Zn
Zn
2+
+
2 e
–
Cl
2
+
2 e
–
2 Cl
–
Zn
+
Cl
2
ZnCl
2
Therefore…
n = 2
–
369,000 J
=
–
(2) (96500) (
E
0
)
Thus…
E
0
=
)
96500
)(
2
(
369000
=
1.91 V
1917
Thermodynamics, Equilibrium and Electrochemistry: A Summary
G
0
=
–
nF
E
0
also
G
0
=
–
RT ln (K)
Therefore…
–
nF
E
0
=
–
RT ln(K)
ln (K)
= nF
E
0
/ RT
At 25
0
C,
F
RT
=
0.025693 J/C
=
0.025693 V
Therefore…
ln (K)
=
V
0.025693
nE
0
or..
E
0
=
n
0.025693
ln (K)=
n
0.0592
log (K)
e.g.
Calculate the equilibrium constant at 25
0
C for the reaction…
Cr
2
O
7
2
–
+
6 Br
–
+
14 H
+
2 Cr
3+
+
3 Br
2
+
7 H
2
O
Given:
3
2 Br
–
Br
2
+
2 e
–
E
0
=
–
1.09 V
1
Cr
2
O
7
2
–
+
14 H
+
+
6 e
–
2 Cr
3+
+
7 H
2
O
E
0
= +1.33 V
Cr
2
O
7
2
–
+
6 Br
–
+
14 H
+
2 Cr
3+
+
3 Br
2
+
7 H
2
O
E
0
= +0.24 V
NOTE:
a)
balancing the equation is necessary to find (n), the number of electrons
transferred.
b)
E
0
is an intensive property.
Its value does not change when the half
reactions are multiplied by small integers to balance the number of
electrons.
Back to our equation…
ln (K)
=
V
0.025693
nE
0
=
V
0.025693
(6)(0.24)
=
56
K
=
e
56
=
2.1 x 10
24
E
0
is positive therefore K is greater than 1.
1918
Spontaneity of Redox Reactions Under Conditions Other Than
Standard State
Consider the reaction:
Zn(s) + Cu
2+
( aq,1M )
Zn
2+
( aq,1M) + Cu(s)
E
0
cell
= 1.103 V
What happens when one increases the concentration of the Cu
2+
ion or decreases
the concentration of the Zn
2+
(ion)? ( nonstandard states)Answer: the equilibrium
will shift to the right i.e. the reaction under these concentrations is even more
spontaneous
E
cell
>
E
0
cell
.
Zn(s) + Cu
2+
( aq,1.5M )
Zn
2+
( aq,0.75M) + Cu(s)
E
0
cell
= 1.142 V
The Nernst Equation
Nonstandard states
G
=
G
0
+
RT ln Q
–
nF
E
=
–
nF
E
0
+
RT ln Q
by dividing both sides by
–nF, one obtains…
E
cell
=
E
0
cell
nF
RT
ln Q
at 25
0
C
R,T, and F are constants
F
RT
=
0.02568
E
cell
=
E
0
cell
n
0.02568
ln Q
Q is the reaction quotient.
NOTE:
E
cell
=
E
0
cell
n
0.02568
ln Q
or
E
cell
=
E
0
cell

n
0592
.
0
log Q
1.
When Q = K (i.e. equilibrium is reached), then
E
cell
= 0.
The components
in the two compartments of the cell have the same free energy.
Thus,
G
= 0.
There is no driving force, the battery is drained (dead)
2.
When we have standard conditions (concentration fo ions = 1M, pressure
= 1atm) then Q = 1 and ln Q = 0.
Thus,
E
cell
=
E
0
cell
.
3.
When Q > 1 (the concentration of product more than those of reactants),
then
E
cell
<
E
0
cell
(check the equation above).
4.
When Q < 1 (the concentration of product less that those of reactants), then
E
cell
>
E
0
cell
(check the equation above).
1919
The relationship between
E
0
and K therefore is…
G
0
=
–
nF
E
0
G
0
=
RT ln K
–
nF
E
0
=
RT ln K
E
0
cell
=
nF
RT
ln K
R, F are constants, T = 25
0
C
E
0
cell
=
n
0.02568
ln K
Does 0.02568 have units?
What is it? It is (V)
e.g.
Consider the reaction…
Cu
2+
(aq)
+
Zn (s)
Zn
2+
(aq)
+
Cu (s)
E
0
cell
=
1.10 V
(calculated from the
E
0
cathode

E
0
anode
)
Calculate the cell potential when [Cu
2+
] = 3.0 M
and
[Zn
2+
]
=
0.10 M
E
cell
=
E
0
cell
–
n
0.02568
ln
]
[Cu
]
[Zn
2
2
(where, n = 2)
E
cell
=
1.10
–
2
0.02568
ln
0
.
3
0.10
=
1.10
+
0.04
=
1.14 V
In other words, the reaction is even more spontaneous under these nonstandard
conditions, and this is consistent with Le Chatelier’s Principle (increasing [Cu
2+
]
and/or decreasing [Zn
2+
] will shift the reaction to the right.
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 Summer '09
 Marky
 Electrochemistry, Electron, Oxidation Number, Reaction, Redox, Cathode, e–