Chapter 11
Electrochemical Cells
work to be done. Zn gets oxidized and its standard reduction potential is −ve (−076 V) while that of Cu is
+ve (+034 V). The standard cell potential is therefore
E =1.10 V.
Electrochemical cells can be divided into two types
a) galvanic, which converts chemical energy (association
sector) into electrical work  →  and b) electrolytic,
in which  →  The former can be used to charge a
battery (it is in fact a battery) and the latter uses electrical energy to drive a reaction. One can move from
a) galvanic to b) electrolytic by simply changing the external potential, and by finding the balance point, determine the cell potential, which is just an electron intensive
version of the free energy change.
Fig. Galvanic (chemical ⇒electrical).
To understand how this is done, we have to understand what happens when, e.g. a metal such as Zn(s)
∼
is immersed in a ZnSO4 solution. A tiny amount of the 11.1
Electrochemical
potential

Zn(s) dissolves leaving the e− ’s on the metal. Thus the
metal builds up a slight −ve charge and the surrounding Consider the following sequence of simple experiments.
solution builds up a slight +ve charge. While the total
1. Bring together two dissimilar metals (A and B).
amount of charge is tiny, the built-up voltages can be
The level diagrams indicate the delocalized electron
significant (∼ 1 ) and thus the work required to move
quantum levels (particle in a box levels). The top
charged species is not the standard chemical potential 
most energy, the energy required to add or subtract
We have to work with a new quantity which accounts for
one e− , is  which is also called the Fermi energy,
the work required to move charge (either e− ’s or charged
  As the metals are different, it is extremely unions) in a medium with a spacially varying electrostatic
likely that the Fermi energies are equal, as a result
potential 
there will be a charge transfer. With a transfer of e−
Consider
the
Daniell
cell:
from A→ , the A side becomes +ve and the B side
Zn(s)|ZnSO4 (aq)||CuSO4 (aq)|Cu.
−ve. This decreases (increases) the (electro) chemiThe “|” denotes a phase boundary and the “||” a salt
cal potential corrected for the electrostatic potential
bridge. The latter is inserted to make the process qua∼
∼
sistationary and thus allow for the maximum (non-PV)
for e− ’s on the A (B) side yielding − = − 
81
3. Now separate the metals
Fig. Dissimilar metals
Fig. Sep. diss. metals+conn.+VM


−  −
∼
−
∼
−

−
∼




≡ 
− + F = − + F ≡ −
As  = −1



∴ 
− − − = F( −  )

−
∼
∼
− = −





− − F = − − F


− = −

∴


− F + F = 
− − F + F



F( −  ) = [
− − − ] − F( −  )

 ≡ ( −  ) =
−[
−
]
−
−
F
+ ( −  )
voltmeter = “chemical” + “electrical”.
The electrochemical potential is
 ≡  + F
;
− F = 
− − F
but as
Lesson: The difference in chemical potentials induces a matter motion which induces electrical po
tential difference. When 
   →     The
(+ve) potential will be greater on side A, thus reducing the energy of e− , exactly compensating for
the greater chemical potential.
∼
∼
= −
4. It is the electrochemical difference which dictates
material flow. The equilibrium condition becomes
P∼

∆  =   = 0
(11.1)
This definition amounts to subsuming the electrical work sector into an enhanced (pseudo) chemical
sector. Note that if the species is neutral ( = 0) or
This can be viewed as the combination of the old
chemical sector  n and the new electrical potential sector F  Of course, the new sector only is
present if the charge (z) is finite.
∼ =0
the potential is zero,  = 
The charge carried by one mole of charge is Faraday’s constant ≡ F =96485 [ 
 ]
2. Now add connections ( and R via identical metal,
thus they posses the same electron chemical potentials) to a high impedance voltmeter.
11.2
Conventions, Notation and
Standard states
1. The work required to move a charge  = F
from a phase  to another  through a potential
difference ∆ is
Fig. Dissimilar metals+connections+VM
 = ( −  )[] = (∆)[F]
2. The electromotive force  is defined as the potential
difference measured between stationary thermodynamic states.


 ≡ ∆ ∴ (recalling: non-PV 
= −∆)
∼
−

−
∼
∼
∆ = −[F− ]
∼
= − = − = −
3. Consider the “Daniell cell”: Zn|Zn++ ||Cu++ |Cu

− F = 
− − F


=
(11.2)

Lesson: Connecting different metals via similar
wires to a voltmeter will read zero (unless the sides
are different temperatures.)
82
(a) Strong electrolytes (sulfates in this case) are
used to from the ions in the half cells.
(b) The “|” signifies a phase boundary across
which charge can be transferred.
(c) Salt bridge ( “||” ), between half cells, allows charge transport without mixing the active electrolytes of the cells. It does this by
having a great reservoir of electrolyte (such as
KCl) suspended in a gel. This electrolyte diffuses into the cells, and the electrolyte in each
half cell, only encroaches slightly into the salt
bridge. (The bridge also removes the so called
“junction potential” caused by the boundary
between dissimilar solutions.)
Fig. SHE
6. (a) The cell is assembled by bubbling H2 (g), with
P (really f) = 1 bar over a Pt electrode immersed in a H+ solution of activity  + = 1.
(d) The circuit is completed with an external wires
(of the same material) connected to a voltmeter.
(b) The reaction and equilibrium condition are:
 + () + − ¿ 12 2 ()
(e) An equilibrium between the metal and the ion
is established. In the case of Zn, some more
of the Zn dissolves leaving the metal -ve and
the solution +ve. This potential (of the 1/2
cell) cannot be measured by sticking the second
electrode in the solution.
∼
=  
+  ln 2 )
(d) Thus the potential becomes
(
+ ln  + )−[ 12 (
2 + ln 2 )]
+
F
12
 − 1 

2
= + F2 2 − 
F ln  + 
 + 2 =
(e) With a reference state of unit activities and
fugacities, the ln term vanishes. Furthermore,
as H2 (g) is the reference state for H2 , the reference potential is
(11.3)
 + 2 =
(11.4)

− 12 
2
+
F
=

+
F
(f) With the reference chemical potential of H+
(c) For metal ions, once eq. has been established,
M¿  + + ze− thus
∼
∼
 ∼
 = ( + ) +  −
=1
[∆ ( +  ) =  + () ≡ 0]
 + 2 ≡ 0

 = ( + + F) − F
 = +
(11.5)
This statement (consequence of − ≡ 0) is
also true for the reference state,  =  + =
0

(11.7)
and thus we are consistent.
11.3
(d) Returning to the electrochemical potential of
the ion,
∼
=0
 =

 1
= 2 (2 ()
(b) For the neutral metal atoms:
∼

∼
(c) Using the standard states and the offsets for
concentrations and nonidealities, the eq. condition becomes
[ + +  ln  + ] − F + 2
(a) If we use identical metals on the two sides
(to connect to the voltmeter) we can take the
reference electron chemical potential as zero,
− ≡ 0 thus
= −F

[ + ()] + − = 12 2 ()
4. Electrochemical references
∼
−
∼
[ + ()] + − = 12 2 () using SHE ref.
(11.6)
5. To set this reference (of  = 0, since the absolute is
unmeasureable) we choose H+ under the conditions
of the Standard Hydrogen Electrode.
83
Reactions & the Nernst Eq.
1. How much  can we get from this cell?
P
 =   −   +  
P
=   −   +   
where
the extent of reaction is defined by  =  
1
 +  =  + (−  +  ) thus
P
  −   +    =  + (−  +  )
P

= ( −  ) +   −   −   
−

=
P
= ( −  ) −
2.  =   −
  
However from the second law (..)5 0 (Recall  =

 ) Therefore


5 −

P
2
 =   − () ln{2± } − () ln{±
}
This can be rearranged to get
 ≡ [ + 2() ln{± }]1 → measure
2
= [  − () ln{±
}]2 → deduce
Noting that in the limit lim2→0 (with no other ions
in solution),
R
√
2 
(a) ln±
→ − ±
(b) lim2→0 []2 =  
 ++ 2− 2
ln{  
}
A plot of  = [ + 2() ln{± }]1
√
vs ± will extrapolate to   as 2 → 0
and as X is 1 in a pure solid,

∆ = ∆
+  ln{++ 2− }
5. With   and (2 ) you can calculate ± (2 )
− F= − F  +  ln{++ 2− }
2
 =   − () ln{2± } − () ln{±
} by recalling that
ln{++ 2− }
More generally, one has the Nernst Equation
 =  −
ln{ }
2
4. Thus,  =   − ( 
) ln{2± ±
}
F

(c) ∆ = ∆
+  ln{}
Y
Y

where  ≡
 
 

 F

 F
2
= 2± ±
(b) Red:2() + 2− → 2() + 2− ;
 ( ) = 022233
 =  −
=  −
 =  + − = ( +  + )(− − )
 → ++ +2− ;  [ ] = −07618

∆ = ∆
+ 
2
3. Therefore, if we know    and measure  ⇒  =
2. The above is true for standard and actual conditions. Thus considering the cell,
P
 
}
12
 

   = − =  F ()
(d) In this case
ln{
as 2 = 2 ∼ 2 = 1
Lesson: the electrical work done BY a cell is equal
to the decrement in the Gibbs F.E. when done reversibly and less otherwise.
(a) Ox :

 F

ln 
 F


1:1
± = (++ −− )1  = 112  = 
(11.8)
Figs.
Just an electron intensive rewrite of
∆ = ∆  +  ln 
3. For future reference,
11.4
 298
=
F
002569 V.
Cell potentials
How do you get the numbers in the table? Consider
combining the AgCl|Ag electrode (used above) with the
SHE. Writing them out in the standard format,
Oxidation|reduction
H2 (g)|HCl(m2 )|AgCl|Ag(s)
11.5
1
2 2 ()
  [  =
→  + () + −
(c) Net:

+ () → (2 ) + 
with  =  − 
 
=

 ]
⇒ e− intensive
3. Thus when adding cells,
(b) Red: () + − → () + −
1
2 2 ()



1. ∆  = − F  and 
= 
+ 


] ⇒ e− extensive
2. ∆  [ 
1. The half reactions are
(a) Ox :
A few pointers & reminders



 →
 Hess’s rule becomes
 →


 
 

  =   +  
84
4. This means when one is dealing with a OX-RED external (− ) circuit one can measure the potential difpair (which must be balanced in the e− transfer)
ference between the LHS and RHS. Recall that the EMF

is just an (− ) intensive version of the ∆ = − F.



+ + − →
Therefore, if we can measure  we can get  Specif

+
−
ically,
we will vary 2 and measure 
against a
 →  + 
2
constant
LHS.
Lets
see
how.



 →
=  
+  



→
= 
+ 

−
The E values ARE per unit e transferred NOT,
like the energy functions, per unit of reaction!
11.6
1. First let me remind you how ∆ is related to 
Consider the ∆ for moving   from the
LHS → 
∆ = (
) − (
)
2
2
The E ( )
+ ln 
)−(
+ ln 
)
= (
2
2
2
2
As the cell EMF’s are just an e− intensive measure of ∆
the increment function for E0  are just a scaled version
of those for ∆   However, one can also get ∆ and
∆ from E ( ) Lets see how.
2. In the amalgam case we have the same reference
states (LHS and RHS).
∆ =  ln

2


2
3. As 2 = 2 2

= − F 
1. ∆

∆
( 
)
2 2
∆ = − F =  ln 
or
2

= − F( 
 ) .
{ln 2 + ln 2 − ln 
}
 = − 
2
F
2. As  = − + 

4. We can obtain values of 2 or 2 from the EMF’s by
a simple graph. Rearranging the above equation,



−∆ = − F( 
 ) or ∆ =  F(  ) 
y≡

3. ∆
= ∆ −  ∆
+ ln 2 = ln 
− ln 2
2
 (2 ) for which the intercept at 2 = 0 (2 = 1
ln 2 = 0) is ln 
and the difference from the
2
intercept at finite 2 is ln 2 

∆ = ∆
+  ∆

= − F  +   F( 
 )


=  F[ ( 
 ) −  ]
11.7
 F

*Another example
Imagine two dilute solid solutions of Pb in Hg (with different mole fractions of Pb (2 ) in contact with a common (but separated by a salt bridge) electrolyte. These
solid solutions are called amalgams. The chemical potential of the − ’s can be (and generally are) different
in the two different solid solutions. This means that
− ’s will move from one side to the other in response to
the generalized force mismatch (on the − ’s) on one side
versus the other. In addition, if the electrolyte consists
of a Pb++ salt, Pb++ will migrate from one amalgam
→ to the solution → to the other amalgam, under the
influence of the different   = 2  Of course, for this
to happen, a path (external circuit) must be provided
for the − ’s. The salt bridge “holds off” attainment of
equilibrium and thus allows for the measurement of the
∆.
Following convention, we always write the half reaction (amalgam in this case) that provides the − ’s on the
LHS. This means that oxidation occurs on the left and
reduction on the RHS. If one adds a galvanometer to the
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