kuliah kf2 (elkim2)
TRANSCRIPT
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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 :