The Nernst Equation
There is a need to be able to calculate cell potentials
under non-standard conditions. The Nernst
equation allows that to be accomplished.
1921
The Nernst Equation
There is a need to be able to calculate cell potentials
under non-standard conditions. The Nernst
equation allows that to be accomplished.
Consider a general redox reaction:
aA + bB
cC + dD
1922
The Nernst Equation
There is a need to be able to calculate cell potentials
under non-standard conditions. The Nernst
equation allows that to be accomplished.
Consider a general redox reaction:
aA + bB
cC + dD
and recall for a species X that GX  G0X  RTln[X] and
hence
1923
The Nernst Equation
There is a need to be able to calculate cell potentials
under non-standard conditions. The Nernst
equation allows that to be accomplished.
Consider a general redox reaction:
aA + bB
cC + dD
and recall for a species X that GX  G0X  RTln[X] and
hence
ΔG  ΔG0  RT ln Q
1924
Now combine ΔG  ΔG0  RT ln Q with the two
results
1925
Now combine ΔG  ΔG0  RT ln Q with the two
results
0
ΔG0  -nFEcell
1926
Now combine ΔG  ΔG0  RT ln Q with the two
results
0
ΔG0  -nFEcell
ΔG  -nFEcell
1927
Now combine ΔG  ΔG0  RT ln Q with the two
results
0
ΔG0  -nFEcell
ΔG  -nFEcell
0  RT ln Q
so that -nFEcell  -nFEcell
1928
Now combine ΔG  ΔG0  RT ln Q with the two
results
0
ΔG0  -nFEcell
ΔG  -nFEcell
0  RT ln Q
so that -nFEcell  -nFEcell
Now divide both sides of the equation by – nF
1929
Now combine ΔG  ΔG0  RT ln Q with the two
results
0
ΔG0  -nFEcell
ΔG  -nFEcell
0  RT ln Q
so that -nFEcell  -nFEcell
Now divide both sides of the equation by – nF
RT
0
Ecell  Ecell 
ln Q
nF
1930
Now combine ΔG  ΔG0  RT ln Q with the two
results
0
ΔG0  -nFEcell
ΔG  -nFEcell
0  RT ln Q
so that -nFEcell  -nFEcell
Now divide both sides of the equation by – nF
RT
0
Ecell  Ecell 
ln Q
nF
This is Nernst equation.
1931
The Nernst equation is often used at 25 oC, and
written in a slightly different form.
1932
The Nernst equation is often used at 25 oC, and
written in a slightly different form.
Since lnX = 2.3026 log X (where log is to the base
10), and
1933
The Nernst equation is often used at 25 oC, and
written in a slightly different form.
Since lnX = 2.3026 log X (where log is to the base
10), and
-1
-1
R
T
2.3026(8.3
14
J
K
mol
)(298.15
K)
1CV
2.3026 
1
1J
F
(96485Cmol )
1934
The Nernst equation is often used at 25 oC, and
written in a slightly different form.
Since lnX = 2.3026 log X (where log is to the base
10), and
-1
-1
R
T
2.3026(8.3
14
J
K
mol
)(298.15
K)
1CV
2.3026 
1
1J
F
(96485Cmol )
= 0.05916 V
1935
The Nernst equation is often used at 25 oC, and
written in a slightly different form.
Since lnX = 2.3026 log X (where log is to the base
10), and
-1
-1
R
T
2.3026(8.3
14
J
K
mol
)(298.15
K)
1CV
2.3026 
1
1J
F
(96485Cmol )
= 0.05916 V
So that the Nernst equation becomes
1936
The Nernst equation is often used at 25 oC, and
written in a slightly different form.
Since lnX = 2.3026 log X (where log is to the base
10), and
-1
-1
R
T
2.3026(8.3
14
J
K
mol
)(298.15
K)
1CV
2.3026 
1
1J
F
(96485Cmol )
= 0.05916 V
So that the Nernst equation becomes
0.05916
0
Ecell  Ecell 
log Q
n
1937
The Nernst equation is often used at 25 oC, and
written in a slightly different form.
Since lnX = 2.3026 log X (where log is to the base
10), and
-1
-1
R
T
2.3026(8.3
14
J
K
mol
)(298.15
K)
1CV
2.3026 
1
1J
F
(96485Cmol )
= 0.05916 V
So that the Nernst equation becomes
0.05916
0
Ecell  Ecell 
log Q
n
and the units of the second term on the right-hand
side are V (volts).
1938
Example: Predict the spontaneity of the following
reaction at 25 OC:
Fe2+(aq) + Co(s)
Co2+(aq) + Fe(s) ,
given that [Co2+] = 0.15 M and [Fe2+] = 0.68 M.
1939
Example: Predict the spontaneity of the following
reaction at 25 OC:
Fe2+(aq) + Co(s)
Co2+(aq) + Fe(s) ,
given that [Co2+] = 0.15 M and [Fe2+] = 0.68 M.
(Note: that the concentrations are not 1 M, so this
is not at equilibrium conditions, and we must
employ the Nernst equation).
1940
Example: Predict the spontaneity of the following
reaction at 25 OC:
Fe2+(aq) + Co(s)
Co2+(aq) + Fe(s) ,
given that [Co2+] = 0.15 M and [Fe2+] = 0.68 M.
(Note: that the concentrations are not 1 M, so this
is not at equilibrium conditions, and we must
employ the Nernst equation).
0
From the table of Ehalf
values the following is
-cell
available:
1941
Example: Predict the spontaneity of the following
reaction at 25 OC:
Fe2+(aq) + Co(s)
Co2+(aq) + Fe(s) ,
given that [Co2+] = 0.15 M and [Fe2+] = 0.68 M.
(Note: that the concentrations are not 1 M, so this
is not at equilibrium conditions, and we must
employ the Nernst equation).
0
From the table of Ehalf
values the following is
-cell
available:
0
Co2+(aq) + 2 eCo(s) Ehalf
-cell = -0.28 V
1942
Example: Predict the spontaneity of the following
reaction at 25 OC:
Fe2+(aq) + Co(s)
Co2+(aq) + Fe(s) ,
given that [Co2+] = 0.15 M and [Fe2+] = 0.68 M.
(Note: that the concentrations are not 1 M, so this
is not at equilibrium conditions, and we must
employ the Nernst equation).
0
From the table of Ehalf
values the following is
-cell
available:
0
Co2+(aq) + 2 eCo(s) Ehalf
-cell = -0.28 V
0
Fe2+(aq) + 2 eFe(s) Ehalf
-cell = -0.44 V
1943
Reverse the Co reaction and then add: so that
0  0.16 V
Ecell
.
1944
Reverse the Co reaction and then add: so that
0  0.16 V
Ecell
.
From the Nernst equation
1945
Reverse the Co reaction and then add: so that
0  0.16 V
Ecell
.
From the Nernst equation
0
Ecell  Ecell

(8.314 JK1mol1)(298.15K) ln [Co2  ]
2(96485Cmol1)
[Fe2 ]
1946
Reverse the Co reaction and then add: so that
0  0.16 V
Ecell
.
From the Nernst equation
0
Ecell  Ecell

(8.314 JK1mol1)(298.15K) ln [Co2  ] 1CV
2(96485Cmol1)
[Fe2 ] 1 J
1947
Reverse the Co reaction and then add: so that
0  0.16 V
Ecell
.
From the Nernst equation
Ecell
1
1
2
(8.314
JK
mol
)(298.15K)
[Co
]
1CV
0
 Ecell 
ln
2(96485Cmol1)
[Fe2 ] 1 J
= -0.16 V + 0.019 V
1948
Reverse the Co reaction and then add: so that
0  0.16 V
Ecell
.
From the Nernst equation
Ecell
1
1
2
(8.314
JK
mol
)(298.15K)
[Co
]
1CV
0
 Ecell 
ln
2(96485Cmol1)
[Fe2 ] 1 J
= -0.16 V + 0.019 V
= -0.14 V
1949
Reverse the Co reaction and then add: so that
0  0.16 V
Ecell
.
From the Nernst equation
1
1
2
(8.314
JK
mol
)(298.15K)
[Co
]
1CV
0
Ecell  Ecell 
ln
2(96485Cmol1)
[Fe2 ] 1 J
= -0.16 V + 0.019 V
= -0.14 V
Since Ecell is negative (therefore ΔG is positive) the
reaction is not spontaneous in the forward
direction.
1950
Equilibrium Conditions
1951
Equilibrium Conditions
At equilibrium: There is no net transfer of
electrons, so that Ecell  0, and also Q = Kc, so that
1952
Equilibrium Conditions
At equilibrium: There is no net transfer of
electrons, so that Ecell  0, and also Q = Kc, so that
0
Ecell
RT
 ln Kc
nF
1953
Equilibrium Conditions
At equilibrium: There is no net transfer of
electrons, so that Ecell  0, and also Q = Kc, so that
RT
 ln Kc
nF
This is an important formula: Given Kc we can
0 .
find Ecell
0
Ecell
1954
Equilibrium Conditions
At equilibrium: There is no net transfer of
electrons, so that Ecell  0, and also Q = Kc, so that
RT
 ln Kc
nF
This is an important formula: Given Kc we can
0 .
find Ecell
0
Ecell
0
n
FE
cell
We can rewrite the equation as: ln Kc 
RT
1955
We can rewrite the preceding equation as:
0
Kc 
n FE cell
e RT
1956
We can rewrite the preceding equation as:
0
Kc 
n FE cell
e RT
This is a particularly important formula.
1957
We can rewrite the preceding equation as:
0
Kc 
n FE cell
e RT
This is a particularly important formula. Many
inorganic redox reactions have huge equilibrium
constants, that would be very difficult or impossible
to determine by direct measurement of reactant
and product concentrations.
1958
We can rewrite the preceding equation as:
0
Kc 
n FE cell
e RT
This is a particularly important formula. Many
inorganic redox reactions have huge equilibrium
constants, that would be very difficult or impossible
to determine by direct measurement of reactant
and product concentrations. However, the
measurement of Ecell turns out to be a very
accurate way to determine Kc.
1959
Example: Determine the equilibrium constant for
the following reaction at 25 oC:
Cu2+(aq) + Zn(s)
Zn2+(aq) + Cu(s)
1960
Example: Determine the equilibrium constant for
the following reaction at 25 oC:
Cu2+(aq) + Zn(s)
Zn2+(aq) + Cu(s)
0
From the table of Ehalf
values the following is
-cell
available
1961
Example: Determine the equilibrium constant for
the following reaction at 25 oC:
Cu2+(aq) + Zn(s)
Zn2+(aq) + Cu(s)
0
From the table of Ehalf
values the following is
-cell
available
0
Zn2+(aq) + 2 eZn(s) Ehalf
= -0.76 V
-cell
1962
Example: Determine the equilibrium constant for
the following reaction at 25 oC:
Cu2+(aq) + Zn(s)
Zn2+(aq) + Cu(s)
0
From the table of Ehalf
values the following is
-cell
available
0
Zn2+(aq) + 2 eZn(s) Ehalf
= -0.76 V
-cell
0
Cu2+(aq) + 2 eCu(s) Ehalf
-cell = 0.34 V
1963
Example: Determine the equilibrium constant for
the following reaction at 25 oC:
Cu2+(aq) + Zn(s)
Zn2+(aq) + Cu(s)
0
From the table of Ehalf
values the following is
-cell
available
0
Zn2+(aq) + 2 eZn(s) Ehalf
= -0.76 V
-cell
0
Cu2+(aq) + 2 eCu(s) Ehalf
-cell = 0.34 V
Reverse the Zn reaction and then add:
1964
Example: Determine the equilibrium constant for
the following reaction at 25 oC:
Cu2+(aq) + Zn(s)
Zn2+(aq) + Cu(s)
0
From the table of Ehalf
values the following is
-cell
available
0
Zn2+(aq) + 2 eZn(s) Ehalf
= -0.76 V
-cell
0
Cu2+(aq) + 2 eCu(s) Ehalf
-cell = 0.34 V
Reverse the Zn reaction and then add: so that
0  1.10 V.
Ecell
0
nFEcell
ln Kc 
From
RT
1965
we have
1
2
(96485
C
mol
)
(1.10
V)
1J
ln Kc 
(8.314 JK1mol1)(298.15K) 1 CV
= 85.6
1966
we have
1
2
(96485
C
mol
)
(1.10
V)
1J
ln Kc 
(8.314 JK1mol1)(298.15K) 1 CV
= 85.6
therefore
Kc = e85.6
= 1.5 x 1037
1967
we have
1
2
(96485
C
mol
)
(1.10
V)
1J
ln Kc 
(8.314 JK1mol1)(298.15K) 1 CV
= 85.6
therefore
Kc = e85.6
= 1.5 x 1037 (Note this calculation loses
a sig. fig. because of the
exponential operation.)
1968
Now,
2
[Zn
]
Kc 
[Cu2  ]
1969
Now,
2
[Zn
]
Kc 
[Cu2  ]
Think about how difficult it would be to measure
the concentrations of Zn2+ and Cu2+ yielding such a
large equilibrium constant.
1970
Batteries
1971
Batteries
Battery: An electrochemical cell, or several
electrochemical cells connected in series, that can
be used as a source of direct electric current at a
constant voltage.
1972
Batteries
Battery: An electrochemical cell, or several
electrochemical cells connected in series, that can
be used as a source of direct electric current at a
constant voltage.
A few different types of batteries will be examined.
1973
Batteries
Battery: An electrochemical cell, or several
electrochemical cells connected in series, that can
be used as a source of direct electric current at a
constant voltage.
A few different types of batteries will be examined.
The chemistry for some of these can be quite
complex.
1974
The Dry Cell Battery (zinc – carbon dry cell).
1975
The Dry Cell Battery (zinc – carbon dry cell).
The most common dry cell is the Leclanché cell,
used in flashlights, etc.
1976
The Dry Cell Battery (zinc – carbon dry cell).
The most common dry cell is the Leclanché cell,
used in flashlights, etc.
The anode of the cell consists of a zinc container
that is in contact with MnO2 and an electrolyte
consisting of ammonium chloride and zinc chloride
in H2O, to which starch is added to make the
solution a thick paste less subject to leakage. A
graphite rod serves as the cathode.
1977
Anode: Zn(s)
Zn2+(aq) + 2e- (simplified)
1978
Anode: Zn(s)
Zn2+(aq) + 2e- (simplified)
Cathode: (simplified reaction)
1979
Anode: Zn(s)
Zn2+(aq) + 2e- (simplified)
Cathode: (simplified reaction)
2 NH4+(aq) + 2MnO2(s) + 2eMn2O3(s) + 2 NH3(aq)
+ H2O
1980