kuliah KF2 (elkim2)

7/29/2019 kuliah KF2 (elkim2)

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Electrochemical thermodynamics


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Electrochemical thermodynamics

elw Q E nF E

The electrical work wel that can be done by

this system is:

el T,pw G

welectrical V Q

since Q n F

n F E


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where n is the number of

electrons transferred and F is

Faradays constant


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Combining Reduction

Half-Equations

Fe3+(aq) + 3e- Fe(s) EFe3+/Fe = ?

Fe2+(aq) + 2e- Fe(s) EFe2+/Fe = -0.440 V

Fe3+(aq) + 1e- Fe2+(aq) EFe3+/Fe2+ = 0.771 V

Fe3+(aq) + 3e- Fe(s)

G= +0.880 (F) J

G= -0.771 (F) J

G= +0.109 (F) JEFe3+/Fe = +0.331 V

G= +0.109 (F) J = -nFE

EFe3+/Fe= +0.109 (F) J /(-3F) = -0.0363 V

but cannot simply add E

can add G


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Hitung E0 untuk reaksi :

Cr2+ + 2e Cr (s) E01

= ?

Diketahui :

Cr3+ + 3e Cr (s) E02 = 0,5 V

Cr3+ + e Cr2+ E03 = 0,41 V

Solution :

G01 = 2FE0

1

G0

2 = 3FE0

2

G03 = FE0

3

G01 = G0

2G0

3

G01 = 2FE0

1


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Spontaneous Change in

Oxidation-Reduction Reactions

G < 0 for spontaneous change.

Therefore Ecell > 0 because Gcell = -nFEcell

Ecell> 0 Reaction proceeds spontaneously as written.

Ecell = 0

Reaction is at equilibrium.

Ecell< 0 Reaction proceeds in the reverse direction

spontaneously.

G = H TS


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The Behavior or Metals Toward Acids

M(s) M2+(aq) + 2 e- E= EM2+/M

2 H+(aq) + 2 e- H2(g) EH+/H2 = 0 V

2 H+(aq) + M(s) H2(g) + M2+(aq)

Ecell = EH+/H2 EM2+/M = EM2+/M

When EM2+/M < 0, Ecell > 0. Therefore G


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Will aluminum metal displace Cu2+ from

aqueous solution ? That is, will a

spontaneous reaction occur in the forward

direction for the following reaction ?

2 Al(s) + 3 Cu2+ (1 M) 3 Cu(s) + 2 Al

3+ (1 M)


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Because E0cell is positive, the

direction of spontaneous change

is that of the forward reaction

Al(s) will displace Cu2+ from

aqueous solution under

standard-state conditions


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Relationship Between Ecell and Keq

G = -RTln Keq = -nFEcell

Ecell =nF

RTln Keq

Ecell =z

0.025693 ln Keq= (0.0592/n ) log Keq


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FIGURE 20-8 A summary of important thermodynamic, equilibriumand electrochemical relationships under standard conditions.


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G = G + RTln Q

nFEcell = nFEcell + RTln Q

Convert to log10 and calculate constants.

Ecell = Ecell log Qz

0.0592 VThe Nernst Equation

Ecell = Ecell ln QnF

RT


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Concentration Cells


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Concentration Cells

Electrolyte concentration cell

the electrodes are identical; they

simply differ in the concentration ofelectrolyte in the half-cells.


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Concentration Cells

Electrode concentration cells

the electrodes themselves have

different compositions. This may bedue to.

Different fugacities of gases involved in

electrode reactions (e.g., The H+(aq)/H2(g)

electrode). Different compositions of metal

amalgams in electrode materials.


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Concentration Cells

A concentration cell

Two half cells with identical electrodes but different

ion concentrations.

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


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Measurement ofKsp

A concentration cell for determining Ksp of AgI

Ag+(0.100 M) Ag+(satd M)

Ag|Ag+(satd AgI)||Ag+(0.10 M)|Ag(s)

Ag+(0.100 M) + e- Ag(s)

Ag(s) Ag+(satd) + e-


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Calculating Redox Equilibrium Constants

Example : Calculate the equilibrium constant

for the reaction

2Fe3+ + 3I- 2Fe2+ + I3-

Sol: 2Fe3+ + 2e- 2Fe2+ E0 = 0.771 V

I3- + 2e- 3I- E0 = 0.536 V

23

22

/FeFe

0

/FeFe ]Fe[

]Fe[

log2

0592.0

EE23

23

][I

]I[log

2

0592.0EE

3

3

/II0

/II 33


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]I[

]I[log

]Fe[

]Fe[log

0.0592

)E-2(E

]I[

]I[log

2

0592.0E

]Fe[

]Fe[log

2

0592.0E

EE

3

3

23

22/II

0/FeFe

03

3

/II0

23

22

/FeFe0

/II/FeFe

--3

23

323

323

323

3

22

]I[]Fe[

]I[][Fe

log

0592.0

)EE(2

]I[]Fe[

]I[][Felog

/II0

/FeFe0

323

3

223

23

7

eq

/II0

/FeFe0

eq

107.894.7loganti

94.70592.0

)536.0771.0(2

0592.0

)EE(2log 3

23

K

K


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Calculate the equilibrium constant for the reaction

2MnO4-

+ 3Mn2+

+ 2H2O

5MnO2(s) + 4H+

Sol: 2MnO4- + 8H+ +6e- 2MnO2(s) + 4H2O E

0 = +1.695 V

3MnO2(s) + 12H+ + 6e- 3Mn2+ + 6H

2O E0 = +1.23 V

EMnO4-/MnO2= EMnO2/Mn2+

12

32

82

4 ]H[

]Mn[log

6

0592.023.1

]H[]MnO[

1log

6

0592.0695.1


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8322

4

12

]H[]Mn[]MnO[

]H[log

0592.0

)23.1695.1(6

eq322

4

4

log

]Mn[][MnO

]H[log1.47 K

47

eq 1011.47loganti K

32

12

82

4]Mn[

]H[log

]H[]MnO[

1log

0592.0

)23.1695.1(6


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Cd CdCl2 25H2O AgCl(s) Ag

E.m.f sel pada 15C = 0,67531 V dan koefisien

temperatur e.m.f = 0,00065 V der 1.

Hitung harga H pada 15C dan aliran panas

jika proses berlangsung reversibel

G = nFE

= (2)(96500 C)(0,67531 V)

S =(2)(96500 C)(0,00065 V der 1)

H = G + TS

Qp = TS

Example :

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