

+

+

+

cellnFEG

E cell with ∆G

F = Faraday constant (96 500 Cmol-1)

n = number electron

cellnFEG

kJJG

G

212212300

10.1965002

∆G –ve, E +ve, K > 1 ∆G <0, E > 0, K > 1

↓ Rxn Spontaneous cKRTG ln

Equilibrium

constant Gas constant, 8.314

∆G with Kc

cKRTG ln 37103.1 cK

Favour products

Page 9: IB Chemistry on Gibbs Free Energy, Equilibrium constant and Cell Potential

Zn ↔ Zn2+ + 2e Eθ = +0.76 2Ag++2e ↔ 2Ag Eθ = +0.80 Zn + Ag+ → Zn 2+ + Ag Eθ = +1.56V

Zn half cell (-ve) Oxidation

Ag half cell (+ve) Reduction

Anode Cathode

Zn(s) | Zn2+(aq) || Ag+

(aq) | Ag (s)

Cell diagram

Anode Cathode

Half Cell Half Cell

(Oxidation) (Reduction)

Salt Bridge Flow

electrons

Zn/Ag Voltaic Cell

-e -e

Zn/Ag half cells

Eθcell = Eθ

(cathode) – Eθ (anode)

Eθcell = +0.80 – (-0.76) = +1.56V

Zn 2+ + 2e ↔ Zn (anode) Eθ = -0.76V Ag + + e ↔ Ag(cathode) Eθ = +0.80V

Std electrode potential as std reduction potential

Find Eθcell (use reduction potential) Find Eθ

cell (use formula)

Eθcell = Eθ

(cathode) – Eθ(anode)

Zn 2+ + 2e ↔ Zn Eθ = -0.76V Ag+ + e ↔ Ag Eθ = +0.80V

Oxidized sp ↔ Reduced sp Eθ/V

Li+ + e- ↔ Li -3.04 K+ + e- ↔ K -2.93 Ca2+ + 2e- ↔ Ca -2.87 Na+ + e- ↔ Na -2.71 Mg 2+ + 2e- ↔ Mg -2.37 Al3+ + 3e- ↔ AI -1.66 Mn2+ + 2e- ↔ Mn -1.19 H2O + e- ↔ 1/2H2 + OH- -0.83

Zn2+ + 2e- ↔ Zn - 0.76

Fe2+ + 2e- ↔ Fe -0.45 Ni2+ + 2e- ↔ Ni -0.26 Sn2+ + 2e- ↔ Sn -0.14 Pb2+ + 2e- ↔ Pb -0.13 H+ + e- ↔ 1/2H2 0.00 Cu2+ + e- ↔ Cu+ +0.15 SO4

2- + 4H+ + 2e- ↔ H2SO3 +0.17 Cu2+ + 2e- ↔ Cu +0.34 1/2O2 + H2O +2e- ↔ 2OH- +0.40 Cu+ + e- ↔ Cu +0.52 1/2I2 + e- ↔ I- +0.54 Fe3+ + e- ↔ Fe2+ +0.77

Ag+ + e- ↔ Ag + 0.80 1/2Br2 + e- ↔ Br- +1.07

+

+1.56 V

Ag

Eθ Zn/Ag = +1.56V

Ag+

-

-

-

-

+

+

+

+

Zn

E cell with ∆G

cellnFEG

n = number electron F = Faraday constant (96 500 Cmol-1)

cellnFEG

kJJG

G

301301000

56.1965002

∆G with Kc

cKRTG ln

Gas constant, 8.314 Equilibrium

constant

cKRTG ln

298314.8

)301000(ln

RT

GK c

52105.3 cK

∆G –ve, E +ve, K > 1 ∆G <0, E > 0, K > 1

↓ Rxn Spontaneous

Favour products

Page 10: IB Chemistry on Gibbs Free Energy, Equilibrium constant and Cell Potential

Mn ↔ Mn2+ + 2e Eθ = +1.19 Ni2+ + 2e ↔ Ni Eθ = -0.26 Mn + Ni2+ → Mn2+ + Ni Eθ = +0.93V

Mn half cell (-ve) Oxidation

Ni half cell (+ve) Reduction

Anode Cathode

Mn(s) | Mn2+(aq) || Ni2+

(aq) | Ni (s)

Cell diagram

Anode Cathode

Half Cell Half Cell

(Oxidation) (Reduction)

Salt Bridge Flow

electrons

Mn/Ni Voltaic Cell

-e -e

Mn/Ni half cells

Eθcell = Eθ

(cathode) – Eθ (anode)

Eθcell = -0.26 – (-1.19) = +0.93V

Mn 2+ + 2e ↔ Mn (anode) Eθ = -1.19V Ni2+ + 2e ↔ Ni (cathode) Eθ = -0.26V

Std electrode potential as std reduction potential

Find Eθcell (use reduction potential) Find Eθ

cell (use formula)

Eθcell = Eθ

(cathode) – Eθ(anode)

Mn 2+ + 2e ↔ Mn Eθ = -1.19V Ni2+ + 2e ↔ Ni Eθ = -0.26V

Oxidized sp ↔ Reduced sp Eθ/V

Li+ + e- ↔ Li -3.04 K+ + e- ↔ K -2.93 Ca2+ + 2e- ↔ Ca -2.87 Na+ + e- ↔ Na -2.71 Mg 2+ + 2e- ↔ Mg -2.37 Al3+ + 3e- ↔ AI -1.66 Mn2+ + 2e- ↔ Mn -1.19

H2O + e- ↔ 1/2H2 -0.83 Zn2+ + 2e- ↔ Zn -0.76 Fe2+ + 2e- ↔ Fe -0.45

Ni2+ + 2e- ↔ Ni - 0.26

Sn2+ + 2e- ↔ Sn -0.14 Pb2+ + 2e- ↔ Pb -0.13 H+ + e- ↔ 1/2H2 0.00 Cu2+ + e- ↔ Cu+ +0.15 SO4

2- + 4H+ + 2e- ↔ H2SO3 + H2O +0.17 Cu2+ + 2e- ↔ Cu +0.34 1/2O2 + H2O +2e- ↔ 2OH- +0.40 Cu+ + e- ↔ Cu +0.52 1/2I2 + e- ↔ I- +0.54

+

+0.93 V

Eθ Mn/Ni = +0.93V

Ni2+

-

-

-

-

Ni Mn

+

+

+

+ Mn2+

E cell with ∆G

cellnFEG

n = number electron F = Faraday constant (96 500 Cmol-1)

cellnFEG

kJJG

G

179179490

93.0965002

cKRTG ln

298314.8

)179000(ln

RT

GK c

cKRTG ln

∆G with Kc

Gas constant, 8.314 Equilibrium

constant

∆G –ve, E +ve, K > 1 ∆G <0, E > 0, K > 1

↓ Rxn Spontaneous

31102.2 cK

Favour products

Page 11: IB Chemistry on Gibbs Free Energy, Equilibrium constant and Cell Potential

Oxidized sp ↔ Reduced sp Eθ/V

Li+ + e- ↔ Li -3.04 K+ + e- ↔ K -2.93 Ca2+ + 2e- ↔ Ca -2.87 Na+ + e- ↔ Na -2.71 Mg 2+ + 2e- ↔ Mg -2.37 Al3+ + 3e- ↔ AI -1.66 Mn2+ + 2e- ↔ Mn -1.19 H2O + e- ↔ H2 + OH- -0.83 Zn2+ + 2e- ↔ Zn -0.76 Fe2+ + 2e- ↔ Fe -0.45 Ni2+ + 2e- ↔ Ni -0.26 Sn2+ + 2e- ↔ Sn -0.14 H+ + e- ↔ H2 0.00 Cu2+ + e- ↔ Cu+ +0.15 SO4

2- + 4H+ + 2e- ↔ H2S +0.17 Cu2+ + 2e- ↔ Cu +0.34

Cu ↔ Cu2+ + 2e Eθ = -0.34 2H+ + 2e ↔ H2 Eθ = +0.00 Cu + 2H+→ Cu2+ +H2

Eθ = -0.34V

Rxn bet Cu + H+

Will it happen ?

Eθ = -0.34V (NON spontaneous) О

Cu(s) | Cu2+

(aq) || H+

H2 | Pt (s)

(Oxidation) (Reduction)

Anode Cathode

Find Eθcell (use formula)

Eθcell = Eθ

(cathode) – Eθ (anode)

Eθcell = 0.00 – (+0.34) = -0.34V

Eθ = -0.34V (NON spontaneous)

О

Rxn not feasible

Determine spontaneity rxn. Will it HAPPEN ?

Find Eθcell (use reduction potential)

Eθ Cu/H+ = - 0.34V

E cell with ∆G

cellnFEG

n = number electron F = Faraday constant (96 500 Cmol-1)

cellnFEG

kJJG

G

6565620

34.0965002

cKRTG ln

Gas constant, 8.314 Equilibrium

constant

∆G with Kc

cKRTG ln

298314.8

)65000(ln

RT

GK c

∆G +ve, E -ve, K < 1 ∆G >0, E < 0, K < 1

↓ Rxn Non Spontaneous

12104 cK

Favour reactants

-0.34 V

acid

copper

Predicting will rxn occur with ΔG, E cell and Kc

+

Page 12: IB Chemistry on Gibbs Free Energy, Equilibrium constant and Cell Potential

Oxidized sp ↔ Reduced sp Eθ/V

Li+ + e- ↔ Li -3.04 K+ + e- ↔ K -2.93 Ca2+ + 2e- ↔ Ca -2.87 Na+ + e- ↔ Na -2.71 Mg 2+ + 2e- ↔ Mg -2.37 Al3+ + 3e- ↔ AI -1.66 Mn2+ + 2e- ↔ Mn -1.19 H2O + e- ↔ H2 + OH- -0.83 Zn2+ + 2e- ↔ Zn -0.76 Fe2+ + 2e- ↔ Fe -0.45 Ni2+ + 2e- ↔ Ni -0.26 Sn2+ + 2e- ↔ Sn -0.14 H+ + e- ↔ H2 0.00 Cu2+ + e- ↔ Cu+ +0.15 SO4

2- + 4H+ + 2e- ↔ H2S +0.17 Cu2+ + 2e- ↔ Cu +0.34 Au3+ + 3e- ↔ Au +1.58

Rxn bet Au + H+

Will it happen ?

Eθ = -1.58 V (NON spontaneous)

О

Au(s) | Au3+(aq) || H

+ H2 | Pt (s)

(Oxidation) (Reduction)

Anode Cathode

Find Eθcell (use formula)

Eθcell = Eθ

(cathode) – Eθ (anode)

Eθcell = 0.00 – (+1.58) = -1.58V

Eθ = - 1.58 V (NON spontaneous)

О

Rxn not feasible

Determine spontaneity rxn. Will it HAPPEN ?

Find Eθcell (use reduction potential)

Eθ Au/H+ = - 1.58V

E cell with ∆G

cellnFEG

n = number electron F = Faraday constant (96 500 Cmol-1)

cellnFEG

kJJG

G

914914820

58.1965006

cKRTG ln

Gas constant, 8.314 Equilibrium

constant

∆G with Kc

cKRTG ln

298314.8

)914000(ln

RT

GK c

∆G +ve, E -ve, K < 1 ∆G >0, E < 0, K < 1

↓ Rxn Non Spontaneous

50104 cK

Kc too small – No reaction at all

-1.58 V

acid

gold

2Au ↔ 2Au3+ + 6e Eθ = -1.58 6H+ + 6e ↔ 3H2 E

θ = 0.00 2Au + 6H+ → 2Au3+ + 3H2

Eθ = -1.58V +

Predicting will rxn occur with ΔG, E cell and Kc

Page 13: IB Chemistry on Gibbs Free Energy, Equilibrium constant and Cell Potential

Eθ = - 0.20 V (NON spontaneous)

(Oxidation) (Reduction)

Anode Cathode

Find Eθcell (use formula)

Eθcell = Eθ

(cathode) – Eθ (anode)

Eθcell = 0.34 – (0.54) = - 0.20V

Eθ = - 0.20 V (NON spontaneous)

Determine spontaneity rxn. Will it HAPPEN ?

Find Eθcell (use reduction potential)

Eθ Cu2+/I- = - 0.20V

E cell with ∆G

cellnFEG

n = number electron F = Faraday constant (96 500 Cmol-1)

cellnFEG

kJJG

G

3838600

20.0965002

cKRTG ln

Gas constant, 8.314 Equilibrium

constant

∆G with Kc

cKRTG ln

298314.8

)38000(ln

RT

GK c

∆G +ve, E -ve, K < 1 ∆G >0, E < 0, K < 1

↓ Rxn Non Spontaneous

7102.2 cK

-1.58 V

Cu2+

I- Rxn bet Cu2+ +I-

Will it happen ?

2I- ↔ I2 + 2e Eθ = -0.54 Cu2+ + 2e ↔ Cu Eθ = +0.34 2I- + Cu2+→ Cu + I2

Eθ = -0.20V

Pt(s) | I-, I2 || Cu2+

(aq) | Cu (s)

Favour reactants

Oxidized sp ↔ Reduced sp Eθ/V

Li+ + e- ↔ Li -3.04 K+ + e- ↔ K -2.93 Ca2+ + 2e- ↔ Ca -2.87 Na+ + e- ↔ Na -2.71 Mg 2+ + 2e- ↔ Mg -2.37 Al3+ + 3e- ↔ AI -1.66 Mn2+ + 2e- ↔ Mn -1.19 Zn2+ + 2e- ↔ Zn -0.76 Fe2+ + 2e- ↔ Fe -0.45 Ni2+ + 2e- ↔ Ni -0.26 Sn2+ + 2e- ↔ Sn -0.14 H+ + e- ↔ 1/2H2 0.00 Cu2+ + e- ↔ Cu+ +0.15 Cu2+ + 2e- ↔ Cu +0.34 1/2O2 + H2O +2e- ↔ 2OH- +0.40 Cu+ + e- ↔ Cu +0.52 I2 + 2e- ↔ I- +0.54 Rxn not feasible

О О

- 0.20 V

Will I- oxidize Cu 2+ to Cu

Predicting will rxn occur with ΔG, E cell and Kc

Page 14: IB Chemistry on Gibbs Free Energy, Equilibrium constant and Cell Potential

Click here to view free energy

Predicting Spontaneity of Rxn

Thermodynamic, ΔG Equilibrium, Kc

1cK

1cK

KRTG lnG

veG

cK

1cK

Energetically favourable

0G

Predicting rxn will occur?

N2(g) + 3H2(g) ↔ 2NH3(g)

H2O(l) ↔ H+(aq)+ OH-

(aq)

Shift toward reactants

Energetically unfavourable

Non spontaneous

Mixture reactant/product Equilibrium

veG Spontaneous Shift toward product

79G

33G

610G

14101 cK

5105cK

Fe(s) + 3O2(g) ↔ 2Fe2O3(s) 261101cK

Shift toward reactants

Energetically unfavourable

Shift toward product

Energetically favourable

Energetically favourable

Kinetically unfavourable/(stable) Rate too slow due to HIGH activation energy

Rusting Process

Energy barrier

Shift toward product

Click here for notes

cellnFEG

Cell Potential

cellE

0cellE

0cellE

0cellE

0cellE

0cellE

0cellE

Page 15: IB Chemistry on Gibbs Free Energy, Equilibrium constant and Cell Potential

Eθ = +0.44V

IB Questions

Esterification produce ethyl ethanoate. ΔG = -4.38kJmol-1 Cal Kc

CH3COOH(l) + C2H5OH(l) ↔ CH3COOC2H5(l) + H2O(l)

Kc = 5.9

cKRTG lnRT

GK c

ln

29831.8

4380ln

cK

2

?cK

NO oxidized to NO2. Kc = 1.7 x 1012. Cal ∆G at 298K 1

3 4

2NO + O2 ↔ NO2 ?G

cKRTG ln

11

12

7.6969772

)107.1ln(298314.8

kJmolJmolG

G

Predict if iron react with HCI in absence air. Cal E cell , ∆G and Kc

Oxidized sp ↔ Reduced sp Eθ/V

Fe2+ + 2e- ↔ Fe -0.44 2H+ + 2e- ↔ H2 0.00 O2 +2H2O+4e ↔ 4OH- +0.40

Fe2+ + 2e- ↔ Fe -0.44 2H+ + 2e- ↔ H2 0.00

О О

Fe ↔ Fe2+ + 2e Eθ = +0.44 2H+ + 2e ↔ H2 E

θ = 0.00V Fe + 2H+ → Fe2+ + H2

Eθ = +0.44V

cellnFEG

kJJG

G

8584900

44.0965002

cKRTG ln

298314.8

)85000(ln

RT

GK c

14108.7 cK

∆G –ve, E +ve, K > 1 ∆G <0, E > 0, K > 1

↓ Rxn Spontaneous

Fe2+ + 2e- ↔ Fe -0.44 O2 +2H2O+4e ↔ 4OH- +0.40

2Fe ↔ 2Fe2+ + 4e Eθ = +0.44 O2+2H2O+4e ↔ 4OH- Eθ = +0.40 2Fe +O2

+2H2O→2Fe2++4OH- Eθ = +0.84V

Eθ = +0.84V

Oxidized sp ↔ Reduced sp Eθ/V

Fe2+ + 2e- ↔ Fe -0.44 2H+ + 2e- ↔ H2 0.00 O2 +2H2O+4e ↔ 4OH- +0.40

Predict iron react HCI in presence of air. Cal E cell , ∆G and Kc

О О

cellnFEG

kJJG

G

324324000

84.0965004

cKRTG ln

298314.8

)324000(ln

RT

GK c

56108.2 cK

∆G –ve, E +ve, K > 1 ∆G <0, E > 0, K > 1

↓ Rxn Spontaneous Rusting is spontaneous

x 2

О О

О О

Page 16: IB Chemistry on Gibbs Free Energy, Equilibrium constant and Cell Potential

Predict if manganate will oxidize chloride ion? MnO2 + 4H+ + 2CI- → Mn2+ + 2H2O + CI2

5

MnO2 +4H+ + 2e- ↔ Mn2+ + 2H2O +1.23

1/2CI2 + e- ↔ CI- +1.36

2CI- ↔ CI2 + 2e Eθ = -1.36 MnO2 + 4H+ + 2e ↔ Mn2+ + 2H2O Eθ = +1.23 MnO2 + 4H++2CI- → Mn2++2H2O+CI2 E

θ= -0.13V

Eθ = -0.13V

Oxidized sp ↔ Reduced sp Eθ/V

Cr2O72-+ 14H+ + 6e- ↔ 2Cr3+ + 7H2O +1.33

MnO2 +4H+ + 2e- ↔ Mn2+ + 2H2O +1.23

1/2CI2 + e- ↔ CI- +1.36 MnO4

- + 8H+ + 5e- ↔ Mn2+ + 4H2O +1.51

Predict if MnO4- able to oxidize aq CI- to CI2

2MnO4 + 16H+ + 10CI- → 2Mn2+ + 8H2O + 5CI2

О

О

Oxidized sp ↔ Reduced sp Eθ/V

Cr2O72-+ 14H+ + 6e- ↔ 2Cr3+ + 7H2O +1.33

MnO2 +4H+ + 2e- ↔ Mn2+ + 2H2O +1.23

1/2CI2 + e- ↔ CI- +1.36 MnO4

- + 8H+ + 5e- ↔ Mn2+ + 4H2O +1.51

О

О

2CI- ↔ CI2 + 2e Eθ = -1.36 MnO4

- + 8H+ + 5e ↔ Mn2+ + 4H2O Eθ = +1.51 2MnO4 + 16H++10CI- → 2Mn2++8H2O+5CI2 E

θ= +0.15V

1/2CI2 + e- ↔ CI- +1.36 MnO4

- + 8H+ + 5e- ↔ Mn2+ + 4H2O +1.51

Eθ = +0.15V

IB Questions

cellnFEG

kJJG

G

2525000

13.0965002

cKRTG ln

298314.8

)25000(ln

RT

GK c

5105.4 cK

∆G +ve, E -ve, K < 1 ∆G >0, E < 0, K < 1

↓ Rxn Non Spontaneous

6

cellnFEG

kJJG

G

144144750

15.09650010

cKRTG ln

298314.8

)144000(ln

RT

GK c

25105.1 cK

∆G –ve, E +ve, K > 1 ∆G <0, E > 0, K > 1

↓ Rxn Spontaneous

x 5 x 2

О

О

О

О