Option C Nernst Equation, Voltaic Cell and Concentration Cell

1. 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)

2. Current– measured Amperesor Coulombs per second
1A = 1 Coulomb charge pass througha 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 – movingcharges 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 

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
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

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 currentflow.
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)
- +

5. 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 – intensiveproperty–independentof 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
Voltmetermeasure energy for 1C – 110J/100C – 1.1V
E cell no change
Current– measured in Amp or Coulomb per s
1A = 1 Coulomb charge pass througha point in 1 s = 1C/s
1 Coulomb charge (elec) = 6.28 x 10 18 elec passing in 1 s
1 electron/protoncarry 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 remainSAME
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
Voltmetermeasure 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

6. Relationship bet ∆G and Kc
cellnFEG  
Relationship bet
Energetics and Equilibrium
cKRTG ln 
STHG 
Enthalpy
change
Entropy
change
Equilibrium
constant
Gibbs free
energy change
H
G
Relationshipbet ∆G, Kc 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, 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)
cellE

G
cK
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

7. E cell/Voltage– dependon natureof 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  nFEG 
QRTnFEnFE ln 
Nernst equation
Work or Free energyto do work
dependon quantitymaterial 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  
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– dependon quantityand surface area

8. 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)
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) Eθ = -0.76V
Cu2+ + 2e ↔ Cu (cathode) Eθ = +0.34V
Eθ
cell = 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




Std electrodepotential- 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





9. 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)
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) Eθ = -0.76V
Ag + + e ↔ Ag(cathode) Eθ = +0.80V
Eθ
cell = 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




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 Eθ = -0.76V
Cu2+ + 2e ↔ Cu Eθ = +0.34V
Zn ↔ Zn2+ + 2e Eθ = +0.76V
Cu2+ + 2e ↔ Cu Eθ = +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
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 + 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

11. Zn half cell (-ve)
Oxidation
Cu half cell (+ve)
Reduction
Zn/Cu Cell
-e -e
Zn 2+ + 2e ↔ Zn Eθ = -0.76V
Cu2+ + 2e ↔ Cu Eθ = +0.34V
Zn ↔ Zn2+ + 2e Eθ = +0.76V
Cu2+ + 2e ↔ Cu Eθ = +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
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 + 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

12. Zn half cell (-ve)
Oxidation
Cu half cell (+ve)
Reduction
Zn/Cu Cell
-e -e
Zn 2+ + 2e ↔ Zn Eθ = -0.76V
Cu2+ + 2e ↔ Cu Eθ = +0.34V
Zn ↔ Zn2+ + 2e Eθ = +0.76V
Cu2+ + 2e ↔ Cu Eθ = +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
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 + 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

13. Cu half cell (-ve)
Oxidation
Cu half cell (+ve)
Reduction
-e
Cu ↔ Cu 2+ + 2e Eθ = - 0.34V
Cu2+ + 2e ↔ Cu Eθ = +0.34V
Cu ↔ Cu2+ + 2e Eθ = - 0.34V
Cu2+ + 2e ↔ Cu Eθ = +0.34V
Cu + Cu2+ → Cu2+ + 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
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 + 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





14. Fe half cell (-ve)
Oxidation
Fe half cell (+ve)
Reduction
-e
Fe ↔ Fe 2+ + 2e Eθ = + 0.45V
Fe2+ + 2e ↔ Fe Eθ = - 0.45V
Fe ↔ Fe2+ + 2e Eθ = + 0.45V
Fe2+ + 2e ↔ Fe Eθ = - 0.45 V
Fe + Fe2+ → Fe2+ +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
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 + 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



