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CHAPTER 6: Chemical Equilibrium
The equilibrium constant and reaction quotient
The response of equilibria to the conditions
Electrical chemical cells
Electrode potentials
This chapter continues demonstration of applying the theory of thermodynamics, now to chemical equilibrium. The relationship
between equilibrium constant (and reaction quotient) and Gibbs free energy and how the equilibrium constant is changed by
reaction conditions (temperature, pressure, concentrations of reactants) are given precise theoretical foundation.
Electrochemistry which is based on redox reactions and thermodynamics is chosen as another best example (in addition to heat
engines) to showcase the power of thermodynamics.
The Extent of Reaction, Reaction Gibbs Free Energy and Equilibrium
The Extent of Reaction ξ
A
B
Suppose an infinitesimal amount d  of A is changed into B, then
dnA  d
dnB  d
When a finite amount  of A is changed into B, then nA  nA,0   , nB  nB ,0  
In general,
 1 R1  2 R2  ...  m Rm
( 1 R1  2 R2  ...  m Rm
dnJ   J d
 m1 P1  m 2 P2  ...   m n Pn
 m1 Rm1  m 2 Rm 2  ...   m n Rm n )
( J is positive for products and negative for reactants)
nJ  nJ ,0   J 
Reaction Gibbs Free Energy
 G 
r G  

   p ,T
Because dG   AdnA  B dnB   Ad  B d  (B   A )d
 G 

  B   A



 p ,T
r G  B   A
The reaction Gibbs free energy is the difference between the chemical potentials of the products and reactants at current
compositions of the reaction mixture. It changes over the course of a reaction before it reaches equilibrium (do not confuse it
with standard Gibbs energies of formation  r G  ).
If  r G  0,  B < A
forward reaction is spontenous;
If  r G  0,  B   A reverse reaction is spontenous.
At equilibrium:
r G  0
Chemical equilibrium is reached when the reaction Gibbs free energy is zero
Forward reaction spontaneous
Reverse reaction spontaneous
Equilibrium
If  r G  0, B < A
forward reaction is spontenous; exergonic (work producing)
If  r G  0, B   A reverse reaction is spontenous; forward reaction is not spontaneous but can occur by force; endergonic (work consuming)
Example: H2 + (1/2)O2  H2O  r G  = - 237 kJ/mol spontaneous, exergonic
Thermodynamically permitted reaction might be hindered by kinetics. A
catalyst might be needed for a reasonable reaction rate for a spontaneous
reaction. Above reaction needs initiation (ignition) or to be put in a fuel cell.
H2O  H2 + (1/2)O2
 r G  = + 237 kJ/mol
not spontaneous, endergonic
To force it occur, at least 237 kJ/mol of work has to be done.
A non-spontaneous reaction can be forced to occur.
A spontaneous reaction can drive another non-spontaneous one.
Equilibrium and Equilibrium Constant
Perfect gases:
A (g )
B (g )
 r G   B   A  (   RT ln pB )  (   RT ln p A )

B

A
r G  GB,m  GA,m  B  A
 r G   r G   RT ln Q, Q 
Reaction quotient
pB
pA
r G   f G ( B)   f G ( A)

p 
RT ln K   r G , K  Q equilibrium   B 
, K  er G / RT
 p A equilibrium

If the partial pressures of the two
gasses at equilibrium are equal (K =1),

then  r G =0 and T has no effect.
If  r G   0, K > 1, the partial pressure of A is larger than that of B – the products are dominant at eq. T  K 
If  r G   0, K < 1, the partial pressure of B is larger than that of A – the reactants are dominant at eq. T  K 
 0 If Q  K
 r G  RT ln Q / K   r G 
 0 If Q  K
The forward reaction is spontaneous if Q < K, the
reverse reaction is spontaneous if Q > K.
These results apply to ALL chemical reactions, not just perfect gases.
The standard Gibbs energy of the isomerization of pentane to 2-methylbutane t 298 K, the reaction CH3(CH2)3CH3 (g) 
(CH3)2CHCH2CH3 (g), is -6.7 kJ/mol (estimated based on enthalpies of formation). Therefore, its equilibrium constant is
K e
 ( 6700 Jmol 1 ) （
/ 8.314 JK 1mol 1 298 K )
 e2.7  15
The Importance of Entropy in Chemical Reaction:
n
A Simple but Important Example
 r H    J  f H m, J  
J 1
 r G = r H  
 r G = r H   mix G
mix G  nRT ( xA ln xA  xB ln xB )= - T Smix
 r G = r H   mix G
 mix G
The presence of mixing entropy creates a minimum of
reaction Gibbs free energy at the intermediate value of the
extent of reaction (red dot) rather than at its maximum
value (green dot). Rare are the cases when only products
or reactants exist in a chemical reaction, owing to the fact
that more reduction of Gibbs energy can be reached when
an appropriate amount of mixing entropy is retained as
compared to ‘pure products’ or ‘pure reactants’ cases.
The general case of a reaction
 1 R1  2 R2  ...  m Rm
 m1 Rm1  m 2 Rm 2  ...   m n Rm n
0   J vJ J
0= R1 R1  R1 R2  ...  Rmn Rm n


dG    J dnJ    J vJ d    vJ  J  d
J
J
 J

N 2 ( g )  3H 2  2 NH 3 ( g ), N2  1, H 2  3, NH3  2
 G 
r G  
   vJ  J
J
   p ,T
 r G   vJ  J = vJ (  J  RT ln aJ )   vJ  J   vJ RT ln aJ )   r G   RT  ln aJvJ
J
J
J
J
  r G   RT ln  aJvJ   r G   RT ln Q
note :  ln xi  ln x1  ln x2  ...  ln xn  ln( x1 x2 ...xn )  ln  xi
J
i
 r G    vJ  J   vJ  f G  ( J )=
J
Q   aJvJ 
J


reactants
J

vi  f G  (i )-
products
v
ai i
products
v
ajj
J
(note: vJ are negative for reactants)

i
v j  f G ( j)
reactants
 aivi
 products

vJ 
K    aJ 

v
 ajj
 J
equilibrium  reactants
Because activities are dimensionless, Q and K must be dimensionless.



equilibrium
RT ln K   r G 
CaCO3 (s)
CaO(s）+CO 2 (g )
1
1
K =aCaCO
a
a
=
3 (s ) CaO(s ) CO2 (g)
aCaO(s ) aCO2 (g)
aCaCO3 (s )
1
 aCO2 (g) 
pCO2
p
State
Measure
Solute
molality
Approximation for aJ
molar concentration
Gas phase
3H 2 (g ) +N 2 (g )
partial pressure
2NH 3 (g )
 r G    vJ  J   vJ  f G  ( J )=
J
J

vi  f G  (i )-
products

bJ / bJ
[ J ] / c
pJ / p 
Definition
bJ  1 molkg 1
c  1 molL1
p   1 bar  105 Pa
N 2 O 4 (g )
v j  f G ( j)
reactants
2NO 2 (g ) at 298 K
 r G  =2 f G  (NO 2 (g ))- f G  (N 2O 4 (g ))}
=2 f G  (NH 3 (g ))-{ f G  (N 2 (g ))  3 f G  (H 2 (g ))}
 2  (51.31 kJmol-1 )-97.89 kJmol-1 （Appendix Table 2C.5）
=2 f G  (NH 3 (g ))  0  0  2  (16.45 kJmol -1 )
 r G  = 4.73 kJmol-1

2( 16.45 kJmol )
rG
ln K   RT
  8.3145
 13.2  K=5.8 105
JK 1mol-1  298K
-1
which is thermodynamically exact and should be equal to
3
1
2
H 2 (g) N 2 (g) NH 3 (g)
K =a
a
At low pressures,
K= (p
a
2
aNH
3 (g)
= a3
H 2 (g) aN 2 (g)
aJ  pJ / p 
( pNH3 / p )2
 3

H2 / p ) pN2 / p

2
pNH
p 2
3
pH3 2 pN 2

Jmol
rG
ln K   RT
 - 8.31454730
 -1.909  K=0.15
JK 1mol-1 298K
-1
Degree of dissociation at equilibrium
=
mole at any dissociaiton stage - mole at dissociation equilibrium n  neq
mole at any dissociaiton stage
n
H 2 (g ) + 12 O 2 (g )
The standard reaction Gibbs free energy of the dissociation reaction H 2 O(g )
is 118.08 kJ/mol at 2300 K. Find the degree of dossociation at 2300 K and 1 bar.
ln K  
r G
RT
-
108080 Jmol-1
8.3145 JK1mol-1298K
 -6.17  K=2.08 10-3
H2O
H2
(1/2)O2
Initial amount
n
0
0
Change to reach
equilibrium
-αn
αn
(1/2) αn
Amount at equilibrium
(1-α)n
αn
(1/2) αn
Mole fraction xJ
(1-α)/(1+α/2)
α/(1+α/2)
α/2)/(1+α/2)
Partial pressure pJ
(1-α)p/(1+α/2)
αp/(1+α/2)
(αp/2)/(1+α/2)
K=
pH2 / p ( p O / p  )1/2
2
pH2O / p

 1/2
pH（
p
/
p
）
2
O
2
pH2O
=
 3/2（p / p）1/2
(1 )(2  )1/2
 1
Total: (1+α/2)n

K 
 3/2（p / p）1/2
(1 0)(2  0)1/2
 =（ 2K） /(p / p ) =（1.414  2.08 10 ） /1=0.0205
 1/2
2/3
O2
-3 2/3

 3/2（p / p )1/2
21/2
Different equilibrium constants
0= R1 R1  R1 R2  ...  Rmn Rm n
Because we have different forms of ‘concentrations’ hence activities, we have different forms of reaction quotient or equilibrium
constant:


aJ = J xJ   B, J bJ / b   c, J [ J ] / c
Q   aJvJ = ( J xJ )vJ   ( B , J bJ / b )vJ   ( c , J [ J ] / c ) vJ
J
J
J
A+B
J
K
Q  K Qx  K , BQb / Kb  K ,c Qc / K c
K     J v J , K  , B    B , J v J K  , c    c , J v J , K b   b  , K c    c 
vJ
J
J
J
J
vJ
C+D
aC aD  C  D bC bD


 K K b
a A aB  A B bAbB
J
K  K  K x  K  , B K b / K b  K  , c K c / K c 
For gas reactions:
K   aJvJ   ( pJ )vJ   (
f
J
J
J p J vJ
p
Perfect gas, J 1
) 
J
vJ
vJ
RT vJ
K perfect gas = ( pp )vJ   [ J ]vJ ( RT

[
J
]
(
 )
  p )
p
J
  ( c )
[ J ] vJ
J

J
(
J
J
)  Kc  (
c RT vJ
p
v
v
T
 K c ( c pRT
=K c ( 12.03
 )
K)
J
c RT vJ
p
J
c RT
c
p
) K (
)
 vJ
J
3H 2 (g ) +N 2 (g )
2NH 3 (g )
 =2-3-1= -2
T  298.15 K
2
298.15 K 2
T
K  K c ( 12.03
)

K

K
(

c 12.03 K )
K
Kc
614.2
The Importance of Entropy in Chemical Reaction:
Why some endothermic reactions can occur or some exothermic reactions are hard
 r H  / RT
r S  / R
K e
e
For endothermic reactions,

r H  0  e
 r H  / RT
 1, but if  r S   0, K still can be >1.
One may say an endothermic reaction can be driven by
entropy and the products dominate.
We will show in chapter 15 (statistical mechanics that
closely spaced energy levels correlate with a high entropy).
Therefore, if the products have denser energy levels, the
reaction entropy will be positive and can drive an
endothermic reaction to K > 1.
3H 2 (g ) +N 2 (g )
2NH 3 (g )
 r G   -33 kJmol-1 ,  r H   -92.2 kJmol-1 ,  r S   -198.8 JK 1mol-1
At 298 K， K  e
  r H  / RT
N 2 O 4 (g )
e
r S  / R
=e
92.2 kJmol-1
8.3145 JK 1mol-1298K
e
-
198.8 Jmol-1
8.3145 JK 1mol-1298K
 e37.2e23.9
2NO 2 (g ) at 298 K
 r H  = 2 f H  (NO2 (g ))- f H  (N 2O 4 (g ))}=66.36-9.16 =57.2 kJmol -1 ,
 r S  =480.12- 304.29 = 175.83 Jmol -1
K e
  r H  / RT
e
r S  / R
=e
-
57200 Jmol-1
8.3145 JK 1mol-1298 K
e
175.83 Jmol-1
8.3145 JK1mol-1298 K
 e26.7 e21.1
Four major types of chemical reactions in terms of driving force
(1)Both enthalpy and entropy promote forward reaction.
e
- r H  / RT r S  / R
，r H   0, r S   0  K  1 (exothermic, entropy increasing)
e
(2) Both enthalpy and entropy discourage forward reaction.
e-  r H

/ RT r S  / R
，r H   0,  r S   0  K  1 (endothermic, entropy decreasing)
e
(3) Enthalpy promotes but entropy discourages forward reaction.
e
- r H  / RT r S  / R
e
，r H   0,  r S   0, (exothermic, entropy decreasing),
both K  1 and K  1 possible, depending on enthalphy/entropy competiton.
(4) Enthalpy discourages but entropy promotes forward reaction.
e
- r H  / RT r S  / R
e
， r H   0,  r S   0 (endothermic, entropy increasing)
both K  1 and K  1 possible, depending on entropy/enthalphy competiton.
The responses of equilibria to the conditions
At a given temperature, K is a constant independent of pressure, but it does not mean the equilibrium compositions are
independent of pressure, but depend on how the pressure is applied. If the partial pressures of the reactants/products are not
changed by increasing pressure (e.g., via filling in noble gas), K is not changed. If the system is compressed, then the
pB2
compositions will be changed.
A (g )
2B (g )
K
pA p
When the gases are compressed, pa and pB are increased at the same rate, that is not enough to keep K constant. The mole
fraction of B has to be decreased to keep K constant.
Le Chatelier’s principle
A
B
Initial amount
n
0
Change to reach
equilibrium
-αn
2αn
Amount at
equilibrium
(1-α)n
2αn
Mole fraction xJ
(1-α)/(1+α)
2α/(1+α)
Total: (1+α)n
Partial pressure pJ (1-α)p/(1+α) 2αp/(1+α)
(1 ) n
1
2
A
B
(1 ) n  2 n
1
1
x 
K=

pB2
p A p

xB2 p 2
x A pp
,x 

p/ p ）
= 4 （
1 2
2

1

 
 
1

4
p
/
Kp


1
2


1
 
 
1

4
p
/
Kp


1
2
3H 2 (g ) +N 2 (g )
K= (p
2NH 3 (g )
( pNH3 / p )2
 3

H2 / p ) pN2 / p
=
2
xNH
p 2 p 2
3
xH3 2 xN2 p 4


2
xNH
p2
3
xH3 2 xN2 p 2
2
pNH
p2
3
pH3 2 pN2
 Kx
p 2
p2
Increasing the pressure must increase Kx
by square to keep K constant. This means
increasing pressure favors product. On the
other hand, if the product is removed, the
pressure must be reduced squarely by reducing
the fraction of reactants.
The van’t Hoff equation
r H 
d ln K r H 
d ln K

or equivalently,

2
dT
RT
d (1/ T )
R
Proof:
rG
d ln K
1 d ( r G  / T )
ln K  


RT
dT
R
dT
d ( r G  / T )
r H 
But

(Gibbs-Helmholtz equation)
2
dT
T
r H 
d ln K  r H  d ln K

，
2
dT
RT
d（1/T）
R
r H 
d ln K

d (1/ T )
R
If  r H   0 (exothermic), T  K 
If  r H   0 (endothermic), T  K 
Exothermic reactions: increasing temperature favors the reactants.
Endothermic reactions: increasing temperature favors the products.
A
B
 r H   0 (endothermic)
 r H   0 (exothermic)
T  K 
T  K 
Measuring a reaction enthalpy
Ag 2CO3 (s)
Ag 2O(s）+CO 2 (g )
T/K
350
400
K
3.98x10-4 1.41x10-2
T/K
450
500
1.86x10-1
1.48
350
400
450
500
103/T 2.86
2.5
2.22
2.00
-lnK
4.26
1.68
-0.39
6.83
 r H  - ln K
3
4.26-6.83
=
=（2.5-2.86
=9.6

10
）10-3
R
(1/ T )
 r H   80 kJmol1
The value of K at different temperatures
r H 
d ln K

d (1/ T )
R
1 1/T2
ln K 2  ln K1     r H  d (1/ T )
R 1/T1
r H   1 1 
ln K2  ln K1  
  
R  T2 T1 
3H 2 (g ) +N 2 (g )
2NH 3 (g )
 r H  =-92.2 kJmol-1
At 298 K, K1 =6.1105
At 500 K,
-1
-92200Jmol
1 
 1
ln K 2 = ln 6.1105 

 = -1.7
-1
-1 
8.32145JK mol  500K 298K 
K 2 =0.18
Electrochemical Cell
• A device in which an electric current is either
produced by a spontaneous chemical reaction or is
used to bring about a nonspontaneous reaction.
A galvanic cell is an electrochemical cell in which a spontaneous
chemical reaction is used to generate an electric current.
In an electrochemical cell, a
reaction takes place in two
separate regions. Oxidation
occurs at one electrode, and
the electrons released travel
through the external circuit to
the other electrode, where they
cause reduction. The site of
oxidation is called the anode,
and the site of reduction is
called the cathode.
negative
positive
Any two objects that have different
(first) ionization energies may function
as a cell.
- 1.234 +
- 0.02 +
When a bar of zinc is placed in a beaker of copper(II) sulfate solution,
copper is deposited on the zinc and the blue copper (II) ions are
gradually replaced by colorless zinc ions. (b) The residue in the
beaker is copper metal. No more copper ions can be seen in solution.
Zn(s)+Cu (aq)  Zn (aq)+Cu(s)
2+
2+
The reaction shown in Fig. 18.3 takes place all over
the surface of the zinc as electrons are transferred
to the Cu2 ions in solution.
Cu 2+ (aq)+2e-  Cu(s)
Zn(s)  Zn (aq)+2e
2+
-
The Daniell cell consists of copper and zinc electrodes dipping into
solutions of copper(II) sulfate and zinc sulfate, respectively. The two
solutions make contact through the porous pot, which allows ions to pass
through to complete the electrical circuit.
Zn(s)|Zn 2+ Cu 2+ |Cu(s)
Zn(s)  Zn 2+ (aq)+2e-
Cu (aq)+2e  Cu(s)
2+
-
Electrodes and Cell Diagram
Zn(s)|Zn 2+ Cu 2+ |Cu(s)
+
H (aq)|H 2 (g)|Pt(s)
3+
2+
Fe (aq),Fe (aq) |Pt(s)
2+
2+
Zn(s)|Zn (aq)|Cu (aq) |Cu(s)
This cell is typical of galvanic cells used in the laboratory. The two
electrodes are connected by an external circuit and a salt bridge. The
latter completes the electrical circuit within the cell.
Pt(s)|Fe3+ (aq),Fe2+ (aq) ||Cu 2+ (aq) |Cu(s)
The cell potential is measured with an electronic voltmeter, a device that
draws negligible current so that the composition of the cell does not
change during the measurement. The display shows a positive value
when the  terminal of the meter is connected to the cathode of the
galvanic cell.
2+
2+
Zn(s)|Zn (aq)||Cu (aq) |Cu(s)
Cell Potential
E = 1.1 V
The cell potential
• An indication of the electron-pulling and –pushing
power of the cell reactions; cell reactions at
equilibrium generate zero potential.
Electrons produced by oxidation leave a galvanic cell at the anode (),
travel through the external circuit, and reenter the cell at the cathode
(), where they cause reduction. The circuit is completed inside the cell
by migration of ions through the salt bridge. A salt bridge is
unnecessary when the two electrodes share a common electrolyte.
positive
negative
This schematic picture of a galvanic cell indicates the identities of the
anode and cathode, displays the oxidation and reduction half-reactions,
and shows the direction of electron flow.
Describing a galvanic cell and identifying the cell reaction
Hg2+ (aq)+2e-  2Hg(l) 2Hg(l)+2Cl- (aq)  Hg 2Cl2 (s)+2e-
 Hg (aq)+2Cl (aq)  Hg 2Cl2 (s)
2+
-
Hg(l)|Hg 2 Cl2 (s)|HCl(aq)||Hg (NO ) (aq) |Hg(l)
2
3 2
(KCl gel)
Exercise: Describing a galvanic cell and identifying the cell reaction
(Assume platinum electrode is used)
H2 (g)  2H+ (aq)+2e-
Co3+ (aq)+e-  Co2+ (aq)
H2 (g)+2Co (aq)  2H (aq)+ 2Co (aq)
3+
+
+
3+
2+
2+
Pt(s)|H 2 (g)||H (aq)||Co (aq),Co (aq) |Pt(s)
The cell potential can be thought of as being the difference of the two
reduction potentials produced by the two electrodes. The cell potential is
positive if the cathode has a higher potential than the anode.
Cell potential and electrode potential
standard cell potential:
E   E  (cathode)  E  (anode)
=[E  (cathode)  E  ( SHE )]  [E  ( anode)  E  ( SHE )]
Fe(s)|Fe 2+ (aq)||Ag + (aq) | Ag(s):
E   1.24 V
Redox couple
+
AgCl(s) 
 Ag（
aq）+Cl ( aq) 
AgCl(s)+e  
 Ag(s)+Cl ( aq)

Ag（aq）+e 
 Ag(s)
+
Ox+ e 
 Red
+
O（
g
）+
4H

 2H2O(l)
2


Cl（
g
）+
2e


2Cl
(aq)
2
Q
Q
aH4 + aO2
p
= 4
aH+ pO2
aCl2 -
aCl2 - p
aCl2
pCl2
aH2 2O
Cell Potential, Electrical Work, and Free Energy
Work
• Work is never the maximum possible if any current is flowing.
• In any real, spontaneous process some energy is always wasted –
the actual work realized is always less than the calculated maximum.
Nernst Equation
 r G   work   Ecell q
Relation between free energy
difference and electric potential
difference:
q  F
 r G   FEcell
dG  r Gd =dwe   FEcell d
The change of free energy is equal to the work done by moving 1 mole of charges (νe-) from anode to cathode.
 r G = r G  +RT ln Q
r G
F
RT
ln Q 
vF
 r G  RT


ln Q
vF
vF
rG

(standard cell potential (emf))
vF
RT
 Ecell 
ln Q （Nernst Equation）
vF

Ecell
Ecell
Ecell
r G
F


Ecell - Ecell
RT / F
=  ln10log Q
Galvanic Cells
1.) Galvanic or Voltaic cell

Example: Calculate the voltage for the following chemical reaction
G = -150kJ/mol of Cd
n – number of moles of electrons
Solution:
G   nFEcell  Ecell  
Ecell
G
F
150  103 J

 0.777 J  0.777V
C

4 C 
(2 mol )  9.649 10

mol


Example: Calculating G for a Cell Reaction
• Using the data in previous Table, calculate G for the reaction
Cu2+(aq) + Fe(s)  Cu(s) + Fe2+(aq)

Is this reaction spontaneous? Ecell
is often denoted as
• The half-reactions are
G  is often denoted as G
• We can calculate G from the equation
G  G   RT ln Q
G  -nFE

cell
Ecell  E
- RT
nF ln Q
Example: Calculating G for a Cell Reaction
• Since two electrons are transferred per atom in the reaction, 2 moles
of electrons are required per mole of reactants and products. Thus n
= 2 mol e-, F = 96,485 C/mol e-, and
= 0.78 V = 0.78 J/C. Therefore,
C 
J

G  (2 mol e )  96, 485
0.78 
 
mol e 
C

 1.5 105 J


• The process is spontaneous, as indicated by both the negative sign of
G and the positive sign of
The relation between the standard
potential of a reaction (reactants,
purple; products, yellow) and the
equilibrium constant.
For a (half) reaction (Mn+ + ne-  M),
the higher the reduction potential (the
more negative ΔGo), the more easily it
proceeds.

Ecell  Ecell
- RT
nF ln K  0

Ecell

RT
nF
ln K
Dependence of Cell Potential on Concentration
A Concentration Cell
Dependence of Cell Potential on Concentration

Nernst Equation
Ecell  Ecell
- RT
nF ln Q
• The relationship between cell potential and
concentrations of cell components
• At 25°C:

Ecell  Ecell -
0.0591

log Q
aA + νe-  bB
RT ABb
EE 
ln a
 F AA

At equilibrium (E=0, Q=K):

Ecell =
0.0591

log K
 E  E 
0.05916 V

[ B]b
log
[ A]a
Dependence of Cell Potential on Concentration
EXERCISE!
A concentration cell is constructed using two nickel
electrodes with Ni2+ concentrations of 1.0 M and
1.00 × 10-4 M in the two half-cells.
Calculate the potential of this cell at 25°C.
0.118 V

1
E  0.0591
log
-4  0.118
2
110
Dependence of Cell Potential on Concentration
CONCEPT CHECK!
You make a galvanic cell at 25°C containing:
• A nickel electrode in 1.0 M Ni2+(aq)
• A silver electrode in 1.0 M Ag+(aq)
Sketch this cell, labeling the anode and cathode,
showing the direction of the electron flow, and calculate
the cell potential.
1.03 V
1.03 V
NO3-
K+
Ni
Ag
Ni2
+
E Ag/Ag  0.78
o
NO3Ag+
E Ni/Ni2  -0.25
o
Example: The Effects of Concentration on
• For the cell reaction
predict whether
following cases.
is larger or smaller than
for the
a. [Al3+] = 2.0 M, [Mn2+] = 1.0 M
b. [Al3+] = 1.0 M, [Mn2+] = 3.0 M

Ecell  Ecell -
0.0591
n

log Q  Ecell -
0.0591
n
[ Al 3 ]2
log [ Mn2 ]3
Example: The Effects of Concentration on
Solution
a. A product concentration has been raised above 1.0 M. This will
oppose the cell reaction and will cause
to be less than
(
< 0.48 V).

E cell  E cell -
0.0591
n
log
22
13
b. A reactant concentration has been increased above 1.0 M, and
will be greater than
(
> 0.48 V).

E cell  E cell -
0.0591
n
log
12
33
18.6 Batteries
• Mercury batteries use either pure mercury(II) oxide (HgO)—also called mercuric oxide—or a
mixture of HgO with manganese dioxide (MnO2) as the cathode. Mercuric oxide is a non-conductor,
so some graphite is mixed with it; the graphite also helps prevent collection of mercury into large
droplets. The half-reaction at the cathode is:
• HgO + H2O + 2e− → Hg + 2OH−
• with a standard potential of +0.0977 V.
• The anode is made of zinc (Zn) and separated from the cathode with a layer of paper or other
porous material soaked with electrolyte; this is known as a salt bridge. Two half-reactions occur at
the anode. The first consists of an electrochemical reaction step:
• Zn + 4OH− → Zn(OH)4−2 + 2e−
• followed by the chemical reaction step:
• Zn(OH)4−2 → ZnO + 2OH− + H2O
• yielding an overall anode half-reaction of:
• Zn + 2OH− → ZnO + H2O + 2e−
• The overall reaction for the battery is:
• Zn + HgO → ZnO + Hg
• In other words, during discharge, zinc is oxidized (loses electrons) to become zinc oxide (ZnO)
while the mercuric oxide gets reduced (gains electrons) to form elemental mercury. A little extra
mercuric oxide is put into the cell to prevent evolution of hydrogen gas at the end of life.
Determination of thermodynamic functions
S  -( )  (
G
T p
 FEcell
T
) p  F (
Ecell
T
)p

E cell  E cell - RT
nF ln Q

dE cell
dT

r S 
F

 r H    r G   T  r S    F ( E cell
T

Ecell
T
)
Determination of thermodynamic functions
The standard cell potential of Pt(s)|H2(g)|HBr(aq)|AgBr(aq)|Ag(s) was measured over a range of temperatures,
and the data were found to fit the following polynomial:
E cell  0.07131  4.99 10 4 (T  298)  3.45 106 (T  298) 2
The cell reaction is AgBr(s)+ 12 H 2 (g)  Ag(s)+HBr(aq). Evaluate the standard Gibbs free energy, enthalpy
and entropy at 298 K.
 r G  =- FEcell  -1 9.6485 104 Cmol-1  0.07131 V =-6.88 103 CVmol-1 =-6.88 103 Jmol-1
dE
cell
dT
 4.99 104  6.9 106 (T  298)
dE
cell
dT T=298 K
|

dEcell
dT

r S 
F
 4.99 104 V/K
  r S = F


dEcell
dT
=1 9.6485 104 Cmol-1 （ 4.99 104 V/K）
=-48.2 J/mol/K
 r H    r G   T  r S   -6.88 103 Jmol-1 +298 K  (-48.2 J/mol/K)
=-21.2 kJ/mol
Hydrogen standard potential
Standard hydrogen electrode (SHE):
Pt(s)|H 2 ( g ) | H  ( aq)
2H +2e  H2 (g): E  0 V
+
-
0
standard electrode potential E (X) :
Pt(s)|H 2 ( g ) | H  ( aq) || X
E  (X) = E cell
Cell: L||R=Ox L / Re d||Ox R / Re d R
E cell  E  (R)  E  (L)
Zn(s)|Zn 2+ (aq)||H + (aq) | H 2 (g)|Pt(s): E 0  0.76 V
E 0  E 0 (H + / H 2 )  E 0 (Zn 2+ / Zn)  0.76 V
 E (Zn / Zn)  0.76 V
0
2+
The variation of standard potentials in the main groups of the periodic
table. Note that the most negative values occur in the s block and the
most positive values occur close to fluorine.
Table 18-1 p856
Example: deducing the standard potential of an electrode
The standard potential of a Zn2+/Zn electrode is -0.76 V and the
standard potential of the cell Zn(s)|Zn2+(aq)||Cu2+(aq)|Cu(s) is
1.10 V.
What is the standard potential of Cu2+/Cu electrode?
E 0 (Cu 2+ /Cu)  E 0 (Zn 2+ /Zn)  1.10 V
E (Cu / Cu )  1.10 V  E (Zn / Zn)
 1.10 V - 0.76 V=+0.34 V
0
2+
0
2+
Exercise: deducing the standard potential of an electrode
The standard potential of a Fe2+/Fe electrode is -0.44 V and the
standard potential of the cell
Fe(s)|Fe2+(aq)||Pb2+(aq)|Pb(s) is 0.31 V.
What is the standard potential of Pb2+/Pb electrode?
E 0 (Pb2+ /Pb)  E 0 (Fe2+ /Fe)  0.31 V
E ( Pb / Pb)  0.31 V+E ( Fe / Fe)
 0.31 V  0.44 V =-0.13 V
0
2+
0
2+
The significance of standard potentials. Only couples with negative
standard potentials (and hence lying below hydrogen) can reduce
hydrogen ions to hydrogen gas. The reducing power increases as the
standard potential becomes more negative.
E 0 (Cu 2+ / Cu)=+0.34 V
E (Cu /Cu)  E (H /H)  0.34 V
0
2+
0
+
Cu (aq)+H 2 (g)  Cu(s)+H (aq)
2+
+
E 0 (Zn2+ / Zn)=-0.76 V
E (Zn /Zn)  E (H /H)  0.76 V
0
2+
0
+
Zn(s)+H (aq)  Zn (aq)+H2 (g)
+
2+
Although aluminum has a negative standard potential, signifying that it can be
oxidized by hydrogen ions (as in the hydrochloric acid, left), nitric acid (right)
stops reacting with it as soon as an impenetrable layer of aluminum oxide has
formed on its surface. This resistance to further reaction is termed passivation of
the metal.
E 0 (Al3+ /Al)=-1.66 V
E (Al /Al)  E (H /H)  1.66 V
0
3+
0
+
Al(s)+H (aq)  Al (aq)+H 2 (g)
+
3+
Measurement of standard electrode potential
Pt|H 2 (g)|HCl(aq, b)|AgCl(s)|Ag(s)
1
2
H 2 +AgCl(s)  HCl(aq)+Ag(s)
Ecell =E (AgCl/Ag,Cl ) - E (SHE)=E (AgCl/Ag,Cl ) ,  1
E

cell


=E (AgCl/Ag,Cl ) 
Ecell  Ecell

RT
F
ln
a
a
H  Cla1/2
H2

2
RT
Ecell =E - RT
ln
a
a
=E
ln(

b)

F
F
H  Cl-
a H =1
2
RT
RT
ln b 2 
ln   2
F
F
Ecell
2 RT
2 RT


ln b  Ecell 
ln  
F
F
At low concentration limit (b  0):
log     z z AI
1
2
 - A(b / b )1/2
ln    ln10 log    - A ln10(b / b )1/2
Ecell 
2 RT
2 RT
2 ART ln10 1/2


ln b  Ecell

[ A ln10(b / b )1/2 ]  Ecell

b
1/2
F
F
Fb
y
slope
intercept
Ecell
x
2 RT
2 ART ln10 1/2 1/2


ln b  Ecell 
b /b
F
F
y
slope
intercept
Ecell 
x
2 RT
2 ART ln10 1/2 1/2

ln b  Ecell

b /b
F
F
The Harned cell at 25 oC:
b/10-3bo
3.215
5.619
9.138
25.63
Ecell/V
0.52053
0.49257
0.46860
0.41824
b/10-3bo
3.215
5.619
9.138
25.63
(b/10-3bo)1/2
1.793
2.370
3.023
5.063
Ecell/V
0.52053 0.49257 0.46860
0.41824
y/V
0.2256
0.2299
0.2263
0.2273

Ecell
=0.2232 V
Determination of the electrochemical series
Redox couple:
Cell: L||R=Ox L / Re d||Ox R / Re d R
Ox R + e   Re d R
Ox L + e   Re d L
E cell  E  (R)  E  (L)
Cell reaction: R - L: Re d L  Ox R  Ox L + Re d R
If Ecell  0, K  1  If E  (L)  E (R ), K  1 
Re d L has a thermodynamic tendency (K>1) to reduce Ox R if E (L)  E  (R ).
(low reduces high)
Determination of the activities coefficient
Ecell 
2 RT
2 RT

ln b  Ecell

ln  
F
F

cell
E  Ecell
ln   =
 ln b
2 RT F
In previous example, Ecell  0.46860 V when b=9.138 10-3b . Because 2RT/F =0.05139 V, and E cell  0.2232 V,
the mean activity coefficient at this molality is
0.2232 V - 0.46860 V
ln   
 ln(9.138 10-3 )  0.0788
0.05139 V
   0.9242
Determination of Equilibrium Constant
The principle: the cell potential = the difference between the electrode potentials of the two electrode.

Ecell
 E  (R)-E  (L)

Ecell
 RTF ln K
ln K 
 F ( E  (R)-E  (L))
RT
Example： disproportionation reaction in which a species is both oxidized and reduced.
2Cu + (aq) = Cu(s) +Cu 2 (aq)
R: Cu(s)|Cu + (aq)
Cu + (aq)+e- (aq) = Cu(s)
L: Pt(s)|Cu 2+ (aq),Cu + (aq)
Find its equilibrium constant at 298 K.
E （R)=+0.52 V
Cu 2+ (aq)+e- (aq) = Cu  (aq)

E cell
 0.52 V - 0.16 V = 0.36 V
 =1
ln K 
0.36V
RT /F
0.36V
= 0.025693V
=14.0  K  1.2  106
E（L)=+0.16 V
Determination of Equilibrium Constant
Example： disproportionation reaction in which a species is both oxidized and reduced.
Sn (s) +Sn 4 (aq)
L: Sn(s)|Sn 2+ (aq)
2Sn 2+ (aq)
Sn 2+ (aq)+2e- (aq) = Sn(s )
R: Pt(s)|Sn 4+ (aq),Sn 2+ (aq)
Find its equilibrium constant at 298 K.
E (L)= -0.14 V
Sn 4+ (aq)+2e- (aq) = Sn 2 (aq)

E cell
 0.15 V - （-0.14）V = 0.29 V
 =2
ln K 
20.29V
RT /F
0.58V
= 0.025693V
=22.57  K  6.5 109
E (R)=+0.15 V