Option C Nernst Equation, Voltaic Cell and Concentration Cell

1. http://lawrencekok.blogspot.com
Prepared by
Lawrence Kok
Tutorial on Voltaic Cell, Nernst Equation and
concentration cell.

2. Types voltaic cell
Conversion electrical energy
to chemical energy
Electrochemistry
Electrolytic cellVoltaic cell
NH4CI and ZnCI2
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 cellDry cell Nickel cadmium cell
Primary cell (Non rechargeable)
MnO2 and KOH
Secondary cell (Rechargeable)

3. 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/WireSolution/Electrolyte
Electron move in random
No current flow cause
No potential difference
Electrons & Protons
-
-
+
+
1A = 6.28 x 1018
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 =××=

4. 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
DElectric 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/VoltagePotential Diff/Voltage

5. 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 Volt2 Volt
Charges passing through wire
Current flow Circuit complete
Internal resistance
Collision bet + ve ions with elec
(drift velocity elec)
- +

6. Eθ
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
Eθ
– intensive property– independent of amt – Ratio energy/charge
Increasing surface area metal will NOT increase E cell
Eθ
Zn/Cu = 1.10V
Surface area - 10 cm2
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 cm2
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
=
=
=
Eθ
Zn/Cu = 1.10V
1000C flow
1100J released
t
Q
I =
t
Q
I =
Surface area exposed 10 cm2
Surface area exposed 100 cm2

7. ∆G θ
=
-nFE θ
cell
Relationship bet ∆G and Kc
cellnFEG −=∆ θ
Relationship bet
Energetics and Equilibrium
cKRTG ln−=∆ θ
STHG ∆−∆=∆
Enthalpy
change
Entropy
change
Equilibrium
constant
Gibbs free
energy change
H∆θ
G∆
Relationship bet ∆G, Kc and E cell
cellnFEG −=∆ θ
STHG ∆−∆=∆ cKRTG ln−=∆ θ
cK
Relationship bet
Energetics and Cell Potential
θ
G∆ cellEθ
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
shift to left (reactant)
shift to right (products)
cellEθ
θ
G∆
cK
∆G
θ =
-RT
ln
K c
K
nF
RT
E cell ln=°
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

8. E cell/Voltage – depend on nature of material
Q
nF
RT
EE ln−= °
T = Temp in K
Q = Rxn Quotient
E0
= 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 −=∆ θnFEG −=∆
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
Eθ
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 −=∆ θ
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

9. Zn Zn↔ 2+
+ 2e Eθ
= +0.76
Cu2+
+ 2e Cu E↔ θ
= +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)
Anode Cathode
Half Cell Half Cell
(Oxidation) (Reduction)
Salt Bridge Flow
electrons
Zn/Cu Cell - 1M std condition
-e -e
Eθ
cell = Eθ
(cathode) – Eθ
(anode)
Eθ
cell = +0.34 – (-0.76) = +1.10V
Zn 2+
+ 2e Zn (anode) E↔ θ
= -0.76V
Cu2+
+ 2e Cu (cathode) E↔ θ
= +0.34V
Eθ
cell = Eθ
(cathode) – Eθ
(anode)
Zn 2+
+ 2e Zn E↔ θ
= -0.76V
Cu2+
+ 2e Cu E↔ θ
= +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
Mg2+
+ 2e- Mg↔ -2.37
Al3+
+ 3e- AI↔ -1.66
Mn2+
+ 2e- Mn↔ -1.19
H2O + e- 1/2H↔ 2 + 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- 1/2H↔ 2 0.00
Cu2+
+ e- Cu↔ +
+0.15
SO4
2-
+ 4H+
+ 2e- H↔ 2SO3 +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 −=∆ θ
E cell with ∆G
F = Faraday constant
(96 500 Cmol-1
)
n = number electron
cellnFEG −=∆ θ
kJJG
G
212212300
10.1965002
−=−=∆
××−=∆
θ
θ
Std electrode potential - std reduction potential
STD CONDITION
Zn/Cu half cellCell 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
=
×
×
−=

10. Zn Zn↔ 2+
+ 2e Eθ
= +0.76
2Ag+
+2e 2Ag E↔ θ
= +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)
Anode Cathode
Half Cell Half Cell
(Oxidation) (Reduction)
Salt Bridge Flow
electrons
-e -e
Eθ
cell = Eθ
(cathode) – Eθ
(anode)
Eθ
cell = +0.80 – (-0.76) = +1.56V
Zn 2+
+ 2e Zn (anode) E↔ θ
= -0.76V
Ag+
+ e Ag(cathode) E↔ θ
= +0.80V
Eθ
cell = Eθ
(cathode) – Eθ
(anode)
Zn 2+
+ 2e Zn E↔ θ
= -0.76V
Ag+
+ e Ag E↔ θ
= +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
Mg2+
+ 2e- Mg↔ -2.37
Al3+
+ 3e- AI↔ -1.66
Mn2+
+ 2e- Mn↔ -1.19
H2O + e- 1/2H↔ 2 + 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- 1/2H↔ 2 0.00
Cu2+
+ e- Cu↔ +
+0.15
SO4
2-
+ 4H+
+ 2e- H↔ 2SO3 +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 −=∆ θ
n = number electron F = Faraday constant
(96 500 Cmol-1
)
cellnFEG −=∆ θ
kJJG
G
301301000
56.1965002
−=−=∆
××−=∆
θ
θ
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

11. Zn half cell (-ve)
Oxidation
Cu half cell (+ve)
Reduction
Zn/Cu Cell
-e -e
Zn 2+
+ 2e Zn E↔ θ
= -0.76V
Cu2+
+ 2e Cu E↔ θ
= +0.34V
Zn Zn↔ 2+
+ 2e Eθ
= +0.76V
Cu2+
+ 2e Cu E↔ θ
= +0.34V
Zn + Cu2+
Zn→ 2+
+ Cu Eθ
= +1.10V
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
Mg2+
+ 2e- Mg↔ -2.37
Al3+
+ 3e- AI↔ -1.66
Mn2+
+ 2e- Mn↔ -1.19
H2O + e- 1/2H↔ 2 + 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- 1/2H↔ 2 0.00
Cu2+
+ e- Cu↔ +
+0.15
SO4
2-
+ 4H+
+ 2e- H↔ 2SO3 + 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
+1.10 V
Cu2+
-
-
-
-
Zn Cu
+
+
+
+
Q
nF
RT
EE ln−= ° 1M 0.1M
Zn2+
10
]1.0[
]1[
][
][
2
2
=
== +
+
c
c
Q
M
M
Cu
Zn
Q
0.1 M 1 M
Using Nernst Eqn
E0
= 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 ↓ [Cu2+
] decrease ↓
↓
Le Chatelier’s principle
Cu2+
+ 2e Cu↔
↓
[Cu2+
] decrease ↓
↓
Shift to left ←
↓
E cell → less ↓ → Cu2+
less able ↓ to receive e-
[Cu2+
] ↓ E cell < Eθ
1.07 < 1.10
Zn/Cu half cellZn +Cu2+
→Zn2+
+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
Cu2+
+ 2e Cu E↔ θ
= +0.34V
Zn Zn↔ 2+
+ 2e Eθ
= +0.76V
Cu2+
+ 2e Cu E↔ θ
= +0.34V
Zn + Cu2+
Zn→ 2+
+ Cu Eθ
= +1.10V
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
Mg2+
+ 2e- Mg↔ -2.37
Al3+
+ 3e- AI↔ -1.66
Mn2+
+ 2e- Mn↔ -1.19
H2O + e- 1/2H↔ 2 + 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- 1/2H↔ 2 0.00
Cu2+
+ e- Cu↔ +
+0.15
SO4
2-
+ 4H+
+ 2e- H↔ 2SO3 + 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
+1.10 V
Cu2+
-
-
-
-
Zn Cu
+
+
+
+
Q
nF
RT
EE ln−= ° 1M 10M
Zn2+
1.0
]10[
]1[
][
][
2
2
=
== +
+
c
c
Q
M
M
Cu
Zn
Q
10 M 1 M
Using Nernst Eqn
E0
=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 ↑ [Cu2+
] increase ↑
↓
Le Chatelier’s principle
Cu2+
+ 2e ↔ Cu
↓
[Cu2+
] increase ↑
↓
Shift to right →
↓
E cell → more ↑→ Cu2+
more able receive e-
[Cu2+
] ↑ E cell > Eθ
1.13 > 1.10
Zn/Cu half cellZn +Cu2+
→Zn2+
+Cu
NON STD CONDITION

13. Zn half cell (-ve)
Oxidation
Cu half cell (+ve)
Reduction
Zn/Cu Cell
-e -e
Zn 2+
+ 2e Zn E↔ θ
= -0.76V
Cu2+
+ 2e Cu E↔ θ
= +0.34V
Zn Zn↔ 2+
+ 2e Eθ
= +0.76V
Cu2+
+ 2e Cu E↔ θ
= +0.34V
Zn + Cu2+
Zn→ 2+
+ Cu Eθ
= +1.10V
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
Mg2+
+ 2e- Mg↔ -2.37
Al3+
+ 3e- AI↔ -1.66
Mn2+
+ 2e- Mn↔ -1.19
H2O + e- 1/2H↔ 2 + 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- 1/2H↔ 2 0.00
Cu2+
+ e- Cu↔ +
+0.15
SO4
2-
+ 4H+
+ 2e- H↔ 2SO3 + 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
+1.10 V
Cu2+
-
-
-
-
Zn Cu
+
+
+
+
Q
nF
RT
EE ln−= ° 0.1M 1M
Zn2+
1.0
]1[
]1.0[
][
][
2
2
=
== +
+
c
c
Q
M
M
Cu
Zn
Q
1 M 0.1 M
Using Nernst Eqn
E0
= 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 ↑ [Zn2+
] decrease ↓
↓
Le Chatelier’s principle
Zn2+
+ 2e ↔ Zn
↓
[Zn2+
] decrease ↓
↓
Shift to left ←
↓
E cell → more ↑→ Zn more able lose elec
[Zn2+
] ↓ E cell > Eθ
1.13 > 1.10
Zn/Cu half cellZn + Cu2+
→ Zn2+
+ Cu
NON STD CONDITION

14. Cu half cell (-ve)
Oxidation
Cu half cell (+ve)
Reduction
-e
Cu Cu↔ 2+
+ 2e Eθ
= - 0.34V
Cu2+
+ 2e Cu E↔ θ
= +0.34V
Cu Cu↔ 2+
+ 2e Eθ
= - 0.34V
Cu2+
+ 2e Cu E↔ θ
= +0.34V
Cu + Cu2+
Cu→ 2+
+ Cu Eθ
= 0V
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
Mg2+
+ 2e- Mg↔ -2.37
Al3+
+ 3e- AI↔ -1.66
Mn2+
+ 2e- Mn↔ -1.19
H2O + e- 1/2H↔ 2 + 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- 1/2H↔ 2 0.00
Cu2+
+ e- Cu↔ +
+0.15
SO4
2-
+ 4H+
+ 2e- H↔ 2SO3 + H2O +0.17
Cu2+
+ 2e- ↔ Cu + 0.34
1/2O2 + H2O +2e- ↔ 2OH-
+0.40
Cu2+
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
E0
= 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
=
+=
×
×
−=
Cu2+/
Cu half cell
Cu + Cu2+
→ Cu2+
+ Cu
-e
Cu2+
0.001M
Cu (s) │Cu2+
(aq) (0.001M) ║ Cu2+
(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/Zn2+
Conc Zn2+
- 0.11M and 0.22M. Find voltage.
Zn (s) │Zn2+
(aq) (0.11M) ║ Zn2+
(aq) (0.22M)│Zn(s)
Zn + Zn2+
→ Zn2+
+ 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
=
×
×
−=

15. Fe half cell (-ve)
Oxidation
Fe half cell (+ve)
Reduction
-e
Fe Fe↔ 2+
+ 2e Eθ
= + 0.45V
Fe2+
+ 2e Fe E↔ θ
= - 0.45V
Fe Fe↔ 2+
+ 2e Eθ
= + 0.45V
Fe2+
+ 2e Fe E↔ θ
= - 0.45 V
Fe + Fe2+
Fe→ 2+
+Fe Eθ
= 0V
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
Mg2+
+ 2e- Mg↔ -2.37
Al3+
+ 3e- AI↔ -1.66
Mn2+
+ 2e- Mn↔ -1.19
H2O + e- 1/2H↔ 2 + 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- 1/2H↔ 2 0.00
Cu2+
+ e- Cu↔ +
+0.15
SO4
2-
+ 4H+
+ 2e- H↔ 2SO3 + H2O +0.17
Fe2+
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
E0
= 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
=
+=
×
×
−=
Fe2+/
Fe half cell
Fe + Fe2+
→ Fe2+
+ Fe
-e
Fe2+
0.01M
Fe(s)│Fe2+
(aq) (0.01M) ║ Fe2+
(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) │Mn2+
(aq) (0.1M) ║ Pb2+
(aq) (0.0001M)│Pb(s)
Mn + Pb2+
→ Mn2+
+ 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
=
×
×
−=

16. Acknowledgements
Thanks to source of pictures and video used in this presentation
Thanks to Creative Commons for excellent contribution on licenses
http://creativecommons.org/licenses/
http://spmchemistry.onlinetuition.com.my/2013/10/electrolytic-cell.html
http://www.chemguide.co.uk/physical/redoxeqia/introduction.html
http://educationia.tk/reduction-potential-table
http://2012books.lardbucket.org/books/principles-of-general-chemistry-v1.0/s23-
electrochemistry.html
Prepared by Lawrence Kok
Check out more video tutorials from my site and hope you enjoy this tutorial
http://lawrencekok.blogspot.com