Option C Nernst Equation, Voltaic Cell and Concentration Cell

Option C Nernst Equation, Voltaic Cell and Concentration Cell

1. Types voltaic cell
Conversion electrical energy
to chemical energy
Electrochemistry
Electrolytic cell Voltaic cell
NH
4CI and ZnCI
2
Redox rxn
(Oxidation/reduction)
Movement electron
Produce electricity
Conversion chemical energy
to electrical energy
Electrodes – different metal (Half cell) Electrodes – same metal (Half cell)
Daniell cell Alkaline cell Dry cell Nickel cadmium cell
Primary cell (Non rechargeable)
MnO
2 and KOH
Secondary cell (Rechargeable)

2. Current – measured Amperes or Coulombs per second
1A = 1 Coulomb charge pass through a point in 1 s = 1C/s
1 Coulomb charge (elec) = 6.28 x 10
18
elec passing in 1 s









1 elec/proton carry charge of – 1.6 x 10 -
19
C ( very small)
6.28 x 10
18
elec carry charge of - 1 C
Electric current
Flow electric charges (elec, -ve)
From High to low electric potential
Potential Diff – measure with ammeter ond
electron
ond
Coulomb
A
sec.1
.1028.6
sec1
1
1
18


Current Electric Current – moving charges in solid wire or solution
Flow of
charges
-
-
-
Solid/Wire Solution/Electrolyte
Electron move in random
No current flow cause
No potential difference
Electrons & Protons
-
-
+
+
1A = 6.28 x 10
18
e
1 s


Potential Difference across wire
Electron move in one direction
Current flow
+ve ions -ve ions
(cations) (anions)
Potential Diff applied/Battery ItQ
t = Time/ s
Find amt charges pass through if
Current is 2.ooA, time is 15 min ItQ
Current flow
Q = Amt Charges/ C I = Current/ A CQ 1800601500.2 

3. Electric Potential C
J
Volt
1
1
-Measured in Volt with Voltmeter
- 1 V = 1 Joule energy released when 1 Coulomb
charge pass through 1 point
- 1 V = 1 J/C
V = Potential Diff
I = Current
R = Resistance
Potential diff bet 2 points is 1 V
↓
1 J energy released when 1 C charge passes through
Voltmeter across
1Volt
1 V

+ -
1 Ω 2 Ω
Charges (-ve)
flow down A
R
V
I
RIV
2
3
6

 VV
RIV
212

-
+
-
+ VV
RIV
422

Total current
Potential Diff(PD) vs Current
PD = Water Pressure
PD = 1.5V – 1.5J energy released 1C charge flow down
PD – cause charge flow = CURRENT
Potential Diff(PD) vs Current
1.5V = 1.5J/C
A
D
Electric potential/PD/Voltage = Electric Pressure = Volt
Electric Current = Charge flow = Amp
Electric Potential Energy = Work done to bring a charge to a point = Joule
Voltage NOT same as energy, Voltage = energy/charge
Battery lift charges, Q to higher potential
Potential Energy bet 2 terminals in battery stored as chemical energy
2A
2A
Potential Diff/Voltage
Potential Diff/Voltage

4. EMF vs PD
V = Potential Diff
I = Current
R = Resistance
Max potential diff bet two
electrodes of battery source.
+ -
1 Ω 2 Ω A
R
V
I
RIV
2
3
6

 VV
RIV
212
 VV
RIV
422

Total current
Current flow Circuit complete
Circuit complete
↓
Current flow
↓
Internal resistance
(battery - 1Ω)
↓
Terminal PD = 8V
(Voltage drop)
Potential Diff/Voltage in Volt
Symbol for EMF = E / ℰ
No Current flow in circuit
EMF (Electromotive Force) Volt
Battery = EMF = 9V
9 Volt ).(9 currentnoVEMFV
IRV


EMF Internal resistance Ir
Place voltmeter across – EMF= 9V
No current flow. A
rR
E
I
rRIE
IrIREMFE
1
9
9
)18(
9
)(
)(
)(






 VV
RIV
881
 VV
RIV
111

EMF = 8V+1V
8 Volt
1 Volt
EMF (6V) = 2V + 4V
4 Volt 2 Volt
Charges passing through wire
Current flow Circuit complete
Internal resistance
Collision bet + ve ions with elec
(drift velocity elec)
- +

5. E
θ
value DO NOT depend surface area of metal electrode.
E cell = Energy per unit charge. (Joule)/C
E cell- 10v = 10J energy released by 1C of charge
= 100J energy released by 10C of charge
E
θ
– intensive property– independent of amt – Ratio energy/charge
Increasing surface area metal will NOT increase E cell
E
θ
Zn/Cu = 1.10V

Surface area - 10 cm
2

Total charge- 100C leave electrode
E
cell = 1.1V = 1.1 J energy for 1 C (charges leaving)
1C release 1.1 J energy
100 C release 110 J energy
Voltmeter measure energy for 1C – 110J/100C – 1.1V
E
cell no change
Current – measured in Amp or Coulomb per s
1A = 1 Coulomb charge pass through a point in 1 s = 1C/s
1 Coulomb charge (elec) = 6.28 x 10
18
elec passing in 1 s









1 electron/proton carry charge of – 1.6 x 10 -
19
C ( very small)
6.28 x 10
18
electron carry charge of - 1 C ond
electron
ond
Coulomb
A
sec.1
.1028.6
sec1
1
1
18


Surface area increase ↑
Total Energy increase ↑
Total Charge increase ↑ Current increase ↑
BUT E cell remain SAME
E cell = (Energy/charge) t
Q
I
tIQ


Q up ↑ – I up ↑
100C flow
110J released VEcell
Ecell
eCh
Energy
Ecell
10.1
100
110
arg



Surface area - 100 cm
2

Total charge 1000C leave electrode
E
cell = 1.1V = 1.1 J energy for 1 C (charges leaving)
1 C release 1.1J energy
1000 C release 1100 J energy
Voltmeter measure energy for 1C – 1100J/1000C – 1.1V
E
cell no change VEcell
Ecell
eCh
Energy
Ecell
10.1
1000
1100
arg



E
θ
Zn/Cu = 1.10V

1000C flow
1100J released t
Q
I t
Q
I

Surface area exposed 10 cm
2

Surface area exposed 100 cm
2

6. Relationship bet ∆G and K
c cellnFEG

Relationship bet
Energetics and Equilibrium c
KRTG ln
 STHG 
Enthalpy
change
Entropy
change
Equilibrium
constant
Gibbs free
energy change H 
G
Relationship bet ∆G, K
c and E
cell cellnFEG
 STHG  cKRTG ln
 cK
Relationship bet
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, 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
shift to left (reactant)
shift to right (products) cellE
 
G cK K
nF
RT
Ecell ln

ΔG
θ
ln K K
c Eq mixture
ΔG
θ
-ve
< 0
Positive
( + )
K
c > 1 Product
(Right)
ΔG
θ
+ve
> 0
Negative
( - )
K
c < 1 Reactant
(left)
ΔG
θ
= 0 0 K
c = 1 Equilibrium

7. E cell/Voltage – depend on nature of material Q
nF
RT
EE ln

T = Temp in K
Q = Rxn Quotient
E
0
= std (1M)
n = # e transfer

F = Faraday constant
(96 500C mol
-1
)

R = Gas constant
(8.31) cKRTQRTG lnln KRTG
KRTQRTG
o
c
ln
lnln


When ratio conc, Q = 1,
all in std conc = 1M
Non std condition 01ln
1


RT
Q Q
nF
RT
EE ln
 QRTGG
o
ln
Non std condition o
nFEG
 nFEG QRTnFEnFE ln

Nernst equation
Work or Free energy to do work
depend on quantity material and surface area
E cell depend
Nature of electrode
Type of metal used Conc of ion Temp of sol
E
θ
Q
T

Current/I depend
Surface area
of contact
Salt bridge conc Size of
cation/anion
Resistance high ↑ – current low ↓
E cell depend
Surface area
of contact Salt bridge conc
Size of
cation/anion cellnFEG

Gibbs free
energy change
do do WORK
n = number
electron
F = Faraday constant
(96 500 Cmol
-1
)
Cell potential
Increasing surface area → increase charge Q and I current - Work increase
Current – depend on quantity and surface area

8. Zn

↔ Zn
2+
+ 2e E
θ
= +0.76
Cu
2+
+ 2e ↔ Cu E
θ
= +0.34
Zn + Cu
2+
→ Zn
2+
+ Cu E
θ
= +1.10V
Zn half cell (-ve)
Oxidation
Cu half cell (+ve)
Reduction
Anode Cathode
Zn
(s) | Zn
2+
(aq) || Cu
2+
(aq) | Cu
(s)
Anode Cathode
Half Cell Half Cell
(Oxidation) (Reduction)
Salt Bridge
Flow
electrons
Zn/Cu Cell - 1M std condition
-e -e
E
θ
cell = E
θ
(cathode) – E
θ

(anode)
E
θ
cell = +0.34 – (-0.76) = +1.10V
Zn
2+
+ 2e ↔ Zn (anode) E
θ
= -0.76V
Cu
2+
+ 2e ↔ Cu (cathode) E
θ
= +0.34V
E
θ
cell = E
θ
(cathode) – E
θ
(anode)
Zn
2+
+ 2e ↔ Zn E
θ
= -0.76V
Cu
2+
+ 2e ↔ Cu E
θ
= +0.34V
Oxidized sp ↔ Reduced sp E
θ
/V
Li
+
+ e- ↔ Li -3.04
K
+
+ e- ↔ K -2.93
Ca
2+
+ 2e- ↔ Ca -2.87
Na
+
+ e- ↔ Na -2.71
Mg
2+
+ 2e- ↔ Mg -2.37
Al
3+
+ 3e- ↔ AI -1.66
Mn
2+
+ 2e- ↔ Mn -1.19
H
2O + e- ↔ 1/2H
2 + OH
-
-0.83
Zn
2+
+ 2e- ↔ Zn - 0.76
Fe
2+
+ 2e- ↔ Fe -0.45
Ni
2+
+ 2e- ↔ Ni -0.26
Sn
2+
+ 2e- ↔ Sn -0.14
Pb
2+
+ 2e- ↔ Pb -0.13
H
+
+ e- ↔ 1/2H
2 0.00
Cu
2+
+ e- ↔ Cu
+
+0.15
SO
4
2-
+ 4H
+
+ 2e- ↔ H
2SO
3 +0.17
Cu
2+
+ 2e- ↔ Cu + 0.34
1/2O
2 + H
2O +2e- ↔ 2OH
-
+0.40
+
+1.10 V
E
θ
Zn/Cu = 1.10V

Cu
2+
-
-
-
-
Zn
Cu

+
+
+
+ cellnFEG

E cell with ∆G
F = Faraday constant
(96 500 Cmol
-1
)
n = number electron cellnFEG
 kJJG
G
212212300
10.1965002




Std electrode potential - std reduction potential
STD CONDITION
Zn/Cu half cell Cell diagram Q
nF
RT
EE ln

Ratio conc, Q = 1,
all in std conc = 1M, T = 298K VE
E
10.1
1ln
965002
298314.8
10.1





9. Zn

↔ Zn
2+
+ 2e E
θ
= +0.76
2Ag
+
+2e ↔ 2Ag E
θ
= +0.80
Zn + Ag
+
→ Zn
2+
+ Ag E
θ
= +1.56V
Zn half cell (-ve)
Oxidation
Ag half cell (+ve)
Reduction
Anode Cathode
Zn
(s) | Zn
2+
(aq) || Ag
+
(aq) | Ag
(s)
Anode Cathode
Half Cell Half Cell
(Oxidation) (Reduction)
Salt Bridge
Flow
electrons
-e -e
E
θ
cell = E
θ
(cathode) – E
θ

(anode)
E
θ
cell = +0.80 – (-0.76) = +1.56V
Zn
2+
+ 2e ↔ Zn (anode) E
θ
= -0.76V
Ag
+
+ e ↔ Ag(cathode) E
θ
= +0.80V
E
θ
cell = E
θ
(cathode) – E
θ
(anode)
Zn
2+
+ 2e ↔ Zn E
θ
= -0.76V
Ag
+
+ e ↔ Ag E
θ
= +0.80V
Oxidized sp ↔ Reduced sp E
θ
/V
Li
+
+ e- ↔ Li -3.04
K
+
+ e- ↔ K -2.93
Ca
2+
+ 2e- ↔ Ca -2.87
Na
+
+ e- ↔ Na -2.71
Mg
2+
+ 2e- ↔ Mg -2.37
Al
3+
+ 3e- ↔ AI -1.66
Mn
2+
+ 2e- ↔ Mn -1.19
H
2O + e- ↔ 1/2H
2 + OH
-
-0.83
Zn
2+
+ 2e- ↔ Zn - 0.76
Fe
2+
+ 2e- ↔ Fe -0.45
Ni
2+
+ 2e- ↔ Ni -0.26
Sn
2+
+ 2e- ↔ Sn -0.14
Pb
2+
+ 2e- ↔ Pb -0.13
H
+
+ e- ↔ 1/2H
2 0.00
Cu
2+
+ e- ↔ Cu
+
+0.15
SO
4
2-
+ 4H
+
+ 2e- ↔ H
2SO
3 +0.17
Cu
2+
+ 2e- ↔ Cu +0.34
1/2O
2 + H
2O +2e- ↔ 2OH
-
+0.40
Cu
+
+ e- ↔ Cu +0.52
1/2I
2 + e- ↔ I
-
+0.54
Fe
3+
+ e- ↔ Fe
2+
+0.77
Ag
+
+ e- ↔ Ag + 0.80
1/2Br
2 + e- ↔ Br
-
+1.07
+
+1.56 V
Ag

E
θ
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




Cell diagram Zn/Ag half cells
Ratio conc, Q = 1,
all in std conc = 1M, T = 298K
Zn/Ag Cell - 1M std condition Q
nF
RT
EE ln
 VE
E
56.1
1ln
965002
298314.8
56.1




STD CONDITION

10. Zn half cell (-ve)
Oxidation
Cu half cell (+ve)
Reduction
Zn/Cu Cell
-e -e
Zn
2+
+ 2e ↔ Zn E
θ
= -0.76V
Cu
2+
+ 2e ↔ Cu E
θ
= +0.34V
Zn

↔ Zn
2+
+ 2e E
θ
= +0.76V
Cu
2+
+ 2e ↔ Cu E
θ
= +0.34V
Zn + Cu
2+
→ Zn
2+
+ Cu E
θ
= +1.10V
Oxidized sp ↔ Reduced sp E
θ
/V
Li
+
+ e- ↔ Li -3.04
K
+
+ e- ↔ K -2.93
Ca
2+
+ 2e- ↔ Ca -2.87
Na
+
+ e- ↔ Na -2.71
Mg
2+
+ 2e- ↔ Mg -2.37
Al
3+
+ 3e- ↔ AI -1.66
Mn
2+
+ 2e- ↔ Mn -1.19
H
2O + e- ↔ 1/2H
2 + OH
-
-0.83
Zn
2+
+ 2e- ↔ Zn - 0.76
Fe
2+
+ 2e- ↔ Fe -0.45
Ni
2+
+ 2e- ↔ Ni -0.26
Sn
2+
+ 2e- ↔ Sn -0.14
Pb
2+
+ 2e- ↔ Pb -0.13
H
+
+ e- ↔ 1/2H
2 0.00
Cu
2+
+ e- ↔ Cu
+
+0.15
SO
4
2-
+ 4H
+
+ 2e- ↔ H
2SO
3 + H
2O +0.17
Cu
2+
+ 2e- ↔ Cu + 0.34
1/2O
2 + H
2O +2e- ↔ 2OH
-
+0.40
Cu
+
+ e- ↔ Cu +0.52
1/2I
2 + e- ↔ I
-
+0.54
+1.10 V
Cu
2+
-
-
-
-
Zn
Cu

+
+
+
+ Q
nF
RT
EE ln

1M 0.1M
Zn
2+ 10
]1.0[
]1[
][
][
2
2




c
c
Q
M
M
Cu
Zn
Q
0.1 M 1 M
Using Nernst Eqn
E
0
= Std condition (1M) – 1.10V
R = Gas constant (8.31)
n = # e transfer (2 e)
F = Faraday constant (96500C mol
-1
)
VE
E
E
07.1
03.010.1
)10ln(
)965002(
)29831.8(
10.1





Non std 0.1M
E cell decrease ↓ [Cu
2+
] decrease ↓
↓
Le Chatelier’s principle
Cu
2+
+ 2e ↔ Cu
↓
[Cu
2+
] decrease ↓
↓
Shift to left ←
↓
E cell → less ↓ → Cu
2+
less able ↓ to receive e-
[Cu
2+
] ↓ E cell < E
θ
1.07 < 1.10
Zn/Cu half cell
Zn +Cu
2+
→Zn
2+
+Cu

NON STD CONDITION

11. Zn half cell (-ve)
Oxidation
Cu half cell (+ve)
Reduction
Zn/Cu Cell
-e -e
Zn
2+
+ 2e ↔ Zn E
θ
= -0.76V
Cu
2+
+ 2e ↔ Cu E
θ
= +0.34V
Zn

↔ Zn
2+
+ 2e E
θ
= +0.76V
Cu
2+
+ 2e ↔ Cu E
θ
= +0.34V
Zn + Cu
2+
→ Zn
2+
+ Cu E
θ
= +1.10V
Oxidized sp ↔ Reduced sp E
θ
/V
Li
+
+ e- ↔ Li -3.04
K
+
+ e- ↔ K -2.93
Ca
2+
+ 2e- ↔ Ca -2.87
Na
+
+ e- ↔ Na -2.71
Mg
2+
+ 2e- ↔ Mg -2.37
Al
3+
+ 3e- ↔ AI -1.66
Mn
2+
+ 2e- ↔ Mn -1.19
H
2O + e- ↔ 1/2H
2 + OH
-
-0.83
Zn
2+
+ 2e- ↔ Zn - 0.76
Fe
2+
+ 2e- ↔ Fe -0.45
Ni
2+
+ 2e- ↔ Ni -0.26
Sn
2+
+ 2e- ↔ Sn -0.14
Pb
2+
+ 2e- ↔ Pb -0.13
H
+
+ e- ↔ 1/2H
2 0.00
Cu
2+
+ e- ↔ Cu
+
+0.15
SO
4
2-
+ 4H
+
+ 2e- ↔ H
2SO
3 + H
2O +0.17
Cu
2+
+ 2e- ↔ Cu + 0.34
1/2O
2 + H
2O +2e- ↔ 2OH
-
+0.40
Cu
+
+ e- ↔ Cu +0.52
1/2I
2 + e- ↔ I
-
+0.54
+1.10 V
Cu
2+
-
-
-
-
Zn
Cu

+
+
+
+ Q
nF
RT
EE ln

1M 10M
Zn
2+ 1.0
]10[
]1[
][
][
2
2




c
c
Q
M
M
Cu
Zn
Q
10 M 1 M
Using Nernst Eqn
E
0
=Std condition (1M) – 1.10V
R = Gas constant (8.31)
n = # e transfer (2 e)
F = Faraday constant (96500C mol
-1
)
VE
E
E
13.1
03.010.1
)1.0ln(
)965002(
)29831.8(
10.1





Non std 0.1M
E cell increase ↑ [Cu
2+
] increase ↑
↓
Le Chatelier’s principle
Cu
2+
+ 2e ↔ Cu
↓
[Cu
2+
] increase ↑
↓
Shift to right →
↓
E cell → more ↑→ Cu
2+
more able receive e-
[Cu
2+
] ↑ E cell > E
θ
1.13 > 1.10
Zn/Cu half cell
Zn +Cu
2+
→Zn
2+
+Cu

NON STD CONDITION

12. Zn half cell (-ve)
Oxidation
Cu half cell (+ve)
Reduction
Zn/Cu Cell
-e -e
Zn
2+
+ 2e ↔ Zn E
θ
= -0.76V
Cu
2+
+ 2e ↔ Cu E
θ
= +0.34V
Zn

↔ Zn
2+
+ 2e E
θ
= +0.76V
Cu
2+
+ 2e ↔ Cu E
θ
= +0.34V
Zn + Cu
2+
→ Zn
2+
+ Cu E
θ
= +1.10V
Oxidized sp ↔ Reduced sp E
θ
/V
Li
+
+ e- ↔ Li -3.04
K
+
+ e- ↔ K -2.93
Ca
2+
+ 2e- ↔ Ca -2.87
Na
+
+ e- ↔ Na -2.71
Mg
2+
+ 2e- ↔ Mg -2.37
Al
3+
+ 3e- ↔ AI -1.66
Mn
2+
+ 2e- ↔ Mn -1.19
H
2O + e- ↔ 1/2H
2 + OH
-
-0.83
Zn
2+
+ 2e- ↔ Zn - 0.76
Fe
2+
+ 2e- ↔ Fe -0.45
Ni
2+
+ 2e- ↔ Ni -0.26
Sn
2+
+ 2e- ↔ Sn -0.14
Pb
2+
+ 2e- ↔ Pb -0.13
H
+
+ e- ↔ 1/2H
2 0.00
Cu
2+
+ e- ↔ Cu
+
+0.15
SO
4
2-
+ 4H
+
+ 2e- ↔ H
2SO
3 + H
2O +0.17
Cu
2+
+ 2e- ↔ Cu + 0.34
1/2O
2 + H
2O +2e- ↔ 2OH
-
+0.40
Cu
+
+ e- ↔ Cu +0.52
1/2I
2 + e- ↔ I
-
+0.54
+1.10 V
Cu
2+
-
-
-
-
Zn
Cu

+
+
+
+ Q
nF
RT
EE ln

0.1M 1M
Zn
2+ 1.0
]1[
]1.0[
][
][
2
2




c
c
Q
M
M
Cu
Zn
Q
1 M 0.1 M
Using Nernst Eqn
E
0
= Std condition (1M) – 1.10V
R = Gas constant (8.31)
n = # e transfer (2 e)
F = Faraday constant (96500C mol
-1
)
VE
E
E
13.1
03.010.1
)1.0ln(
)965002(
)29831.8(
10.1





Non std 0.1M
E cell increase ↑ [Zn
2+
] decrease ↓
↓
Le Chatelier’s principle
Zn
2+
+ 2e ↔ Zn
↓
[Zn
2+
] decrease ↓
↓
Shift to left ←
↓
E cell → more ↑→ Zn

more able lose elec
[Zn
2+
] ↓ E cell > E
θ
1.13 > 1.10
Zn/Cu half cell Zn + Cu
2+
→ Zn
2+
+ Cu

NON STD CONDITION

13. Cu half cell (-ve)
Oxidation
Cu half cell (+ve)
Reduction
-e
Cu ↔ Cu
2+
+ 2e E
θ
= - 0.34V
Cu
2+
+ 2e ↔ Cu E
θ
= +0.34V
Cu

↔ Cu
2+
+ 2e E
θ
= - 0.34V
Cu
2+
+ 2e ↔ Cu E
θ
= +0.34V
Cu + Cu
2+
→ Cu
2+
+ Cu E
θ
= 0V
Oxidized sp ↔ Reduced sp E
θ
/V
Li
+
+ e- ↔ Li -3.04
K
+
+ e- ↔ K -2.93
Ca
2+
+ 2e- ↔ Ca -2.87
Na
+
+ e- ↔ Na -2.71
Mg
2+
+ 2e- ↔ Mg -2.37
Al
3+
+ 3e- ↔ AI -1.66
Mn
2+
+ 2e- ↔ Mn -1.19
H
2O + e- ↔ 1/2H
2 + OH
-
-0.83
Zn
2+
+ 2e- ↔ Zn -0.76
Fe
2+
+ 2e- ↔ Fe -0.45
Ni
2+
+ 2e- ↔ Ni -0.26
Sn
2+
+ 2e- ↔ Sn -0.14
Pb
2+
+ 2e- ↔ Pb -0.13
H
+
+ e- ↔ 1/2H
2 0.00
Cu
2+
+ e- ↔ Cu
+
+0.15
SO
4
2-
+ 4H
+
+ 2e- ↔ H
2SO
3 + H
2O +0.17
Cu
2+
+ 2e- ↔ Cu + 0.34
1/2O
2 + H
2O +2e- ↔ 2OH
-
+0.40
Cu
2+
Zn
Cu

+
+
+
+ Q
nF
RT
EE ln

0.1M
01.0
]1.0[
]001.0[
][
][
2
2




c
cathode
anode
c
Q
Cu
Cu
Q
0.1 M 0.001 M
Using Nernst Eqn
E
0
= Std condition (1M) – 1.10V
R = Gas constant (8.31)
n = # e transfer (2 e)
F = Faraday constant (96500C mol
-1
)
VE
E
E
0285.0
0285.00
)01.0ln(
)965002(
)29831.8(
0





Cu
2+/
Cu half cell
Cu + Cu
2+
→ Cu
2+
+ Cu

-e
Cu
2+
0.001M
Cu
(s) │Cu
2+
(aq) (0.001M) ║ Cu
2+
(aq) (0.1M)│Cu
(s)
-
-
-
-
Concentration cell
Electrode same - diff conc
Oxi cell – anode – lower conc
Red cell – cathode – higher conc
cathode anode
Cu

Conc cell made of Zn/Zn
2+

Conc Zn
2+
- 0.11M and 0.22M. Find voltage.
Zn
(s) │Zn
2+
(aq) (0.11M) ║ Zn
2+
(aq) (0.22M)│Zn
(s)
Zn + Zn
2+
→ Zn
2+
+ Zn

cathode anode
0.22M 0.11 M 5.0
]22.0[
]11.0[
][
][
2
2




c
cathode
anode
c
Q
Zn
Zn
Q Q
nF
RT
EE ln
 VE
E
0089.0
)5.0ln(
)965002(
)29831.8(
0





14. Fe half cell (-ve)
Oxidation
Fe half cell (+ve)
Reduction
-e
Fe ↔ Fe
2+
+ 2e E
θ
= + 0.45V
Fe
2+
+ 2e ↔ Fe E
θ
= - 0.45V
Fe

↔ Fe
2+
+ 2e E
θ
= + 0.45V
Fe
2+
+ 2e ↔ Fe E
θ
= - 0.45 V
Fe + Fe
2+
→ Fe
2+
+Fe E
θ
= 0V
Oxidized sp ↔ Reduced sp E
θ
/V
Li
+
+ e- ↔ Li -3.04
K
+
+ e- ↔ K -2.93
Ca
2+
+ 2e- ↔ Ca -2.87
Na
+
+ e- ↔ Na -2.71
Mg
2+
+ 2e- ↔ Mg -2.37
Al
3+
+ 3e- ↔ AI -1.66
Mn
2+
+ 2e- ↔ Mn -1.19
H
2O + e- ↔ 1/2H
2 + OH
-
-0.83
Zn
2+
+ 2e- ↔ Zn -0.76
Fe
2+
+ 2e- ↔ Fe -0.45
Ni
2+
+ 2e- ↔ Ni -0.26
Sn
2+
+ 2e- ↔ Sn -0.14
Pb
2+
+ 2e- ↔ Pb -0.13
H
+
+ e- ↔ 1/2H
2 0.00
Cu
2+
+ e- ↔ Cu
+
+0.15
SO
4
2-
+ 4H
+
+ 2e- ↔ H
2SO
3 + H
2O +0.17
Fe
2+
Zn
Fe

+
+
+
+ Q
nF
RT
EE ln

0.1M
1.0
]1.0[
]01.0[
][
][
2
2




c
cathode
anode
c
Q
Fe
Fe
Q
0.1 M 0.01 M
Using Nernst Eqn
E
0
= Std condition (1M) – 1.10V
R = Gas constant (8.31)
n = # e transfer (2 e)
F = Faraday constant (96500C mol
-1
)
VE
E
E
029.0
029.00
)1.0ln(
)965002(
)29831.8(
0





Fe
2+/
Fe half cell
Fe + Fe
2+
→ Fe
2+
+ Fe

-e
Fe
2+
0.01M
Fe
(s)│Fe
2+
(aq) (0.01M) ║ Fe
2+
(aq) (0.1M)│Fe
(s)
-
-
-
-
Concentration cell
Electrode same - in diff conc
Oxi cell – anode – lower conc
Red cell – cathode – higher conc
cathode anode
Fe

Find cell potential
Mn
(s) │Mn
2+
(aq) (0.1M) ║ Pb
2+
(aq) (0.0001M)│Pb
(s)
Mn + Pb
2+
→ Mn
2+
+ Pb

0.0001M 0.1 M
cathode anode 001.0
]0001.0[
]1.0[
][
][
2
2




c
cathode
anode
c
Q
Pb
Mn
Q Q
nF
RT
EE ln
 VE
E
96.0
)001.0ln(
)965002(
)29831.8(
05.1



