Transcript Redox equilibrium - SALEM-Immanuel Lutheran College

Redox Equilibrium
1.
Electrochemical Cells
2.
Electromotive Force (e.m.f.) of
Electrochemical Cells
3.
Standard Electrode Potentials (Eo)
4/5. Use of Eo Values and Its Limitations
6.
Electrode Potentials at Non-standard
Conditions - Nernst Equation
7.
Commercial Electrochemical Cells
Electrochemical Cells
An electrochemical cell is a device that
converts chemical energy into electrical
energy.
Spontaneous redox reactions are involved
in the operation of electrochemical cells.
G < 0
E > 0
Electrochemical cells
Chemical
energy
Electrical
energy
Electrolytic cells
(electrolysis)
Important terms about electrochemical cells
To
To
balance
direct
the
charge
reaction
ofby
ions
between
in each
R.A.
half-cell
and
O.A.
Toprevent
complete
the
circuit
allowing
flow
of ions
A
salt
bridge
serves
three
purposes
e
e
e
e
e
Loss
replenished
anion
Electrolytes
Gain of
of –ve
-ve charge
is
cancelled
Inert
Zn,
Pt
Mg,
or
or
active
…etc.
C by cations
Oxidation
half-cell
e
Reduction
half-cell
Anions
Cations
()
(+)
Daniell Cell
e
e
V
e
Porous
partition
e
Cu
1M CuSO4
Cu2+
SO42
Zn
Cu2+
Zn2+
e
e
1M ZnSO4
Direct reaction is prevented
Conduction is completed by flow of ions
Q.1 (a)
Anode
Cathode
Active electrode : Zn(s)  Zn2+(aq) + 2e
Inert electrode : Cu
e
Q.1 (b)
e
e
Anode
e
e
Cathode
Q.1 (c)
Anode :
Zn(s)  Zn2+(aq) + 2e
Cathode :
Cu2+(aq) + 2e  Cu(s)
Overall : Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s)
Q.1 (d)
Cl
Zn2+
No. of +ve charges  due to formation of Zn2+(aq)
Cl ions from salt bridge moves in the left
half-cell to balance the extra +ve charges.
Q.1 (e)
K+
Cu2+
No. of +ve charges  due to discharge of Cu2+(aq)
K+ ions from salt bridge moves in the right
half-cell to replenish the +ve charges lost.
Electromotive Force of Electrochemical Cells
Electrons generated at the anode of a cell are
thought to be “pushed” toward the cathode by an
electromotive force, or e.m.f..
This force is due to the difference in electric
potential energy of an electron at the two electrodes.
Electrons move from an electrode of higher electric
potential (anode) to one of lower electric potential
(cathode).
e
e
e
e
e
Work done
Higher electric
potential
Lower electric
potential
e
Electric work = charge  potential difference
(Joule)
(Coulomb)
(Volt)
1 Joule = 1 Coulomb  1 Volt
1 Joule
1 Volt 
1 Coulomb
1 Joule = 1 Coulomb  1 Volt
1 Joule
1 Volt 
1 Coulomb
One Joule of work is performed by a
cell when one Coulomb of charge
passes through a potential difference
of one Volt across the electrodes
In practice, the potential difference of a cell
is measured by a voltmeter of high resistance
 the current (i) that flows through the
cell and the external circuit is minimized.
 the p.d. measured is maximized.
 e.m.f. is measured.
By definition, the electromotive force is the
p.d. across a cell in open circuit,
i.e. when i = 0.
Examples of high resistance voltmeter which
draw only a negligible current from the cell : -
Valve voltmeter
Transistor voltmeter
Digital multimeter
E.m.f. can also be measured using a
potentiometer.
E.m.f. can also be measured using a potentiometer.
E.m.f.
 Length of PR 2(volts)
Length of PQ
The e.m.f. is independent of the
physical dimensions of the cell
It depends only on
1.
the concentrations of the OA/RA
involved;
2.
the nature of the electrodes and
3.
the temperature.
All give the same voltage of 1.5 V
AAA AA
C
size size size
D
size
9V rectangular battery
= 6  1.5V AAAA cylindrical cells
Standard Electrode Potential
E.m.f. is the potential difference between two
electrodes.
It means each electrode has its own potential.
This potential is called the electrode potential.
Origin of Electrode potential
When a metal electrode, M(s), is dipped into
a solution of its ions, Mn+(aq), the following
equilibrium is set up.
Mn+(aq)
+
ne
M(s)
There are TWO ways for the system
to attain equilibrium.
At the start,
rate of losing e > rate of gaining e
Mn+(aq)
Case (A),
Mn+(aq)
+
ne
M(s)
M(s)
Metal M tends to lose electrons until
the rate of losing e = rate of gaining e
Mn+(aq)
Case (A),
Mn+(aq)
+
ne
M(s)
M(s)
Metal M tends to lose electrons until
the rate of losing e = rate of gaining e
Mn+(aq)
Case (A),
+
M(s)
ne
M(s)
Extra Mn+(aq)
released from M(s)
Mn+(aq)
Surplus of e
left on electrode
At equilibrium, a potential difference is set up
between the electrode M and Mn+(aq)
Mn+(aq) +
M(s)
Case (A),
Mn+(aq)
ne
M(s)
Since the electrode is negatively charged, it
is said to have a negative potential (E < 0).
Case (A),
M(s)
Mn+(aq)
Mn+(aq)
+
ne
M(s)
At the start,
rate of gaining e > rate of losing e
Mn+(aq)
Case (B),
Mn+(aq)
+
ne
M(s)
M(s)
Mn+ ion tends to gain electrons from M until the
rate of gaining e = rate of losing e
Mn+(aq)
+
M(s)
Case (B),
Mn+(aq)
ne
M(s)
Extra M(s)
deposited
from Mn+(aq)
Surplus of
+ve charge
on electrode
Due to loss of
electrons in the
metallic structure
Mn+ ion tends to gain electrons from M until the
rate of gaining e = rate of losing e
Mn+(aq)
+
ne
M(s)
M(s)
Case (B),
Mn+(aq)
Surplus of –ve charge
in the solution
Due to discharge
of Mn+(aq)
At equilibrium, a potential difference is set up
between the electrode M and Mn+(aq)
Mn+(aq)
+
M(s)
Case (B),
Mn+(aq)
ne
M(s)
Since the electrode is positively charged, it
is said to have a positive potential (E > 0).
Mn+(aq)
Case (B),
Mn+(aq)
+
M(s)
ne
M(s)
Zn tends to lose electrons
Zn2+(aq) +
Zn(s)
2e
Zn(s)
Extra Zn2+(aq)
released from Zn(s)
Q.2 (a)
Zn2+(aq)
Surplus of e left
on electrode
Zn electrode has a negative potential.
Zn2+(aq) +
Zn(s)
2e
Zn(s)
Extra Zn2+(aq)
released from Zn(s)
Q.2 (a)
Zn2+(aq)
Surplus of e left
on electrode
Cu2+ tends to gain electrons
Cu2+(aq) +
Cu(s)
Q.2(b)
2e
Cu(s)
Extra Cu(s)
deposited
from Cu2+(aq)
Cu2+(aq)
Surplus of
+ve charge
on electrode
Cu electrode has a positive potential.
Cu2+(aq) +
Cu(s)
Q.2(b)
2e
Cu(s)
Extra Cu(s)
deposited
from Cu2+(aq)
Cu2+(aq)
Surplus of
+ve charge
on electrode
The absolute electric potential (or simply electrode
potential) of an electrode cannot be measured.
Only the potential difference (or e.m.f.) between
two electrodes can be measured.
e
e
e
e
e




+
+
+
+
To find the relative electrode potentials, we
arbitrarily assign an electrode potential of
zero to one particular half-cell and compare
the electrode potentials of all other half-cells
with this standard.
By international agreement, the standard
hydrogen electrode was chosen as the
reference electrode.
The standard electrode potential of a
half-cell is the e.m.f. generated by the
cell consisting of a standard hydrogen
electrode and the half-cell under standard
conditions.
E
0
H / H2
0
E
0
M n / M
 e.m.f .
Metal M
1.0 M Mn+
Q.3
H2(g) at 1 atm
and 298 K
Platinum electrode
coated with
‘platinum black’
Solution containing
1M H+(aq) at 298K
Outlet
for H2(g)
Q.3(b)
2H+(aq) + 2e
H2(g)
Eo = 0.00V
(i)  in pressure of H2 shifts the equilibrium
position to the left
 more e left on the electrode
 the electrode potential becomes
negative.
Q.3(b)
2H+(aq) + 2e
H2(g)
Eo = 0.00V
(ii)  in [H+(aq)] shifts the equilibrium
position to the right
 less e left on the electrode
 the electrode potential becomes
positive.
Since the equilibrium position and thus
the electrode potential is affected by
temperature as well as pressure and
concentration,
comparisons of electrode potentials
have to be made under
standard conditions.
Q.4
Standard conditions:-
T = 298 K.
Gases : partial pressure = 1 atm
Ionic species : concentration = 1.0 mol dm3
Solids : in their most stable form at the stated
conditions
Q.5
The electrode used is a piece of platinum
coated with finely divided ‘platinum black’
which catalyses the half-cell reaction and
provides a surface on which the hydrogen can
be adsorbed.
Equilibrium is then established between the
adsorbed layer of hydrogen gas and hydrogen
ions in solution :
2H+(aq) + 2e
H2(g)
Eo = 0.00V
Measuring Standard Electrode Potentials
Eo of the half-cell Mn+(aq)M(s) = the e.m.f.
generated by the cell shown below.
Metal
M
1.0 M Mn+
The Eo values are for reduction reactions
written in the form :
oxidized form + electrons  reduced form.
Li+(aq) + e
Li(s)
Eo = 3.05 V
Li+ / Li = 3.05 V
Oxidizing
Reducing
agent
agent
So standard electrode potentials are also
called standard reduction potentials.
When writing the reaction
reduced form  oxidized form + electron
Li(s)
Li+(aq) + e Eo = + 3.05 V
The sign of Eo is reversed,
but the magnitude of Eo is unaffected.
Oxidation potential
Reduction half-Reaction
Eo(V)
Li(s)
3.05
Mg2+(aq) +
2e-
Mg(s)
2.37
Zn2+(aq) +
2e-
Zn(s)
0.76
Fe2+(aq)
+
2e-
Fe(s)
0.44
Pb2+(aq) +
2e-
Pb(s)
0.13
2H+(aq) +
2e
H2(g)
0.00
Cu2+(aq) +
2e-
Cu(s)
+0.34
Ag+(aq)
e-
Ag(s)
+0.80
3e-
Au(s)
+1.50
+
+
Au3+(aq) +
 strength of reducing agents
 strength of oxidizing agents
e-
Li+(aq)
+(aq)
3+ the
Stronger
+the
Stronger

because
E
of2+
reaction
isagent
always
> table
0
2Li(s)
+R.A
Zn
(aq)
 2LiO.A
+ feasible
Zn(s)
Au
Li
is
is the
strongest
strongest
reducing
oxidizing
agent
in
the
in the
table
Eo = 0.76V – (3.05)V = +2.27V
Reduction half-Reaction
Eo(V)
Li(s)
3.05
Mg2+(aq) +
2e-
Mg(s)
2.37
Zn2+(aq) +
2e-
Zn(s)
0.76
Fe2+(aq)
+
2e-
Fe(s)
0.44
Pb2+(aq) +
2e-
Pb(s)
0.13
2H+(aq) +
2e
H2(g)
0.00
Cu2+(aq) +
2e-
Cu(s)
+0.34
Ag+(aq)
e-
Ag(s)
+0.80
3e-
Au(s)
+1.50
+
+
Au3+(aq) +
Zn(s) + Fe2+(aq)  Zn2+(aq) + Fe(s)
Eo = 0.44V – (0.76)V = +0.32V
 strength of reducing agents
 strength of oxidizing agents
e-
Li+(aq)
Reduction half-Reaction
Eo(V)
Li(s)
3.05
Mg2+(aq) +
2e-
Mg(s)
2.37
Zn2+(aq) +
2e-
Zn(s)
0.76
Fe2+(aq)
+
2e-
Fe(s)
0.44
Pb2+(aq) +
2e-
Pb(s)
0.13
2H+(aq) +
2e
H2(g)
0.00
Cu2+(aq) +
2e-
Cu(s)
+0.34
Ag+(aq)
e-
Ag(s)
+0.80
3e-
Au(s)
+1.50
+
+
Au3+(aq) +
Cu(s) + 2Ag+(aq)  Cu2+(aq) + 2Ag(s)
Eo = +0.80V – (0.34)V = +0.46V
 strength of reducing agents
 strength of oxidizing agents
e-
Li+(aq)
Reduction half-Reaction
Eo(V)
Li(s)
3.05
Mg2+(aq) +
2e-
Mg(s)
2.37
Zn2+(aq) +
2e-
Zn(s)
0.76
Fe2+(aq)
+
2e-
Fe(s)
0.44
Pb2+(aq) +
2e-
Pb(s)
0.13
2H+(aq) +
2e
H2(g)
0.00
Cu2+(aq) +
2e-
Cu(s)
+0.34
Ag+(aq)
e-
Ag(s)
+0.80
3e-
Au(s)
+1.50
+
+
Au3+(aq) +
 strength of reducing agents
 strength of oxidizing agents
e-
Li+(aq)
weaker R.A + weaker O.A  NOT feasible
because E of the reaction is always < 0
Reduction half-Reaction
Eo(V)
Li(s)
3.05
Mg2+(aq) +
2e-
Mg(s)
2.37
Zn2+(aq) +
2e-
Zn(s)
0.76
Fe2+(aq)
+
2e-
Fe(s)
0.44
Pb2+(aq) +
2e-
Pb(s)
0.13
2H+(aq) +
2e
H2(g)
0.00
Cu2+(aq) +
2e-
Cu(s)
+0.34
Ag+(aq)
e-
Ag(s)
+0.80
3e-
Au(s)
+1.50
+
+
Au3+(aq) +
2H+(aq) + Cu(s)  H2(g) + Cu2+(aq)
Eo = +0.00V – (0.34)V = 0.34V
 strength of reducing agents
 strength of oxidizing agents
e-
Li+(aq)
The algebraic sign of the potential is the sign of the electrode
when it is connected to the standard hydrogen electrode.
Li+ / Li = 3.05 V
H+ / H2 = 0.00 V
3.05 V
+

e
Li loses e more
readily than H2
Li
1.0 M Li+
Zn2+ / Zn = 0.76 V
H+ / H2 = 0.00 V
0.76 V
+

e
Zn loses e more
readily than H2
Zn
1.0 M Zn2+
The algebraic sign of the potential is the sign of the electrode
when it is connected to the standard hydrogen electrode.
Cu2+ / Cu = +0.34 V
H+ / H2 = 0.00 V
0.34 V

+
e
H2 loses e more
readily than Cu
Cu
1.0 M Cu2+
The algebraic sign of the potential is the sign of the electrode
when it is connected to the standard hydrogen electrode.
Au3+ / Au = +1.50 V
H+ / H2 = 0.00 V
1.50 V

+
e
H2 loses e more
readily than Au
Au
1.0 M Au3+
The algebraic sign of the potential is the sign of the electrode
when it is connected to the standard hydrogen electrode.
?
The e.m.f. of the cell,, is the difference of
the electrode potentials of the two half-cells.
o
E
=?
o
 ΔE  E
o
ΔE  E
o
o
cell
E
o
cell
E
o
cathode
anode
E
E
o
o
anode
cathode
Which one is correct ?
?
The e.m.f. of the cell,, is the difference of
the electrode potentials of the two half-cells.
The redox reaction involved in a
chemical cell is always spontaneous
 e.m.f. of a cell is always positive
E
o
cell
E
o
cathode
E
o
anode
0
0.76 V
+

e
Zn
1.0 M Zn2+
Cathode :
2H+(aq) + 2e
Anode :
Zn(s)
H2(g)
Zn2+(aq) + 2e-
Eo = 0.00V
Eo = +0.76V
oxidation potential
(1) Cathode : 2H+(aq) + 2e
(2) Anode :
Zn(s)
Eo = 0.00V
H2(g)
Zn2+(aq) + 2e-
Eo = +0.76V
Overall reaction : (1) + (2)
2H+(aq) + Zn(s)
E
0
cell
H2(g) + Zn2+(aq)
 0.00V  (0.76V)  0.00V  (0.76V)
E
o
cathode
E
o
anode
Both are reduction potentials
(1) Cathode : 2H+(aq) + 2e
(2) Anode :
H2(g)
Zn2+(aq) + 2e-
Zn(s)
Eo = 0.00V
Eo = +0.76V
Overall reaction : (1) + (2)
2H+(aq) + Zn(s)
if E
0
cell
E
E
0
cell
 (0.76V)  (0.00V)  0.76V  0
∵
0
anode
E
H2(g) + Zn2+(aq)
0
cathode
Spontaneous redox reaction always
has a positive cell e.m.f.
0.34 V
+

e
Cu
1.0 M Cu2+
Cathode :
Cu2+(aq) + 2e
Anode :
H2(g)
Cu(s)
Eo = +0.34V
2H+(aq) + 2e-
Eo = 0.00V
(1) Cathode : Cu2+(aq) + 2e
(2) Anode :
H2(g)
Cu(s) Eo = +0.34V
2H+(aq) + 2e-
Eo = 0.00V
Overall reaction : (1) + (2)
Cu2+(aq) + H2(g)
E
0
cell
Cu(s) + 2H+(aq)
 0.34V  (0.00V)  0.34V  (0.00V)
E
o
cathode
E
o
anode
Both are reduction potentials
Q.6
Which of the following half-cells, when coupled
with a Zn2+(aq)Zn(s) half-cell would give the
greatest e.m.f. ?
Predict the direction of electron flow and
calculate the e.m.f. generated in each case.
Al3+(aq)Al(s);
Cu2+(aq)Cu(s);
Ag+(aq)Ag(s)
Al3+(aq)Al(s)
Eo = 1.66 V
Zn2+(aq)Zn(s)
Eo = 0.76 V
 Al loses electrons more readily
 Al is the anode (electrons flow from Al to Zn)
At the anode : Al(s)  Al3+(aq) + 3e
(1)
At the cathode : Zn2+(aq) + 2e  Zn(s)
(2)
Overall reaction = 2(1) + 3 (2)
2Al(s) + 3Zn2+(aq)  2Al3+(aq) + 3Zn(s)
E
0
cell
 E cathode  E anode
 (0.76V)  (1.66V)  0.90V
F
Electrode potentials depend on the nature of the O.A. and
R.A. and their concentrations,
NOT on the quantities of material used.
Changing the stoichiometric coefficients of a halfreaction therefore does not change the value of Eo.
For example, the reduction of Ag+(aq) to Ag(s) has an Eo
of +0.80 V, whether the reaction is written as
Ag+(aq) + e
Ag(s)
Eo = +0.80V
2Ag(s)
Eo = +0.80V
or as
2 Ag+(aq) + 2e
Volt is defined as “energy per charge”.
1 Joule
1 Volt 
1 Coulomb
Multiplying a reaction by some number causes
both the energy(work done) and the charge to
be multiplied by that number.
Thus, the ratio “energy per charge = volt” does
not change.
On the contrary, the enthalpy change of a reaction
depends on the stoichiometry of the equation
involved. (Refer to notes on “Energetics”, p.8).
Q.7
Q.7
1
2
Given : CO(g) + O2(g)  CO2(g)
Hco[CO(g)] = -283 kJ mol-1
Calculate the standard enthalpy change of the following
reaction,
2CO(g) + O2(g)  2CO2(g)
H of the reaction
= -283 kJ mol1  2 mol
= -566 kJ
Cu2+(aq)Cu(s)
Eo = +0.34 V
Zn2+(aq)Zn(s)
Eo = 0.76 V
 Zn loses electrons more readily
 Zn is the anode (electrons flow from Zn to Cu)
At the anode : Zn(s)  Zn2+(aq) + 2e
(1)
At the cathode : Cu2+(aq) + 2e  Cu(s)
(2)
Overall reaction = (1) + (2)
Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s)
E
0
cell
 E cathode  E anode
 (0.34V)  (0.76V)  1.10V
Ag+(aq)Ag(s)
Eo = +0.80 V
Zn2+(aq)Zn(s)
Eo = 0.76 V
 Zn loses electrons more readily
 Zn is the anode (electrons flow from Zn to Ag)
At the anode : Zn(s)  Zn2+(aq) + 2e
(1)
At the cathode : Ag+(aq) + e  Ag(s)
(2)
Overall reaction = (1) + 2 (2)
Zn(s) + 2Ag+(aq)  Zn2+(aq) + 2Ag(s)
E
0
cell
Greatest e.m.f.
 E cathode  E anode
 (0.80V)  (0.76V)  1.56V
Extension of Electrochemical Series to Include
Other Redox Systems
F
The electrode processes we have discussed so far have
been limited to metal-metal ion systems,
i.e. systems consisting of a metal in contact with a solution
of its ions.
In other electrode systems without a metal,
an inert electrode such as platinum or graphite is needed
for transfer of electrons.
A.
Non-metal non-metal ion system
 more =easily
I2 gains
Eoe
0.54 V than H+
cell(e.m.f.)
V
H2 at 1 atm
and 298 K
Pt coated with
platinum black
1M
H+(aq)
e
e
Salt bridge
Outlet
of H2
Positive electrode
e
1M I2 / 1M I
Pt
Reduction half reaction :
I2(aq) + 2e
2I(aq)
Eo = +0.54 V
B.
Metal ion metal ion system
Fe3+ gains
e more
easily
Eocell(e.m.f.)
= 0.77
V than H+
V
H2 at 1 atm
and 298 K
Pt coated with
platinum black
1M
H+(aq)
e
e
Salt bridge
Outlet
of H2
Positive electrode
e
1M Fe3+ / 1M Fe2+
Pt
Reduction half reaction :
Fe3+(aq) + e
Fe2+(aq)
Eo = +0.77 V
C. Half-cells involving more than two chemical species
o (e.m.f.)
MnO4 Egains
e more
than H+
= 1.51 easily
V
cell
V
H2 at 1 atm
and 298 K
e
e
Salt bridge
Pt coated with
platinum black
1 M H+(aq)
Outlet
of H2
Positive electrode
e
1M MnO4 / 1M
Mn2+ / 1M H+
Pt
Reduction half reaction :
MnO4(aq)+8H+(aq)+ 5e
Mn2+(aq)+4H2O(l) Eo = +1.51V
D.
Metal insoluble metal salt system
H+ gains
e(e.m.f.)
more=readily
Eocell
0.35 V than PbSO4
V
H2 at 1 atm
and 298 K
Pt coated with
platinum black
1 M H+(aq)
e
e
Salt bridge
Outlet
of H2
Reduction half reaction :
PbSO4(s) + 2e
Negative electrode
e
Saturated
PbSO4(aq)
Pb coated with
PbSO4(s)
from saturated PbSO4(aq)
Pb(s) + SO42(aq) Eo = 0.35 V
Extended Electrochemical Series at 298 K
Q.8
A
B
B
C
B
C
C
C
C
Extended Electrochemical Series at 298 K
-
C
A
C
C
C
C
A
C
C
Extended Electrochemical Series at 298 K
C
O
B
B
C
O
A
-
Extended Electrochemical Series at 298 K
C
D
D
C
B
A/R
-
Extended Electrochemical Series at 298 K
-
D
B
Hydrogen electrode at pH 14
C/R*
2H+ + 2e  H2Hydrogen electrode at pH 0 0.00
Extended Electrochemical Series at 298 K
-
A
-
-
Short-hand Representation of half-cells
(IUPAC Conventions)
1. Metal – metal ion system
E.g.
Li+(aq) + e  Li(s)
Li+(aq)  Li(s)
Represents a phase boundary
Zn2+(aq) + 2e  Zn(s)
Zn2+(aq)  Zn(s)
2. Non-metal – non-metal ion system
E.g.
I2(aq) + 2e  2I(aq)
I2(aq), 2I(aq)  Pt(s)
Br2(aq) + 2e  2Br(aq)
Br2(aq), 2Br(aq)  Pt(s)
The reduced form is always written next to
the inert electrode.
3. Metal ion – metal ion system
E.g.
Fe3+(aq) + e  Fe2+(aq)
Fe3+(aq), Fe2+(aq)  Pt(s)
Cu2+(aq) + e  Cu+(aq)
Cu2+(aq), Cu+(aq)  Pt(s)
2Hg2+(aq) + 2e  Hg22+(aq)
2Hg2+(aq), Hg22+(aq)  Pt(s)
4. Half-cells involving more than two chemical species
MnO4(aq) + 8H+(aq) + 5e  Mn2+(aq) + 4H2O(l)
[MnO4(aq) + 8H+(aq)], [Mn2+(aq) + 4H2O(l)]  Pt(s)
Cr2O72(aq) + 14H+(aq) + 6e  2Cr3+(aq) + 7H2O(l)
[Cr2O72(aq) + 14H+(aq)], [2Cr3+(aq) + 7H2O(l)]  Pt(s)
5. Metal – insoluble metal salt system
PbSO4(s) + 2e  Pb(s) + SO42(aq)
PbSO4(s), [Pb(s) + SO42(aq)]  Pt(s)
AgCl(s) + e  Ag(s) + Cl(aq)
AgCl(s), [Ag(s) + Cl(aq)]  Pt(s)
Hg2Cl2(s) + 2e  2Hg(l) + 2Cl(aq)
Hg2Cl2(s), [2Hg(l) + 2Cl(aq)]  Pt(s)
6. Others
ClO2(aq) + e  ClO2(aq)
ClO2(aq), ClO2(aq)  Pt(s)
MnO4(aq) + e  MnO42(aq)
MnO4(aq), MnO42(aq)  Pt(s)
Q.9
PbO2(s) + 4H+(aq) + SO42(aq) + 2e  PbSO4(s) + 2H2O(l)
[PbO2(s) + 4H+(aq) + SO42(aq)], [PbSO4(s) + 2H2O(l)]  Pt(s)
Or
[PbO2(s) + 4H+(aq) + SO42(aq)], [PbSO4(s) + 2H2O(l)]  Pb(s)
Or
H2SO4(aq)  PbO2(s)  PbSO4(s)  Pb(s)
Cathode : H2SO4(aq)  PbO2(s)  PbSO4(s)  Pb(s)
Anode : Pb(s)  PbSO4(s)  H2SO4 (aq)
lead-acid accumulator
Pb(s)  PbSO4(s)  H2SO4(aq)  PbO2(s)  PbSO4(s)  Pb(s)
Salt bridge or porous partition is NOT
required since the O.A. and R.A. are on
different electrodes
Pb(s) + SO42(aq)  PbSO4(s) + 2e
PbO2(s) + 4H+(aq) + SO42(aq) + 2e  PbSO4(s) + 2H2O(l)
Zn(s)  Zn2+(aq, 1.0 M)
Cu2+(aq, 1.0 M)  Cu(s)
Zn(s)  Zn2+(aq, 1.0 M)
Cu2+(aq, 1.0 M)  Cu(s)
digital multimeter
(as a voltmeter)
electron flow
1.0 M zinc
sulphate solution
copper container
(positive electrode)
zinc (negative
electrode)
porous pot
1.0 M copper(II)
sulphate solution
Use of Standard Electrode Potential
1.
To compare the strength of
oxidizing agents and reducing agents
2.
To calculate cell e.m.f.(Eocell)
3.
To predict the feasibility of redox
reactions
To compare the strength of oxidizing
agents and reducing agents
Oxidized form
+
ne
Reduced form
Eo
A more positive Eo value

species on the left gain electrons more
readily.

species on the
oxidizing agent.
left
is
a
stronger
To compare the strength of oxidizing
agents and reducing agents
Oxidized form
+
ne
Reduced form
Eo
A more negative Eo value

species on the right loses electrons more
readily.

species on the
reducing agent.
right
is
a
stronger
Q.10 Consider the electrochemical series on p.10.
Give the strongest reducing agent
and the strongest oxidizing agent.
Strongest reducing agent : Li
Strongest oxidizing agent : F2
To calculate cell e.m.f.(Eocell)
F
Cell e.m.f.(Eocell) can be calculated
using the formula,
E
F
o
cell
E
o
cathode
E
o
anode
Note that the Eo values always
refer to the reduction electrode
potentials of the half-cells.
To predict the feasibility of redox reactions
A + B
C + D
Eocell > 0

the forward reaction is
feasible
spontaneous
energetically favourable
under standard conditions.
Eocell
To predict the feasibility of redox reactions
A + B
C + D
Eocell < 0

the backward reaction is
feasible
spontaneous
energetically favourable
under standard conditions.
Eocell
Q.11 (a) Predict the feasibility of the following reactions
based on the reduction potential values given.
(i) 4Ag(s) + O2(g) + 2H2O(l)  4Ag+(aq) + 4OH(aq)
Oxidation : Ag(s)  Ag+(aq) + e
(1)
Eo = 0.80 V
Reduction : O2(g) +2H2O(l) +4 e  4OH(aq) (2)
Eo = 0.40 V
4(1) + (2)
4Ag(s) + O2(g) + 2H2O(l)  4Ag+(aq) + 4OH(aq)
Eocell = 0.80V + 0.40V = 0.40V < 0
Or
Eocell = Eocathode – Eoanode = (0.40V) – (0.80V) = 0.40V
 Not feasible
Reduction potential
(ii) 4Ag(s) + 4Cl(aq) + O2(g) + 2H2O(l)  4AgCl(s) + 4OH(aq)
Oxidation : Ag(s) + Cl(aq)  AgCl(s) + e
(1) Eo = 0.22 V
Reduction : O2(g) +2H2O(l) +4 e  4OH(aq)
(2) Eo = 0.40 V
4(1) + (2)
4Ag(s) + 4Cl(aq) + O2(g) + 2H2O(l)  4AgCl(s) + 4OH(aq)
Eocell = 0.22V + 0.40V = +0.18V > 0 Or
Eocell = Eocathode – Eoanode = (0.40V) – (0.22V) = +0.18V
 Feasible
(iii)
MnO42(aq)  MnO4(aq) + MnO2(s)
( in acidic solution, pH 0)
Oxidation : MnO42(aq)  MnO4(aq) + e
(1) Eo =-0.56 V
Reduction : MnO42(aq)+4H+(aq)+2e  MnO2(s)+2H2O(l) (2) Eo = 2.26 V
2(1) + (2)
3MnO42(aq) + 4H+(aq)  2MnO4(aq) + MnO2(s) + 2H2O(l)
Eocell = -0.56V + 2.26V = +1.70V > 0 Or
Eocell = Eocathode – Eoanode = (2.26V) – (0.56V) = +1.70V
 Feasible
(iv)
MnO42(aq)  MnO4(aq) + MnO2(s)
( in alkaline solution, pH 14)
Oxidation : MnO42(aq)  MnO4(aq) + e
(1) Eo = -0.56V
Reduction : MnO42(aq)+2H2O(l)+2e  MnO2(s)+4OH(aq) (2) Eo =0.60 V
2(1) + (2)
3MnO42(aq) + 2H2O(l)  2MnO4(aq) + MnO2(s) + 4OH(aq)
Eocell = 0.56V + 0.60V = +0.04V > 0 Or
Eocell = Eocathode – Eoanode = (0.60V) – (0.56V) = +0.04V
 Much less feasible than in acidic solution
Equilibrium position shifts to the left
Ag+(aq) + e
Ag(s)
Eoanode = 0.80V
Addition of Cl
[Ag+(aq)]  by forming AgCl ppt
Eanode = 0.22V < 0.80 V
AgCl(s) + e
Ag(s) + Cl(aq)
Eoanode = 0.22V
4Ag(s) + 4Cl(aq) + O2(g) + 2H2O(l)  4AgCl(s) + 4OH(aq)
Ecell = Eocathode – Eoanode = (0.40V) – (0.22V) = +0.18V > 0
in standard conditions as far as the
half reaction below is concerned
AgCl(s) + e
because
[Cl(aq)] = 1.0 M
Ag(s) + Cl(aq)
4Ag(s) + 4Cl(aq) + O2(g) + 2H2O(l)  4AgCl(s) + 4OH(aq)
Ecell = Eocathode - Eanode = (0.40V) – (0.22V) = +0.18V > 0
Not in standard conditions as far as
the half reaction below is concerned
Ag+(aq) + e
Ag(s)
because
[Ag+(aq)] << 1.0 M
[Ag (aq)][Cl (aq)] Ksp (AgCl(s)) 1.8  10 10
10


1.8

10
M




1.0
[Cl (aq)]
[Cl (aq)]
Equilibrium position shifts to the left
MnO42(aq) + 4H+(aq) + 2e
MnO2(s) + 2H2O(l)
Eocathode = 2.26V
Addition of OH
[H+(aq)]  by forming H2O
Ecathode = 0.60V < 2.26V
MnO42(aq) + 2H2O(l) + 2e
MnO2(s) + 4OH(aq)
Eocathode = 0.60V
3MnO42(aq) + 2H2O(l)  2MnO4(aq) + MnO2(s) +4OH(aq)
Eocell = Eocathode – Eoanode = (0.60V) – (0.56V) = 0.04V
in standard conditions as far as the
half reaction below is concerned
MnO42(aq) + 2H2O(l) + 2e
MnO2(s) + 4OH(aq)
because [OH(aq)] = 1.0 M
3MnO42(aq) + 2H2O(l)  2MnO4(aq) + MnO2(s) +4OH(aq)
Ecell = Ecathode – Eoanode = (0.60V) – (0.56V) = 0.04V
Not in standard conditions as far as
the half reaction below is concerned
MnO42(aq) + 4H+(aq) + 2e
MnO2(s) + 2H2O(l)
because [H+(aq)] << 1.0 M
[H (aq)][OH (aq)]
Kw
1.0  10 14
14


1.0

10
M




1.0
[OH (aq)]
[OH (aq)]
Q.12 Predict the effects of the following changes on the
cell e.m.f.(Ecell) of the reaction :
Zn(s) + 2Ag+(aq)  Zn2+(aq) + 2Ag(s) Eocell  1.56 V
Eocell = Eocathode – Eoanode = (0.80V) – (0.76V) = 1.56 V
Equilibrium position shifts to the right
Zn2+(aq) + 2e
 in
Zn(s)
forward reaction becomes
2+
[Zn (aq)]
 Eanode > Eoanode
more feasible
i.e. Eanode > 0.76V
 Ecell = Eocathode – Eanode < Eocell i.e. Ecell < 1.56V
Q.12 Predict the effects of the following changes on the
cell e.m.f.(Ecell) of the reaction :
Zn(s) + 2Ag+(aq)  Zn2+(aq) + 2Ag(s) Eocell  1.56 V
Eocell = Eocathode – Eoanode = (0.80V) – (0.76V) = 1.56V
Equilibrium position shifts to the right
Ag+(aq) + e
 in
Ag(s)
forward reaction becomes
+
[Ag (aq)]
more feasible
 Ecathode > Eocathode i.e. Ecathode > 0.80V
 Ecell = Ecathode – Eoanode > Eocell i.e. Ecell > 1.56V
Eocell = Eocathode – Eoanode = (0.80V) – (0.76V) = 1.56V
Equilibrium position shifts to the left
Ag+(aq) + e
Ag(s)
Addition of Cl to Ag+(aq)Ag(s)
[Ag+(aq)]  by forming AgCl ppt
forward reaction becomes less feasible
 Ecathode < Eocathode i.e. Ecathode < 0.80V
 Ecell = Ecathode – Eoanode < Eocell i.e. Ecell < 1.56V
Eocell = Eocathode – Eoanode = (0.80V) – (0.76V) = 1.56V
Equilibrium position shifts to the left
Zn2+(aq) + 2e
Zn(s)
Addition of S2 to Zn2+(aq)Zn(s)
[Zn2+(aq)]  by forming ZnS ppt
forward reaction becomes less feasible
 Eanode < Eoanode i.e. Eanode < 0.76V
 Ecell = Eocathode – Eanode > Eocell i.e. Ecell > 1.56V
Zn2+(aq) + 2e
Zn(s)
Doubling the size of Zn electrode
Increase both forward and backward
reactions to the same extent.
 No effect on the equilibrium position
 No effect on cell e.m.f.
Doubling the size of Zn electrode does
increase the current produced by the cell.
Shortening the distance between the electrodes
No effect on [Zn2+(aq)] and [Ag+(aq) ]
 No effect on the equilibrium position
 No effect on cell e.m.f.
Shortening the distance between the
electrodes does increase the current
produced by the cell.
Let the cell run for 30 minutes
Zn(s) + 2Ag+(aq)  Zn2+(aq) + 2Ag(s)
[Zn2+(aq)]  and [Ag+(aq) ] 
Zn2+(aq) + 2e
Zn(s)
forward reaction becomes more feasible
Eanode > Eoanode i.e. Eanode > 0.76V
Ag+(aq) + e
Ag(s)
forward reaction becomes less feasible
Ecathode < Eocathode i.e. Ecathode < 0.80V
Zn(s) + 2Ag+(aq)  Zn2+(aq) + 2Ag(s)
Zn2+(aq) + 2e
Zn(s)
Eanode > 0.76V
Ag+(aq) + e
Ag(s)
Ecathode < 0.80V
Ecell = Ecathode – Eanode < 1.56V
The cell e.m.f. decreases on discharging
Zn(s) + 2Ag+(aq)  Zn2+(aq) + 2Ag(s)
Ecell may ,  or remain unchanged
depending on the reaction quotient
[Zn (aq)]
Q

2
[Ag (aq)]
2
Refer to Nernst equation p.16
RT
EE 
lnQ
nF
o
[Zn (aq)]
Q

2
[Ag (aq)]
2
T
If Q > 1
 lnQ > 0
RT
 (
lnQ ) is more negative
nF
E
RT
EE 
lnQ
nF
o
[Zn (aq)]
Q

2
[Ag (aq)]
2
T
If Q = 1
 lnQ = 0
 E = Eo
Changing T has no effect on Eo
RT
EE 
lnQ
nF
o
[Zn (aq)]
Q

2
[Ag (aq)]
2
T
If Q < 1
 lnQ < 0
RT
 (
lnQ ) is more positive
nF
E
RT
EE 
lnQ
nF
o
[Zn (aq)]
Q

2
[Ag (aq)]
T
If Q > 1
 lnQ > 0
RT
 (
lnQ ) is less negative
nF
E
2
RT
EE 
lnQ
nF
o
[Zn (aq)]
Q

2
[Ag (aq)]
2
T
If Q = 1
 lnQ = 0
 E = Eo
Changing T has no effect on Eo
RT
EE 
lnQ
nF
o
[Zn (aq)]
Q

2
[Ag (aq)]
T
If Q < 1
 lnQ < 0
RT
 (
lnQ ) is less positive
nF
E
2
Limitations of using Eo values in predicting
the feasibility of redox reactions
• The predictions are only valid under
standard conditions
• The predictions are about energetics
but not kinetics
Q.13(a)(i)
+161 k J mol1
Li(s)

+519 k J mol1
Li(g)

499 k J mol1
Li+(g)

Li+(aq)
H = +161 + 519 – 499 = +181 k J mol1
Q.13(a)(ii)
+90 k J mol1
K(s)

+418 k J mol1
K(g)

K+(g)
305 k J mol1

K+(aq)
H = +90 + 418 – 305 = +203 k J mol1
Q.13(b)(i)
Since the oxidation of Li(s) is less endothermic,
(less positive Ho)
the reaction is energetically more favourable
than that of K(s)

Lithium has a more positive oxidation
electrode potential.
Q.13(b)(ii)
Energy is needed to separate the metal atoms
from the crystal lattice before the reaction
with oxygen can occur. M(s)  M(g)
Since Li forms stronger metallic bonds than K,
as seen in the more positive standard enthalpy
change of atomization, more energy is required
for lithium to start the reaction.
In other words, the activation energy for the
reaction of lithium with oxygen is higher than
that of potassium with oxygen.
Electrode Potentials at Non-standard Conditions
RT
EE 
lnQ
nF
o
Nernst equation
E is electrode potential at non-standard conditions
Eo is electrode potential at standard conditions
Electrode Potentials at Non-standard Conditions
RT
EE 
lnQ
nF
o
Nernst equation
R is the gas constant (8.314 J K1 mol1)
T is temperature in Kelvin
Electrode Potentials at Non-standard Conditions
RT
EE 
lnQ
nF
o
Nernst equation
F is the Faraday constant (96485 coulomb mol1)
Quantity of electricity (in coulomb) carried by
one mole of electrons
Electrode Potentials at Non-standard Conditions
RT
EE 
lnQ
nF
o
Nernst equation
Q is the reaction quotient of the reaction concerned
n is the number of moles of electrons transferred in
the reaction concerned
RT
EE 
lnQ
nF
o
At standard conditions,
lnQ = 0
E=
o
E
Q.14
RT
EE 
lnQ
nF
o
(8.314)(298)
E 
lnQ
n(96485)
o
(8.314)(298)
E 
(2.303log10Q)
n(96485)
o
0.059
E 
log10Q
n
o
Q.15(a)
0.059
1
EE 
log10
2
2
[Ni (aq)]
o
Q.15(b)
[I (aq)]
0.059
EE 
log10
2
[I2 (aq)]
o

2
Q.15(c)
[Fe (aq)]
0.059
EE 
log10
3
1
[Fe (aq)]
o
2
Q.15(d)
2
[Mn
(aq)]
0.059
o
EE 
log10


8
5
[MnO4 (aq)][H (aq)]
Q.15(e)
0.059

EE 
log10[Cl (aq)]
1
o
Calculate the e.m.f. of the following cells at 298 K.
Zn(s)  Zn2+(aq, 1.0 M)
Cu2+(aq, 1.0 M)  Cu(s)
The cell is at standard conditions.
Eocell = Eocathode – Eoanode
= +0.34V – (0.76V) = +1.10V
Calculate the e.m.f. of the following cells at 298 K.
Zn(s)  Zn2+(aq, 0.010 M)
Zn2+(aq) + 2e
Cu2+(aq, 1.0 M)  Cu(s)
Zn(s)
0.059
1
E1  E 
log10
2
[Zn2 (aq)]
o
1
Cu2+(aq) + 2e
Cu(s)
0.059
1
E2  E 
log10
2
[Cu2 (aq)]
o
2
Zn2+(aq) + 2e
Zn(s)
0.059
1
E1  E 
log10
2
[Zn2 (aq)]
o
1
Cu2+(aq) + 2e
Cu(s)
0.059
1
E2  E 
log10
2
2
[Cu (aq)]
o
2
[Zn (aq)]
0.059
Ecell  E2  E1  E  E 
log10
2
2
[Cu (aq)]
2
[Zn
(aq)]
0.059
o
Ecell  Ecell 
log10
2
[Cu2 (aq)]
o
2
o
1
2
Ecell  E
o
cell
[Zn (aq)]
0.059

log10
2
2
[Cu (aq)]
2
Reaction quotient of the overall cell reaction
Zn(s) + Cu2+(aq)
(i)
Ecell
Zn2+(aq) + Cu(s)
0.059
0.010
 1.10 
log10
2
1.0
= 1.10 V + 0.059 V = 1.159 V
Ecell  E
o
cell
[Zn (aq)]
0.059

log10
2
2
[Cu (aq)]
2
Reaction quotient of the overall cell reaction
Zn(s) + Cu2+(aq)
(ii)
Ecell
Zn2+(aq) + Cu(s)
0.059
1.0
 1.10 
log10
2
0.010
= 1.10 V – 0.059 V = 1.041 V
Calculate the e.m.f. of the following cells at 298 K.
Zn(s)  Zn2+(aq, 0.010 M)
Cu2+(aq, 1.0 M)  Cu(s)
The left half-cell is at non-standard conditions
Ecell = Eocathode – Eanode
 o

0.059
1

log10
= +0.34V –  Eanode 
2
2
[Zn (aq)] 

0.059
1
log10
= + 0.34V –(0.76V) 
2
0.010
= 1.10V + 0.059V = 1.159V
Calculate the e.m.f. of the following cells at 298 K.
Zn(s)  Zn2+(aq, 1.0 M)
Cu2+(aq, 0.010 M)  Cu(s)
The right half-cell is at non-standard conditions
Ecell = Ecathode – Eoanode
 o

0.059
1
 –(0.76V)
  Ecathode 
log10
2
2
[Cu (aq)] 

0.059
1 

  0.34V 
log10
 –(0.76V)
2
0.010 

= 1.10 V – 0.059 V = 1.041 V
Q.16(i)
Cu(s)  Cu2+(aq, 0.010 M)
Ag+(aq, 1.0 M)  Ag(s)
Cu(s) + 2Ag+(aq)  Cu2+(aq) + 2Ag(s)
Ecell  E
o
cell
Ecell
[Cu (aq)]
0.059

log10
2
[Ag (aq)]2
2
0.059
0.010
 (0.80 - 0.34) 
log10
2
(1.0)2
= 0.519 V
Electrons flow from Cu(s) to Ag(s)
Q.16(ii)
Zn(s)Zn2+(1.0 M) Fe3+(0.10 M), Fe2+(0.010 M)  Pt(s)
Zn(s) + 2Fe3+(aq)  Zn2+(aq) + 2Fe2+(s)
Ecell
2
2
[Zn
(aq)][Fe
(aq)]
0.059
o
 Ecell 
log10
3
2
2
[Fe (aq)]
Ecell
0.059
(1.0)(0.010)
 (0.77  0.76) 
log10
2
(1.0)2
= 1.589 V
Electrons flow from Zn(s) to Pt(s)
Q.17
Pb(s)  Pb2+(aq, x M)
Cu2+(aq, 1.0 M)  Cu(s)
Pb(s) + Cu2+(aq)  Pb2+(aq) + Cu(s)
Ecell  E
o
cell
[Pb (aq)]
0.059

log10
2
[Cu2 (aq)]
2
0.059
x
0.529  (0.34  0.13) 
log10
2
1.0
x = 1.0102 M
Zn(s)  Zn2+(aq, 1.0 M)
Zn2+(aq, 1.0 M)  Zn(s)
Both electrode systems are at
standard conditions
No potential difference across the two
electrodes
Not a workable cell
Example (i)
Zn(s)  Zn2+(aq, 0.010 M)
Zn2+(aq, 1.0 M)  Zn(s)
The left half-cell is NOT at standard
conditions
There is a potential difference across
the two electrodes.
It is a workable cell
Concentration Cells
Zn(s)  Zn2+(aq, 0.010 M)
Zn2+(aq, 1.0 M)  Zn(s)
A concentration cell is
an electrochemical cell with
two equivalent half-cells differing only
in concentrations.
Zn(s)  Zn2+(aq, 0.010 M)
Zn2+(aq, 1.0 M)  Zn(s)
For the half-cell on the right
Zn2+(aq, 1.0 M) + 2e
Zn(s)
The system is at standard conditions
ER = Eo = 0.76 V
Zn(s)  Zn2+(aq, 0.010 M)
Zn2+(aq, 1.0 M)  Zn(s)
For the half-cell on the left
Zn2+(aq, 0.010 M) + 2e
The forward reaction is less feasible
than that at standard conditions
EL < 0.76 V (more negative than ER)
 The left electrode is the anode(-)
 Electrons flow from left to right
Zn(s)
Zn(s)  Zn2+(aq, 0.010 M)
Zn2+(aq, 1.0 M)  Zn(s)
For the half-cell on the left
Zn2+(aq, 0.010 M) + 2e
0.059
1
EL  E 
log10
2
2
[Zn (aq)]
0.059
1
 0.76 
log10
2
0.010
0.059
 0.76 
(2)  0.819 V
2
o
Zn(s)
Zn(s)  Zn2+(aq, 0.010 M)
Ecell  E
o
cathode
Zn2+(aq, 1.0 M)  Zn(s)
 Eanode
 0.76 V  (0.819 V) = 0.059 V
Eanode = 0.819V
(1) Zn2+(aq, 0.010 M) + 2e
Zn(s)
Eocathode = 0.76V
(2) Zn2+(aq, 1.0 M) + 2e
Zn(s)
Overall reaction : (2) – (1)
Zn2+(aq, 1.0 M)
Zn2+(aq, 0.010 M)
Zn2+(aq, 1.0 M)
Zn2+(aq, 0.010 M)
e
Zn(s)  Zn2+(aq, 0.010 M)
Zn2+(aq, 1.0 M)  Zn(s)
In a concenration cell,
electrons always flow from the less
concentrated half-cell to the more
concentrated half-cell until the
concentrations of the two half-cells
become equal.
Q.18(i)
e
Cu(s)  Cu2+(aq, 0.050 M)
Cu2+(aq, 0.50 M)  Cu(s)
Cu2+(aq, 0.50 M)  Cu2+(aq, 0.050 M)
Ecell
0.059
0.050
E 
log10
2
0.50
0.059
0.050

log10
2
0.50
o
cell
 0.030 V
Q.18(ii)
e
Pt(s)Fe2+(1.0 M), Fe3+(0.10 M) Fe3+(1.0 M), Fe2+(0.10 M)Pt(s)
Fe2+(aq) + Fe3+(aq)  Fe3+(aq) + Fe2+(aq)
(1.0 M)
(1.0 M)
Ecell  E
o
cell
(0.10 M)
(0.10 M)
0.059
(0.10)(0.10)

log10
1
(1.0)(1.0)
Ecell  0.059log101  10
2
= 0.12 V
Zn(s) + Cu2+(aq)
Ecell  E
o
cell
Zn2+(aq) + Cu(s)
RT [Zn (aq)]

ln
2
2F [Cu (aq)]
2
At equilibrium, Ecell = 0
The cell reaction has stopped
No more work done by the cell.
No more energy released by the cell.
Zn(s) + Cu2+(aq)
Ecell  E
o
cell
Zn2+(aq) + Cu(s)
RT [Zn (aq)]

ln
2
2F [Cu (aq)]
2
At equilibrium, Ecell = 0
0E
o
cell
0E
o
cell
RT [Zn (aq)]equil

ln
2
2F [Cu (aq)]equil
2
RT

lnKc
2F
Zn(s) + Cu2+(aq)
Ecell  E
o
cell
0E
o
cell
Zn2+(aq) + Cu(s)
RT [Zn (aq)]

ln
2
2F [Cu (aq)]
2
RT

lnKc
2F
2FE
lnKc 
RT
o
cell
2  96485 1.10

8.314  298
Kc (at 298 K) = 1.6  1037
For any redox reaction at equilibrium
nFE
lnKc 
RT
o
cell
Redox reaction with a more positive Eocell is
more complete at equilibrium.
nFE
o
cell
a potential difference of Eocell
Total number of charge
moving across
nFE
o
cell
= maximum electric work done
by the cell
= the maximum free energy
available for the system to do
work
Zn(s) + Cu2+(aq)
Zn2+(aq) + Cu(s)
Go of the cell reaction = 
nFE
It has a negative sign because
this amount of energy is to be
released to the surroundings
o
cell
For any redox reaction at equilibrium
nFE
 ΔG
lnKc 

RT
RT
o
cell
o
For spontaneous processes,
ΔG  0, E
o
o
cell
 0, Kc  1
For any redox reaction at equilibrium
nFE
 ΔG
lnKc 

RT
RT
o
cell
o
For non-spontaneous processes,
ΔG  0, E
o
o
cell
 0, Kc  1
Q.19(i)
o
ΔG o  nFEcell
 (2)(96485)(1.5621)  301 kJ mol 1
nFE
(2)(96485)(1.5621)
lnKc 

RT
(8.314)(298)
o
cell
Kc  6.9  10
52
mol
1
dm
3
Q.19(ii)
o
ΔG o  nFEcell
 (2)(96485)( 2.707)  522 kJ mol 1
nFE
(2)(96485)(2.707)
lnKc 

RT
(8.314)(298)
92
Kc  2.7  10
o
cell
Q.19(iii)
o
ΔG o  nFEcell
 (2)(96485)(0.040)  7.7 kJ mol 1
nFE
(2)(96485)(0.040)
lnKc 

RT
(8.314)(298)
o
cell
Kc  22.5