Electrochemistry I
Electrical Work
Whenever we wish to add some new phenomenon, like electrical energy,
into thermodynamics we must go all the way back to the combined first and
second laws of thermodynamics and insert the new form of energy as a work
term. That is, we write in general,
dU  TdS  pdV  dwother . (1)
As we have been doing all along, we make a Legendre transformation to
obtain the Gibbs free energy,
G = U + pV  TS,
(2)
and we obtain,
dG   SdT  Vdp  dwother . (3)
Then, for processes at constant temperature and pressure, we get
dGT , p  dwother , (4)
so,
GT , p  wother,max  reversible work, (5)
(We have indicated in Equation 5 that the work is the maximum work. This
is because G is a state function and is independent of path. Work is not
independent of path, but thermodynamics here relates it to a quantity that is
independent of path. The only way to realize this maximum work is to do
the process reversibly. Any irreversibility will give an amount of work that
is less in magnitude than indicated in Equation 5.)
For electrochemistry we want this "other work" to be electrical work and we
appeal to the equations of physics to find out what it should look like.
Physical work is force times distance. That is, a force, F, acting through a
distance dx gives an amount of work Fdx (in this case it is work done on the
surroundings, not on the system). In the case of electrical work the force is
produced by the action of an electric field, E , on a quantity of charge Q and
so we write,
dwsurr,elec  Fdx  QE( x)dx. (6)
But we know that an electric field is derivable as the derivative (gradient) of
an electrical potential, ,
E

, (7)
x
so,

dx  Qd (8)
x
   Qd  Q  d
dwsurr,elec  Q
wsurr,elec
 Q(final  initial )
(9a, b, c)
 QE ,
where we have written
E  final  initial , (10)
from which we conclude that
wsys  wsurr  QE. (11)
Notice that E is the potential difference through which the charge moves.
The charges that are moving here are electrons and it is more convenient to
use moles of electrons rather than individual electrons. For n moles of
electrons the charge, Q, can be written,
Q  nF, (12)
(this is negative because electrons have a negative charge)
coul
where F  96,485
is the Faraday number. The Faraday number is the
mol
number of coulombs of charge in one mole of (positive) electron charges.
We could say that the Faraday number is the number of coulombs in one
mole of protons. That way we have to keep track of whether we are talking
about electrons or protons by specifying the sign of the charge. Then
wsys  nFE  G p ,T . (13)
The summary equation of this section is then,
Gp ,T  nFE. (14)
Equation 14 is the fundamental equation of electrochemistry because it
relates the electromotive force (emf) of an electrochemical cell to the
thermal properties of the chemical reaction carried out in that cell.
Recall from our previous studies that we can write the Gibbs free energy for
the reaction when the reactants and products are not in their standard states
(they are in some state specified by the activity, ai) as
aCc aDd
G  G  RT ln a b . (15)
aA aB
0
We insert the fundamental equation of electrochemistry into this equation to
get
aCc aDd
0
nFE  G  RT ln a b . (16)
aA aB
If we define
Go  nFE o (17)
and divide by nF to get
RT aCc aDd
(18)
EE 
ln
nF aAa aBb
o
we get
E  Eo 
RT
ln Q. (19)
nF
These equations are different forms of the Nernst equation. In Equation 19
we have used Q not as charge but as a shorthand way to write the product
and quotient of activities appearing in Equation 18.
Let’s apply these ideas to a chemical reaction. Consider the reaction,
Cu(s) + Cl2 (g)  CuCl2 (aq). (20)
We can get Go from tables of thermodynamic properties (in this case we
look for the Gibbs free energy of formation for CuCl2(aq)),
G o  nFE o
Eo  
G o
.
nF
(21a, b)
Then,
E  Eo 
RT aCuCl2 (aq )
ln
p
nF
1  Cl2
1 atm
(22a, b)
RT
4m3 3
E 
ln
.
pCl2
nF
1
1 atm
o
We can determine the value of n by breaking the reaction into two halfreactions, one a reduction and the other an oxidation.
R
Cl2 + 2e  2 Cl
L
Cu
 Cu2+ + 2e
-------------------------------Cu + Cl2  Cu2+ + 2 Cl
(23)
The value of n is given by the number of electrons that must be cancelled
when the two half-reactions are added together to give an overall reaction.
In this case n = 2. (We have labeled the half-reactions “R” and “L” because
in the cell notation (picture) of the cell we will place the substances involved
in the Reduction half of the reaction on the Right-hand side.)
Cell notation
The cell notation for an electrochemical cell is intended to take the place of
(and be easier to write than) a picture or actual diagram of the cell. The
notation for a cell has a right-hand side and a left-hand side. There are also
indicators for phase boundaries. For example, the phase boundary between a
metal electrode and a solution, or a metal and a gas, or a gas and a solution,
etc is indicated by a vertical bar, |. The cell which would give the reaction
discussed above is,
Cu(s)|CuCl2(aq)|Cl2(g)|Pt(s). (24)
The vertical line “|” indicates a phase boundary (boundary between phases).
In this case there is one boundary between the Cu metal and the CuCl2
solution, one between the solution and the Cl2 gas, and one between the gas
and the platinum electrode. (You probably wouldn’t want to use platinum in
contact with chlorine gas because chlorine will attack platinum, but we’re
only doing this on paper so there’s no great loss.)
In general a cell diagram would have the form,
A,B|C,D,E|F,G . . .|J…., |X,Y, (25)
and so on. A phase might have more than one substance in it so we separate
different components in the same phase with commas.
Liquid junctions and the Salt Bridge
Consider the cell
Cu(s)|CuSO4(aq)|ZnSO4(aq)|Zn(s). (26)
This cell diagram describes a cell with a copper rod immersed in a CuSO4
solution and a zinc rod immersed in a ZnSO4 solution, but it also appears
that we have two different solutions in direct contact with each other. This
could probably be done by placing a felt divider (or some other barrier
which will allow the movement of electricity, but not allow the solutions to
mix) between the two solutions, but it would create problems. The Cu2+ and
Zn2+ ions will diffuse at different rates and create a potential drop across the
boundary. This potential is called a “junction potential.” Since we are
mainly interested in the potential (emf) created by the chemical reaction this
junction potential is a nuisance. There are several ways to eliminate or, at
least greatly reduce this junction potential, and we will give them all the
generic name “salt bridge.” The salt bridge is so named because one
(historical) way to greatly reduce the junction potential is to separate the two
offending solutions into two different beakers and then provide a electrical
connection by an inverted “U” tube filled with KCl(aq) solution. One leg of
the inverted “U” tube goes in each beaker. Since K+(aq) and Cl(aq) ions are
about the same size and diffuse at very nearly the same rate the junction
potential is greatly reduced if not entirely eliminated. (There are more
sophisticated methods which, in special cases, can completely eliminate the
junction potential.) We will symbolize a salt bridge by two verticle lines, ||.
By this symbol we mean only that a way has been found to eliminate the
junction potential, whether by an actual salt bridge or by some other method.
Using this notation we can write our electrochemical cell as,
Cu(s)|CuSO4(aq)||ZnSO4(aq)|Zn(s). (27)