Transcript Lectures 28-31

Electrochemistry
Reduction-Oxidation (REDOX) Reactions
In chemical reaction bonds, both covalent and ionic, are made and broken by moving electrons.
So far all in reactions considered the number of electrons on each atom has been preserved
0
0
1+
2-
2 Na + ½ O2 → Na2O 2 Na+ + O22 Na → 2 Na+ + 2 e
Oxidized
½ O2 + 2 e → O2-
Loss e’s Oxidation = LEO
Gain e’s Reduction = GER
Reduced
In REDOX reaction electrons are transferred between atoms as well as bonds being broken and
formed.
The balance between ionic/covalency of the bonds in the reactants is different from that in the
product.
We therefore need to keep track of the number of electrons of each atom.
The oxidation state of an atom is used for this purpose, which is related to the idea of the formal
charge.
Oxidation State
The oxidation state of an element is its charge assuming ionic bonding
Electrons are not shared they are placed on the more electronegative element
Rules for Assigning Oxidation States
1) Pure elements have oxidation states of 0
2) Ions have oxidation states that add up to the charge of the ion
3) Hydrogen has an oxidation state of +1 unless bonded to a less electronegative atom.
When bound to metals or boron it has an oxidation state of -1.
4) Fluorine has an oxidation state of -1
5) Oxygen has an oxidation state of -2 unless bonded to fluorine or another oxygen.
6) Halogens other than fluorine have oxidation states of -1 unless bonded to oxygen or a more
electronegative halogen
7) The rest are determined by the process of elimination, where the oxidation states must add
up to the total charge of the molecule or ion.
Oxidation State
1)
Determine the oxidation state of all the element in the following molecules:
a) H2
H=0
b) CO2
O = -2
0 = C + 2(-2)
c) BF3
F= -1
0 = B + 3(-1)
d) H2SO4
H = 1 O = -2
e) SO2ClF
O = -2 F = -1 Cl = -1
f) IO2F2
_
g) HPO42-
O = -2
O = -2
F = -1
H=1
C=4
B=3
S=6
0 = S +(2(+1) + 4(-2)) = S - 6
0 = S +(2(-2) -1 -1) = S - 6
S=6
-1 = I + (2(-2) + 2(-1)) = I - 6
I=5
-2 = P + (4(-2) + 1) = P - 7
P=5
Oxidation States of Poly Atomic Ions
CO324
NO35
NO23
PO4
5
3-
PO333
SO42-
ClO3-
6
5
SO324
ClO23
Recognizing Redox Reactions
Consider the following reaction:
Cu(s) + 4HNO3 (aq) 
Cu(NO3 )2 (aq) + N2O4 (g) + 2H2O(l)
Reactants
Products
Cu
N=0
Cu2+
HNO3
N=5
N2O4
N=2
N=4
Cu → Cu2+ + 2 e
Oxidation
4 HNO3 + 2 e →
2NO3- + N2O4 + 2 H2O
Reduction
REDOX equation seem difficult to balance
# of e’s gained = # of e’s lost
Recognizing Redox Reactions
The species that is oxidized in a Redox reaction is called the reducing agent
In the previous example Cu is the reducing agent
The species that is reduced in a Redox reaction is called the oxidizing agent
In the previous example HNO3 is the oxidizing agent
Redox Reactions
The term Oxidation originates from the fact that many of these reactions involve oxygen where
the element loses electrons to it.
For example,
Burning magnesium:
Mg
 Mg2+ + 2 e–
Oxidation Half Reaction
Oxygen gains electrons:
½ O2 + 2 e–  O2–
and the net reaction is:
1 Mg
+ ½ O2
 1 MgO
Reduction Half Reaction
Redox Reactions
The term Reduction originates from the concept of reducing an ore, usually a metal oxide, to its
elemental form, usually the pure metal.
For example:
Iron is made from iron ore, the iron reaction is:
Fe2+ + 2 Fe3+ + 8 e–  3 Fe
Reduction Half Reaction
and the carbon monoxide reaction is:
4 CO  4 CO2+ + 8 e–
Oxidation Half Reaction
Here the oxide ions are spectators, transferring from the iron to the oxidized carbon. The net
reaction therefore becomes:
1 Fe3O4 + 4 CO  3 Fe + 4 CO2
Balancing Redox Reactions in Solution
The method of half reactions
Step 1. Recognize the reaction as an oxidation-reduction.
Step 2. Separate the overall process into half-reactions.
Step 3. Balance each half-reaction by mass.
i) In acid solution,
a) add H2O to the side requiring O atoms.
b) add H+ to balance any remaining unbalanced H atoms.
ii) In basic solution,
a) add 2 OH– to the side requiring O atoms, and H2O to the other
b) add H2O to the side requiring H atoms, and one OH– to the other side.
Step 4. Balance the half-reactions by charge.
Step 5. Multiply the balanced half-reactions by appropriate factors to achieve common whole
number of electrons being transferred.
Step 6. Add the balanced half-reactions.
Step 7. Eliminate common reactants and products.
Step 8. Check the final result for mass and charge balance.
Balancing Redox Reactions Sample Problems
2. Balance equations for the half-reactions. Assume these reactions occur in acidic solution,
meaning that H+ or H+ and H2O may be used to balance the equation.
(a)
Ox. St.
Mass balance:
Charge balance:
(b)
Ox. St.
O balance:
H balance:
Charge balance:
Br2(l) 
Br–(aq)
0
-1
Br2(l)  2 Br–(aq)
Br2(l) + 2e-  2 Br–(aq)
reduction
VO2+(aq)  V 3+(aq)
+4
+3
reduction
VO2+(aq)  V 3+(aq) + H2O
VO2+(aq) + 2 H+  V 3+(aq) + H2O
VO2+(aq) + 2 H+ + e-  V 3+(aq) + H2O
Balancing Redox Reactions Sample Problems
3. The half-reactions here are in basic solution. You may need to use OH– or the OH– / H2O
pair to balance the equation.
(a)
Ox. St.
O balance:
Charge balance:
(b)
Ox. St.
H balance:
Charge balance:
CrO2– (aq)

Ni(OH)2(s)

3+
CrO2– (aq) + 4 OH- 
CrO2– (aq) + 4 OH- 
2+
Ni(OH)2(s) + 2 OH- 
Ni(OH)2(s) + 2 OH- 
CrO42– (aq)
6+
oxidation
CrO42– (aq) + 2 H2O
CrO42– (aq) + 2 H2O + 3 e-
NiO2(s)
4+
oxidation
NiO2(s) + 2 H2O
NiO2(s) + 2 H2O + 2 e-
Balancing Redox Reactions Sample Problems
4. The reactions here are in acid solution, meaning that H+ or H+ and H2O may be used to balance the
equation.
Ox. St.
MnO4– (aq) + HSO3– (aq)  Mn2+ (aq) + SO42– (aq)
+7
+4
2+
Reduction half reaction: MnO4– / Mn2+
Mass Balance MnO4– (aq) +

Mn2+ (aq)
O balance:
MnO4– (aq) + 8H+
 Mn2+ (aq) + 4 H2O
Charge balance: MnO4– (aq) + 8H+ + 5 e– 
Mn2+ (aq) + 4 H2O
Oxidation half reaction: HSO3– / SO42–
Mass balance: HSO3– (aq) 
SO42– (aq)
O balance:
HSO3– (aq) + H2O
 SO42– (aq) + 3 H+
Charge balance: HSO3– (aq) + H2O
 SO42– (aq) + 3 H+ + 2 e–
6+
x2
x5
2 MnO4– (aq) + 5 HSO3– (aq) + 16H
H2O + 15XH+
X + + 5 HX2O  2 Mn2+ (aq) + 5 SO42– (aq) + 8 X
2 MnO4– (aq) + 5 HSO3– (aq) + 1 H+  2 Mn2+ (aq) + 5 SO42– (aq) + 3 H2O
Balancing Redox Reactions Sample Problems
4. The reactions here are in acid solution, meaning that H+ or H+ and H2O may be used to
balance the equation.
Cr2O72– (aq) + Fe2+ (aq)
Ox. St.
+6
 Cr3+ (aq)
+2
+
3+
Fe3+ (aq)
3+
Reduction half-reaction: Cr2O72–/ Cr3+
Mass balance:
Cr2O72– (aq)
O balance:
Cr2O72– (aq) + 14H+
Charge balance: Cr2O72– (aq) + 14H+ + 6 e–



2Cr3+ (aq)
2 Cr3+ (aq) + 7 H2O
2 Cr3+ (aq) + 7 H2O
Oxidation half-reaction: Fe2+/ Fe3+
Mass balance: Fe2+ (aq) 
Charge balance: Fe2+ (aq)

Fe3+ (aq)
Fe3+ (aq) + 1 e–
1 Cr2O72– (aq) + 6 Fe2+ (aq) + 14H+  2 Cr3+ (aq) + 6 Fe3+ (aq) + 7 H2O
x6
Balancing Redox Reactions Sample Problems
5. The reactions here are in basic solution. You may need to add OH– or the OH– / H2O pair to
balance the equation.
Fe(OH)2(s)
Ox. St.
+2
+ CrO42– (aq)
+6
 Fe2O3(s) + Cr(OH)4– (aq)
3+
Oxidation half-reaction: Fe(OH)2/Fe2O3
Mass balance : 2 Fe(OH)2(s) + 
Fe2O3(s)
O balance:
2 Fe(OH)2(s) + 2 OH–  Fe2O3(s) + 3 H2O
Charge balance: 2 Fe(OH)2(s) + 2 OH–  Fe2O3(s) + 3 H2O + 2 e–
Reduction half-reaction: CrO42–/ Cr(OH)4–
Mass balance: CrO42– (aq)  Cr(OH)4– (aq)
O balance:
CrO42– (aq) + 4 H2O  Cr(OH)4– (aq) + 4 OH–
Charge balance: CrO42– (aq) + 4 H2O + 3 e–  Cr(OH)4– (aq) + 4 OH–
3+
x3
x2
X
X
6 Fe(OH)2(s) + 2 CrO42–(aq) + 6 OH– + 8 H2O  3 Fe2O3(s) + 2 Cr(OH)4–(aq) + 9 H2O + 8 OH-
X
X
6 Fe(OH)2(s) + 2 CrO42–(aq)  3 Fe2O3(s) + 2 Cr(OH)4–(aq) + 1 H2O + 2 OH-
Balancing Redox Reactions Sample Problems
5. The reactions here are in basic solution. You may need to add OH– or the OH– / H2O pair to
balance the equation.
(b) PbO2(s) +
Cl– (aq) 
ClO– (aq) +
Pb(OH)3– (aq)
Ox. St. 4+
11+
3+
Reduction half-reaction: PbO2/ Pb(OH)3–
Mass balance: PbO2 (s) + 
Pb(OH)3– (aq)
O balance:
PbO2(s) + 2 H2O  Pb(OH)3– + OH–
Charge balance: PbO2(s) + 2 H2O + 2 e–  Pb(OH)3– + OH–
Oxidation half-reaction: Cl–/ ClO–
Mass balance: Cl– (aq)  ClO– (aq)
O balance:
Cl– (aq) + 2 OH–  ClO– (aq) + H2O
Charge balance: Cl– (aq) + 2 OH–  ClO– (aq) + H2O + 2 e–
– + 2 H O  Pb(OH) – (aq) + ClO–(aq) + OH– + H O
PbO2(s) + Cl–(aq) + 2 OH
X X2
3
X X2
1 PbO2(s) + 1 Cl–(aq) + 1 OH– + 1 H2O  1 Pb(OH)3– (aq) + 1 ClO–(aq)
Gibbs Free Energy and Cell Potential
The Gibbs Free Energy of reaction tells us whether a certain reaction is product favored or reactantfavoured in the forward direction
The electrochemical analogue is the cell potential given by the symbol E
E is expressed in the common electrical unit of volts.
The exact relationship between cell potential and Gibbs energy is given by:
ΔG= -nFE
in general, and for standard conditions
where F = 96485 C/mol
Faradays Constant
The cell voltage E has the opposite sign convention to that of G:
i) Product Favoured Reaction:
ii) Reactant Favoured Reaction:
–ve G
+ve G
or
or
ΔG0 = -nFE0
+ve E
–ve E
A reaction which is product favoured produces a voltage, and is therefore called a Voltaic cell, in
honor of Alessandro Volta.
A reaction which is reactant favoured can be driven forward by the application of a greater opposite
voltage; such reactions are called electrolytic cells, and the process is named electrolysis.
Voltaic Cells
Consider the reaction:
Gf (kJ mol–1)
Zn(s) +Cu2+(aq) 
Zn2+(aq) + Cu(s)
0
–147.0
65.52
Grxn = –212.5 kJ mol–1 < 0 &
n=2
Anode:
Ox
Zn(s)  Zn2+(aq) + 2 e–
Cathode:
Cu2+(aq) + 2 e– 
Red.
Cu(s)
ΔG= -nFE
-ΔG0
-212.5kJ
E =
=
nF
2×96.485kC
= +1.10V
0
> 0 under zero load conditions
0
Reversible cell reactions and cell notation
Although the reaction only requires a copper(II) salt and metallic zinc, the cell is built in such a way
that both metallic zinc and copper are present, and copper and zinc sulfate.
This ensures that the reactions are reversible as accurate potentials can only be measured for
reversible electrochemical cells.
A short-hand notation indicates the complete composition of a given electrochemical cell.
The notation utilizes the balanced redox reaction, including both the reactants and the products.
The notation for the previous example would be:
Zn(s) | Zn2+(1 M) || Cu2+(1 M) | Cu(s)
Left Hand Side
Right Hand Side
Anode reaction
Oxidation reaction.
Cathode reaction
Reduction reaction
The vertical lines, |, indicate a phase boundary, such as between a solid and a liquid.
Double lines, ||, indicate double boundaries, such as commonly occur when a salt bridge is placed
between the two half cells.
Electrochemical Cell Conventions
Standard Half-Cell Reduction Potentials
The Cu/Zn cell has a standard potential of +1.10 V
Can we relate this to
the voltage of the two
“half-cells”?
Choose an arbitrary
reference point:
Standard Hydrogen
Electrode, or SHE
Thus Zn/Zn2+ has
a voltage of +0.76 V
against a SHE
Standard Half-Cell Reduction Potentials
The Cu/Cu2+ has half cell has a voltage of +0.34 V against a SHE
Combining the two
half-cell voltages gives
+0.76 + +0.34 = +1.10 V
This is in good agreement
with the measured voltage
These values have been
compiled into a general
table
To account for the sign
convention, the table is
written only as reduction
reactions
Standard Hydrogen Electrode
Pt( s) | H2(
g)
| H3O+(
aq)
∥
The glass tube is a hydrogen electrode
that can be used as real reference electrode
The picture below is a diagram showing the
function of this device.
Strongest oxidizers
Reduction Half-Reaction
E (V)
F2(g) + 2 e –
 2 F – (aq)
+2.87
H2O2(aq) + 2 H3O + (aq) + 2 e –
 4 H2O(l)
+1.77
PbO2(s) + SO4 2– (aq) + 4 H3O + (aq) + 2 e –
 PbSO4(s) + 6 H2O(l)
+1.685
MnO4 – (aq) + 8 H3O + (aq) + 5 e –
 Mn 2+ (aq) + 12 H2O(l)
+1.52
Au 3+ (aq) + 3 e –
 Au(s)
+1.50
Cl2(g) + 2 e –
 2 Cl – (aq)
+1.360
Cr2O7 2– (aq) + 14 H3O + (aq) + 6 e –
 2 Cr 3+ (aq) + 21 H2O(l)
+1.33
O2(g) + 4 H3O + (aq) + 4 e –
 6 H2O(l)
+1.229
Br2(l) + 2 e –
 2 Br – (aq)
+1.08
NO3 – (aq) + 4 H3O + (aq) + 3 e–
 NO(g) + 6 H2O(l)
+0.96
OCl – (aq) + H2O(l) + 2 e –
 Cl – (aq) + 2 OH – (aq)
+0.89
Hg 2+ (aq) + 2 e –
 Hg(l)
+0.855
Ag + (aq) + e –
 Ag(s)
+0.80
Hg2 2+ (aq) + 2 e –
 2 Hg(l)
+0.789
Fe 3+ (aq) + e –
 Fe 2+ (aq)
+0.771
I2(s) + 2 e –
 2 I – (aq)
+0.535
O2(g) + 2 H2O(l) + 4 e –
 4 OH – (aq)
+0.40
Cu 2+ (aq) + 2 e –
 Cu(s)
+0.337
Sn 4+ (aq) + 2 e –
 Sn 2+ (aq)
+0.15
2 H3O + (aq) + 2 e –
 H2(g) + 2 H2O(l)
0.00
Reduction Half-Reaction
E (V)
2 H3O + (aq) + 2 e –
 H2(g) + 2 H2O(l)
0.00
Sn 2+ (aq) + 2 e –
 Sn(s)
–0.14
Ni 2+ (aq) + 2 e –
 Ni(s)
–0.25
V 3+ (aq) + e –
 V 2+ (aq)
–0.255
PbSO4(s) + 2 e –
 Pb(s) + SO4 2– (aq)
–0.356
Cd 2+ (aq) + 2 e –
 Cd(s)
–0.40
Fe 2+ (aq) + 2 e –
 Fe(s)
–0.44
Zn 2+ (aq) + 2 e –
 Zn(s)
–0.763
2 H2O(l) + 2 e –
 H2(g) + 2 OH – (aq)
–0.8277
Al 3+ (aq) + 3 e –
 Al(s)
–1.66
Mg 2+ (aq) + 2 e –
 Mg(s)
–2.37
Na + (aq) + e –
 Na(s)
K + (aq) + e –
 K(s)
Li + (aq) + e –
 Li(s)
† In
Strongest
reducing
agents
–2.714
–2.925
–3.045
volts (V) versus the standard hydrogen electrode.
Such tables may have the strongest reducing agent at the top, or at the bottom, as is the case in
this table. Reactions are always written as reductions, but this is correlated to the voltages. For
negative potentials, the reverse reaction is predicted.
Examples
Copper/Lithium cell
From the table we get the standard reduction potentials of the half-reactions:
Cu 2+ (aq) + 2 e –  Cu(s)
Li + (aq) + e –  Li(s)
E = +0.337 V
E = –3.045 V
Reduction
We now combine them in such a way as to get a positive overall cell potential. This means we
must reverse the lithium equation, making it the anode (where the oxidation will take place.)
Li(s)
 Li + (aq) + e –
E = +3.045 V
Anode
now we have the cell potential for the overall reaction
Cu 2+ (aq)
+ 2 Li(s)
 Cu(s)
+ 2 Li + (aq)Ecell = +3.382 V
Note here very carefully, that the voltage was not doubled when the coefficients are doubled.
However, n for this reaction = 2. Cell voltages are therefore independent of stoichiometry. The
stoichiometric information is stored in the value of n.
Some examples of using the Tables
Silver/zinc cell
From the table we get the standard reduction potential for the half-reactions:
Zn2+ (aq) + 2 e–  Zn(s)
Ag+ (aq) + e–  Ag(s)
E = –0.763 V
E = +0.80 V
Reduction
We now combine them in such a way as to get a positive overall cell potential. This means
we must reverse the zinc equation, making it the anode (where the oxidation will take place.)
Zn(s) 
Zn 2+ (aq)
+
2 e–
E = +0.763 V
Oxidation
now we have the cell potential for the overall reaction
Zn (s)
+ 2 Ag+ (aq)
 Zn2+ (aq)
+ 2 Ag (s)
Ecell = +1.563 V
Note here very carefully, that the voltage was not doubled when the coefficients are doubled.
However, n for this reaction = 2. Cell voltages are therefore independent of stoichiometry. The
stoichiometric information is stored in the value of n.
Nernst Equation - Cells Non-Standard Conditions
Voltaic cells eventually get depleted and stop producing a voltage.
In fact, as the reactants in the cell are consumed, the voltage produces decreases.
Temperature effects cell potential – the potential is reduced upon cooling.
Walter Nernst developed an expression, the Nernst equation, defined for the general equilibrium
equation:
aA + bB
cC + dD
RT C D
o RT
Eactual = Eo ln
=
E
lnQ
a
b
nF  A  B
nF
c
d
Frequently the equation is given with all the constants applied out at 298.15K
Eactual = Eo -
0.0257V
lnQ
n
pH meter: the glass electrode
A “pH meter” is actually a sensitive voltmeter attached to a glass
electrode
The electrode response to changes in [H3O+] by altering its
voltage as described by the Nernst equation
The key to its function as a meter is the very thin and fragile glass
membrane made of special materials that allow [H3O+] but not
other cations or anions to be sensed on the outer surface of the
membrane
The meter is calibrated in pH units by rewriting the Nernst
equation in base-10 logarithms
+

H
0.0592
 inside
Eglass_ electrode = 0 log +
1
H 
outside
Filling in the constants at 298.15K this becomes:
pH =
Eglass_ electrode
0.0592
- const.
Note that calibration is required to determine the constant
Silver wire
AgCl (s)
KCl (saturated)
Porous membrane
0.1 M HCl
Glass membrane
The Glass Electrode
Electrochemical Cells and Equilibrium Constants
We have already seen that ΔG and E are related:
We also know that ΔG and K are related:
ΔG0 = -nFE0
ΔG0 = -RTlnK
So its stands to reason that E and K are related, and you already know the relationship:
An electrochemical reaction is at equilibrium when the cell voltage is zero
So using the Nernst equation:
RT
Eactual = E lnQ
nF
o
when E= 0 V
0 V = Eo -
RT
lnK
nF
Gives the relationship between cell potential and equilibrium constant
Many equilibria are actually determined in this way
nF o
lnK =
E
RT
Note that “nF” is on
top in this equation!
Electrochemistry Example Problems
1. Calculate the Ecell of a cell formed from a Zn2+/Zn half-cell in which [Zn2+] = 0.0500 M and a
Cl2/Cl– half cell in which the [Cl–] = 0.0500 M and P(Cl2) = 1.25 atm. In addition, provide a
cell notation.
From the Standard Reduction Potentials Table:
 2 Cl – (aq)
 Zn(s)
Cl2(g) + 2 e –
Zn 2+ (aq) + 2 e –
Rev.#2:
Add:
Zn(s)
 Zn 2+ (aq) + 2 e –
Cl2(g) + Zn(s)

+1.360 V
–0.763 V
+0.763 V
Zn 2+ (aq) + 2 Cl –(aq) +2.123 V
RT [Zn2+ ][Cl- ]2
Eactual = E ln
nF
p  Cl2 
o
8.314 JKmol ×298.15K ( 0.0500) ( 0.0500)2
Eactual = +2.123V ln
= +2.203V
C
2×96485 mol
( 1.25atm)
Voltage higher
than standard!
∥
Pt( s) | Cl2( g) | Cl(- 0.0500M) Zn2+(0.0500M)
| Zn
( s)
Electrochemistry Example Problems
More oxidizing
2. Consider the following half-reactions:
Half-Reaction
E (V)
Ce 4+ (aq) + e –
 Ce 3+ (aq)
+ 1.61
Ag + (aq) + e –
 Ag(s)
+ 0.80
Hg2 2+ (aq) + 2e –
 2 Hg(l)
+ 0.79
Sn 2+ (aq) + 2 e –
 Sn(s)
– 0.14
Ni 2+ (aq) + 2 e –
 Ni(s)
– 0.25
Al 3+ (aq) + 3 e –
 Al(s)
– 1.66
(a) Which is the weakest oxidizing agent in the list?
(b) Which is the strongest oxidizing agent?
(c) Which is the strongest reducing agent?
(d) Which is the weakest reducing agent?
(e) Does Sn(s) reduce Ag + (aq) to Ag(s)?
(f) Does Hg(l) reduce Sn 2+ (aq) to Sn(s)?
(g) Name the ions that can be reduced by Sn(s).
(h) What metals can be oxidized by Ag + (aq)?
More reducing
Al3+
Ce4+
Al
Ce3+
+0.80 V + 0.25 V = +0.94 V: YES
-0.79 V + -0.14 V = -0.93 V: NO
Hg22+, Ag+, Ce4+
Hg, Sn, Ni, Al
Electrochemistry Example Problems
3. Calculate equilibrium constant for the following reaction:
Zn 2+ (aq) + Ni(s)  Zn(s) + Ni 2+ (aq)
Locate in tables:
Zn 2+ (aq) + 2 e - (aq)  Zn(s)
Ni(s)  Ni 2+ (aq) + 2 e - (aq)
-0.763 V
+0.25 V (reversed, oxidation)
So Ecell = -0.763 + 0.25 = -0.51 V (two sig. fig. from the subtraction)
We use the equation:
K=e
nF o
lnK =
E
RT
which becomes:


2×96485C mol
×-0.51V


J
8.314
×298.15K
Kmol


K=e
nF o
E
RT
= 5.7×10
-18
Commercial Voltaic Cells and Storage Batteries
Strictly speaking, a “battery” refers to an assembly, or battery, of 2 or more individual voltaic cells.
In this sense, a typical flashlight battery is misnamed: it is really just a single cell.
If cells are arranged in series their voltages add together.
Thus for example a 12 V car battery is an assembly of six cells producing about 2 V each.
A 9 V radio battery is also an assembly of 6 cells, each delivering about 1.5 V.
Commercial voltaic cells are divided into two main categories:
i) primary cells which are essentially the throw-away single use type, and
ii) secondary cells which are capable of multiple recharges.
In charging mode, these cells function as electrolysis cells.
Nowadays there are many different cell technologies in the marketplace.
Let us consider just a few examples of both categories of cells
We will first look at several primary cells, then after consider a few secondary cells
Some typical Primary and Secondary Cells
Battery
Primary Battery
and cell (throwaway)
Battery
Secondary Battery
and cell (rechargeable)
Battery
Battery
Cell?
Cell
Battery
Cells
Cells
Cells
Zinc-carbon cell (Leclenché cell)
CATEGORY:
(Zn/MnO2 in NH4Cl)
Primary (Throwaway) Zinc Family
CONSTRUCTION:
The Leclanché dry cell or ordinary "flashlight battery." The anode is
a zinc container covered with an insulating wrapper, the cathode is
the carbon rod at the center. The electrolyte paste contains ZnCl2,
NH4Cl, MnO2, starch, graphite, and water. A fresh dry cell delivers
about 1.5 V.
REDOX REACTIONS
Nominal cell voltage = +1.5 V
Positive terminal:
2MnO2(s) + 2NH4+(aq) + 2e–  Mn2O3(s) + 2NH3(aq) + H2O (l) 0.74 V
Negative terminal:
Zn (s) ––––> Zn2+ (aq) + 2e–
0.76 V
COMMENTS
The original "flashlight battery" was developed in 1865 by Georges
Leclanché. It is often called a dry cell because the electrolyte is
incorporated into a non-flowing paste.
Alkaline (Zinc/MnO2 in KOH)
CATEGORY:
Primary (Throwaway) Zinc Family
(+)
CONSTRUCTION:
The alkaline dry cell. The anode is a paste of zinc,
KOH, and water, which donates electrons to the
cell base via a brass collector. The cathode is a Outer steel jacket
paste of MnO2, graphite, and water, which takes Plastic sleeve
electrons from the inner steel case. A plastic
sleeve separates the inner steel case from the Inner steel jacket
outer steel jacket.
REDOX REACTIONS Nominal cell voltage = +1.5 V
Positive terminal:
2MnO2(s) + H2O(l) + 2e–  Mn2O3(s) + 2OH-(aq)
0.15 V
Negative terminal:
Zn (s) + 2OH-(aq)  Zn2+ (aq) + H2O(l) + 2e–
1.25 V
COMMENTS
The alkaline dry cell is more expensive than the Leclanché cell, but
it is also more efficient. Again, zinc is the anode and manganese
dioxide the oxidizing agent. The electrolyte is 40% KOH saturated
with zinc oxide (ZnO).
Cathode(+): paste containing
MnO2, graphite, and water
Anode(-): Paste containing
powdered zinc, KOH, and water
Brass collector
(-)
Cell base
Alkaline Button Cell (Zinc/MnO2 in KOH)
CATEGORY:
Primary (Throwaway) Zinc Family
CONSTRUCTION:
The alkaline dry cell. The anode is a paste of zinc,
KOH, and water, which donates electrons to the
cell base via a brass collector. The cathode is a
paste of MnO2, graphite, and water, which takes
electrons from the inner steel case. A plastic
sleeve separates the inner steel case from the
outer steel jacket.
REDOX REACTIONS Nominal cell voltage = +1.5 V
Positive terminal:
2MnO2(s) + H2O(l) + 2e–  Mn2O3(s) + 2OH-(aq)
0.15 V
Negative terminal:
Zn (s) + 2OH-(aq)  Zn2+ (aq) + H2O(l) + 2e–
1.25 V
COMMENTS
Apart from the different type of container, the chemistry of this cell is
the same as that of the standard alkali dry cell.
Lithium (Li/MnO2 in KOH)
CATEGORY:
Primary (Throwaway) Lithium Family
CONSTRUCTION:
Very similar to alkaline cell in design, except much more MnO2
paste is used compared to Li due to the light weight of Lithium
metal.
REDOX REACTIONS Nominal cell voltage = +3.0 V
Positive terminal:
2MnO2(s) + H2O(l) + 2e–  Mn2O3(s) + 2OH-(aq)
0.15 V
Negative terminal:
Li (s)  Li+ (aq) + –
3.04 V
(+)
Outer casing
Positive terminal
MnO 2 , graphite
KoH, and water
Absorbent fabric
sat'd in KOH and
moistened
Wire mesh cont.
lithium metal
COMMENTS
The electrolyte consists of between 20 and 40% by mass of KOH,
Inner steel
which is impregnated on the absorbent material between the two
jacket
half-cells. This is only one of a whole family of Li based batters.
Others are:
Li/SO2, Li/SOCl2, Li/CuO, Li-poly(vinyl pyridine)/I2 solid electrolyte.
Thus the lithium batteries can replace the zinc family of batteries
with a whole new range of high-power, low mass cells. The next 1020 years should see substantial development in this area.
(-)
Mercury (Zn/HgO in KOH)
CATEGORY: Primary (Throwaway) Zinc Family
CONSTRUCTION:
A mercury battery. A zinc-mercury
amalgam is the anode; the cathode
is a paste of HgO, graphite, and
water, Mercury batteries, some of
which are smaller than a pencil
eraser, deliver about 1.34 V.
REDOX REACTIONS
Positive terminal:
HgO (s) + H2O(l) + 2e–  Hg(l) + 2OH-(aq)
Negative terminal:
Zn (s) + 2OH-(aq)  ZnO(s) + H2O(l) + 2e–
COMMENTS
0.09 V
1.25 V
The mercury cell, developed in 1942, is another zinc dry cell devised
for use in small appliances such as watches, and us usually
produced as a button battery. It delivers only 1.34 V, which is
significantly less than the 1.5 V of a standard dry-cell. It has the
advantage of maintaining a fairly constant voltage during its lifetime.
Nominal cell voltage = +1.34 V
Silver oxide (Zn/Ag2O in KOH)
CATEGORY:
Primary (Throwaway) Zinc Family
CONSTRUCTION:
A silver oxide button battery, similar to the mercury and alkaline
manganese cell. Also a reliable source of voltage. Used in medical
devices, calculators, older cameras.
REDOX REACTIONS
Positive terminal:
Ag2O (s) + H2O(l) + 2e–  2 Ag(s) + 2OH-(aq)
Negative terminal:
Zn (s) + 2OH-(aq)  ZnO(s) + H2O(l) + 2e–
0.34 V
1.25 V
COMMENTS
Because no solution species is involved in the net reaction, the
quantity of electrolyte is very small, and the electrodes can be
maintained very close together. The storage capacity of a silverzinc cell is about six times as great as a lead-acid cell of the same
size. They are expensive because of the use of silver oxide.
Nominal cell voltage = +1.5 V
Zinc-Air (Zn/O2 in KOH)
CATEGORY:
Primary (Throwaway) Zinc Family
CONSTRUCTION:
Constructions very similar to
mercury or silver oxide button
cells. But there are holes in
the base to allow air in!
REDOX REACTIONS
Positive terminal:
½ O2 (s) + H2O(l) + 2e–  2OH-(aq)
Negative terminal:
Zn (s) + 2OH-(aq)  ZnO(s) + H2O(l) + 2e–
COMMENTS
0.40 V
1.25 V
Mercury is added to the zinc, forming a liquid mixture called an
amalgam, which prevents the formation of an insulating layer of zinc
oxide. The zinc-air cell is sold with a patch covering the air holes.
This patch must be removed before the battery develops a voltage.
When oxygen gas from the air percolates into the hydroxide solution in
the positive half-cell, reaction commences and a voltage develops in
the cell. Can you see why the voltage is less than predicted by E?
Nominal cell voltage = +1.4 V
p(O2) ≈ 0.20 atm
Lead Acid (PbO2/Pb)
CATEGORY:
Secondary/Rechargeable
CONSTRUCTION:
Shown in the picture is a 6-V (i.e. motorcycle) battery, which
contains 3 individual cells in series. The anode (negative terminal)
consists of a lead grid filled with spongy lead, and the cathode
(positive terminal) is a lead grid filled with lead dioxide. The cell
also contains 38% (by mass) sulfuric acid.
REDOX REACTIONS
Nominal cell voltage = +2.0 V
Positive terminal:
PbO2(s) + HSO4-(aq) + 3H+ + 2e–  PbSO4(s) + 2H2O (l)
1.69 V
Negative terminal:
Pb (s) + HSO4-(aq)  PbSO4(s) + H+ + 2e–
0.34 V
COMMENTS
The cell equations shows that discharging depletes the sulfuric acid
electrolyte and deposits solid lead sulfate on both electrodes. The
electrolyte density can be measured by a hydrometer; it is charged
when d = 1.28 (37% H2SO4) and discharged when it is below 1.15
g/mL (21% H2SO4). The lead acid battery is extremely reliable, but
heavy. Electrical cars using such batteries are too inefficient.
How many cells are there in
a 12 V automotive battery?
How can you tell?
Nickel-Cadmium (Nicad)
CATEGORY:
Secondary/Rechargeable
CONSTRUCTION:
The Nicad cell consists of many layers of alternating Ni and Cd
electrodes. The actual electrodes consist of steel foil with lots of
holes in them, onto which the active components are smeared.
Each electrode is surrounded by an absorbent pad which contains
the electrolyte. The nickel hydroxide electrodes are slightly longer
at the top, and are soldered to a collector along the top of the cell;
the cadmium electrodes are similarly joined along the bottom.
REDOX REACTIONS
Nominal cell voltage = +1.2 V
Positive terminal:
NiO(OH)(s) + H2O(l) + e–  NiO2(s) + OH-(aq) 0.82 V
Negative terminal:
Cd (s) + 2 OH-(aq  Cd(OH)2 (s)
0.49 V
COMMENTS
The nickel-cadium cell, for example, is used in hand calculators,
electronic flash units, and other small battery-operated devices. The
nickel-cadmium cell has electrodes of cadmium and NiO(OH), and
the electrolyte is a 20% aqueous KOH solution.
Positive terminal
collector
Concentric rings of
mesh covered alternatively
with NiO(OH) and Cd
Negative terminal
collector
Electrolysis Reactions
So far, voltaic cells, which are product favoured reactions doing useful work
Many electrochemical reactions are truly reversible, and can be driven against the spontaneous
direction by applying a greater voltage of opposite polarity
Driven electrochemical reactions are called electrolytic reactions, and the process is know as
electrolysis.
Important industrial electrolytic reactions:
Aluminum
Magnesium
Sodium
Many of the rarer metals
Electro-refining of copper depends
in part on electrolysis
Elemental chlorine
Sodium hydroxide
Electroplating technologies,
e.g. chrome plating
The aluminum refining industry is of
great significance in Canada
Electro-Refining of Copper
Electrolytic Production of Magnesium
Counting Electrons
It is both desirable and straightforward to relate the electrical current in an electrochemical process
to the quantity of materials produced and/or consumed during an electrochemical reaction.
This is best understood as a branch of stoichiometry in which the number of moles of electrons
is used as a stoichiometric factor.
The basis of this is the Faraday constant – which we have already dealt with – and the equation
relating current to charge from basic physics.
This equation is:
I=
q
t
in which I is the current in Amperes, q is the charge in
Coulombs and t is the time in seconds.
Note that these precise units must be used.
Rearrangment allows the calculation of the number of Coulombs of electrical charge that is
consumed in any given process.
Then we remember that Faraday’s constant is 96 485 C/mol of electrons
Once we have the number of moles of electrons, and knowing the value of n in the balanced redox
reaction, the whole problem boils down to simple stoichiometry.
Counting Electron Example Problems
1. If you electrolyze a solution of Ni 2+ (aq) to form Ni(s), and use a current of 0.15 amp for 10
min, how many grams of Ni(s) is produced?
Ni2+ (aq) + 2 e-  Ni (s)
n=2
q
I = ; ⇒ q = I× t
t
q = 0.15amp ×10min×60 s min = 90C
90C
#mol e =
= 9.327×10-4 mol
96485 C mol
-
1
#mol Ni = #mol e - = 0.5×9.327×10 -4 mol = 4.66×10 -4 mol
2
#g Ni = 4.66×10-4 mol×58.69 g mol = 0.027g
Counting Electron Example Problems
2. An electrolysis cell for aluminum production operates at 5.0 V and a current of 1.0  10 5
amp. Calculate the number of kilowatt-hours of energy required to produce 1 tonne
(1.0  10 3 kg) of aluminum.
1 tonne = 1.0 x 106 g
Al3+ (aq) + 3 e-  Al (s)  n = 3
1.0×106 mol
4
#mol Al =
=
3.706×10
mol
g
26.9815 mol
#mol e- = 3× #mol Al = 3×3.706×104 mol =1.11×105 mol
#C e- =1.11×105 mol×96485 C mol =1.07×1010 C
J = C× V; W = J×s; 1kWh = 3.60×106 J
1.07×1010 C×5.0V
4
#kWh =
=1.5×10
kWh
6 J
3.60×10 kWh
Important Concepts In Electrochemistry
Assigning oxidation states, redox reactions, etc.
Using half-reactions to balance redox reactions
Voltaic cells vs. electrolytic cells
Cell potential (and how it relates to Gibbs free energy)
Components of a voltaic cell
Electrodes (cathode and anode)
Salt bridge
Cell notation
Cells under nonstandard conditions (the Nernst equation)
Relationship between E˚ and equilibrium constant
Practical considerations for batteries, pH meters, etc.
Applications of electrolytic cells
Determining efficiency of electrolytic cells