Chapter 18
Electrochemistry
Hill, Petrucci, McCreary & Perry 4th Ed.
Oxidation Reduction Reactions
•Oxidation = Loss of electrons from a
chemical species = Increase in Oxidation
State.
Na
Na
+ 1 electron
a "+1" oxidation state
•Reduction = Gain of electrons by a
chemical species = Decrease in Oxidation
State.
Cl2 + 2 electrons
2 Cl
a "-1" oxidation state
Balancing RedOx Equations
The “Oxidation Number” Method
0
I2
a 5 electron loss per iodine
+
HNO3
+5
+5
2 HIO3
+
a 1 electron gain per nitrogen
NO2
+4
Can you balance this equation?
You must balance the electron transfer and the atoms.
This reaction occurs in aqueous acidic solution.
1
Balancing RedOx Equations
The “Oxidation Number” Method
0
I2
a 5 electron loss per iodine
+
+5
2 HIO3
HNO3
+
NO2
+4
a 1 electron gain per nitrogen
The atoms which change oxidation state are balanced first!
+5
The Nitrogen change must occur 10 times to change one
Iodine Molecule:
I2 + 10 HNO3
2 HIO3 + 10 NO2
Now we must balance the atoms which do not change
oxidation state!
I2 + 10 HNO3
2 HIO3 + 10 NO2 + 4 H2O
Being in acidic aqueous solution means we have lots
of H+ and H2O to balance hydrogens and oxygens!
“Rules” for Oxidation Number Method
1. Assign Oxidation Numbers.
2. Initially Balance atoms that change oxidation
state.
3. Determine the electron transfer for like
atoms.
4. Alter coefficients (only!) so electron transfer
balances for the entire equation.
5. Balance Hydrogen and Oxygen and other
atoms or ions which do not change oxidation
state.
Balancing RedOx Equations
“Half-Reaction” Method
0
Zn
a 2 electron loss per Zn
+
NO3-
+5
+2
Zn2+
+
NH4+
-3
an 8 electron gain per nitrogen
Balance this reaction by first dividing the reaction into an
oxidation half-reaction and a reduction half-reaction.
Using coefficients only, balance all atoms, using H2O and
H+/OH- to balance H & O, balance charge using electrons
in the oxidation half-reaction.
Balance all atoms and charge in the reduction half-reaction
the same way.
Multiply each balanced half-reaction by coefficients which
cause the electrons to cancel, add the half-reactions together
finally reducing any species occurring on both sides of arrow.
2
“Half-Reaction” Method
a 2 electron loss per Zn
0
+
Zn
NO3-
+5
+2
Zn2+
+
an 8 electron gain per nitrogen
NH4+
-3
Balance this reaction by first dividing the reaction into an
oxidation half-reaction and a reduction half-reaction.
Oxidation:
Zn
Reduction:
NO3- + 8 e
Zn2+ + 2 e
+ 10 H+
x4
NH4+ + 3 H2O
Using coefficients only, balance all atoms, using H2O and
H+/OH- to balance H & O, balance charge using electrons
in the oxidation half-reaction /the reduction half-reaction.
Multiply each half-reaction by coefficients to cancel electrons, add
the half-reactions together, reduce species occurring on both sides
4 Zn + NO3- + 10 H+
4 Zn2+ + NH4+ + 3 H2O
“Rules” for Half-Reaction Method
1. Assign Oxidation Numbers.
2. Separate into Oxidation and Reduction halfreactions.
3. Initially balance atoms that change oxidation
state by changing only coefficients.
4. Balance any spectator ions.
5. Balance Hydrogen & Oxygen: H+/H2O acidic;
H2O/OH- basic solution.
“Rules” for Half-Reaction Method
continued
6. Balance the charges on each half-reaction by
using electrons.
7. Multiply half-reactions by whole numbers to
balance the electron transfer.
8. Add the half-reactions together and cancel
species occurring on both sides of the
chemical equation.
3
Electrochemical Cells
An application of Oxidation/Reduction
Reactions
Voltaic Cell – Produces electric current by
means of a spontaneous RedOx reaction.
Electrolytic Cell – Uses an electric current to
cause a non-spontaneous reaction to occur.
A Voltaic Cell
Zn +
Zn2+ + - Cu
Cu2+
e-
e
salt bridge
Cu2+
Zn2+
Anode Compartment
Zn2+ + 2 e-
Zn
Oxidation Half-reaction
Cathode Compartment
Cu2+ + 2 e-
Cu
Reduction Half-reaction
Voltaic Cell Notation
ANODE
CATHODE
salt bridge
Anode
Terminal
Anode
Electrolyte
Cathode
Electrolyte
Cathode
Terminal
phase separator
Zn
Zn2+
Cu2+
Cu
4
Standard Hydrogen Electrode
Oxidation
Reduction
ANODE
CATHODE
salt bridge
Pt
Anode
Terminal
Anode
Electrolyte
Gas
salt bridge
Cathode
Electrolyte
phase separator
Pt
o
Half Reaction
H2(g)
2
phase separator
H+(aq)
H2(g)
Pt
Cathode
Terminal
Gas
Ered
0
H+(aq)
+ 2e-
H+(aq)
Half Reaction
H2(g)
Pt
o
0
Ered
2 H+(aq) + 2e-
H2(g)
Standard Reduction Potentials
Cathode Potentials
Table 18.1 p 762: emf in Volts vs SHE
Table Layout: Red. Half-Reactions vs SHE
Strongest Oxidizing Agents at Top
Strongest Reducing Agents at Bottom
For any pair of couples (half-reactions) the
Anodes will always be lower in the table in
the text.
Reduction Half-Reaction - Acidic Solution - Eo, volt
H2O2(aq) + 2 H+(aq) + 2 e- => 2 H2O(l)
Cl2(g) + 2 e- => 2 Cl-(aq)
O2(g) + 4 H+(aq) + 4 e- => 2 H2O(I)
Ag+(aq) + e- => Ag(s)
I2(s) + 2 e- => 2 I-(aq)
Cu2+(aq) + 2 e- => Cu(s)
2 H+(aq) + 2 e- => H2(g)
Pb2+(aq) + 2 e- => Pb(s)
Sn2+(aq) + 2 e- => Sn(s)
Zn2+(aq) + 2 e- => Zn(s)
Al3+(aq) + 3 e- => Al(s)
Mg2+(aq) + 2 e- => Mg(s)
Na+(aq) + e- => Na(s)
Li+(aq) + e- => Li(s)
+1.763
+ 1.358
+1.229
+0.800
+0.535
+0.340
0
-0.125
-0.137
-0.763
-1.676
-2.356
-2.713
-3.040
5
Cell Potentials
Eo
cell
= Eored cathode Higher in Table
Eored anode
Lower in Table
The Anode Reaction is the Reverse of the
reaction in the Table.
The Anode potential is opposite in sign of
that tabulated, hence the subtraction.
Zn2+(aq) + 2e-
Zn(s)
Zn2+(aq) + 2e-
Zn(s)
Eored = Eocathode = -0.76 V
Eoox = Eoanode = +0.76 V
Half-Reactions are commonly
represented by Reaction “Couples”
Half Reaction
Zn2+(aq)
Zn(s)
+ 2e-
Abbreviated "Couple"
Zn(s)
Zn2+(aq) Zn(s)
Zn2+(aq) + 2e-
Zn(s) Zn2+(aq)
The species Oxidized or Reduced are
shown in the couples in the same order
they occur in the reaction.
Reduction Half-Reaction - Acidic Solution – Couple
H2O2(aq) + 2 H+(aq) + 2 e- => 2 H2O(l)
Cl2(g) + 2 e- => 2 Cl-(aq)
O2(g) + 4 H+(aq) + 4 e- => 2 H2O(I)
Ag+(aq) + e- => Ag(s)
I2(s) + 2 e- => 2 I-(aq)
Cu2+(aq) + 2 e- => Cu(s)
2 H+(aq) + 2 e- => H2(g)
Pb2+(aq) + 2 e- => Pb(s)
Sn2+(aq) + 2 e- => Sn(s)
Zn2+(aq) + 2 e- => Zn(s)
Al3+(aq) + 3 e- => Al(s)
Mg2+(aq) + 2 e- => Mg(s)
Na+(aq) + e- => Na(s)
Li+(aq) + e- => Li(s)
H2O2(aq)/ H2O(l)
Cl2(g)/ Cl-(aq)
O2(g)/ H2O(I)
Ag+(aq)/ Ag(s)
I2(s)/ I-(aq)
Cu2+(aq)/Cu(s)
H+(aq)/ H2(g)
Pb2+(aq)/ Pb(s)
Sn2+(aq)/ Sn(s)
Zn2+(aq)/ Zn(s)
Al3+(aq)/ Al(s)
Mg2+(aq)/ Mg(s)
Na+(aq)/ Na(s)
Li+(aq)/ Li(s)
6
Gibb’s Free Energy & Cell Potential
o
∆G o = - n F Ecell
A positive Ecell
n = # electrons transferred
F = Faraday Constant
spontaneous
= 96,500 Coulombs/mole
A negative ∆G
"charge of a mole of electrons"
"one Faraday of charge"
o
A negative Ecell
non-spontaneous
A positive ∆G
o
Cell Potentials & the Equilibrium
Constant
o
∆G o = - n F Ecell = - R T ln K
Rearranging:
o
Ecell
=
R T ln K
nF
Nernst Equation:
An Ecell < 0.104 V
n
will give an equilibrium
o
mixture: ∆G < -+ 10 kJ
R T ln Q
nF
[product]
Q =
[reactant]
Zn2+ + Cu
[Zn2+]
Q =
[Cu2+]
o
Ecell = Ecell
-
For
Zn + Cu2+
Calculating Cell Potentials
Calculate the cell potential and write the cell notation
and net chemical equation for the half reactions
below.
I2(s) + 2 e- => 2 I-(aq)
+0.535
-0.137
Sn2+(aq) + 2 e- => Sn(s)
7
Calculating a Cell Potential -
Process Specified
Given the following reaction, determine the cell
potential and whether the reaction is spontaneous.
Write the cell notation for the reaction as written.
Sn(s) + Cu2+(aq) => Sn2+(aq) + Cu(s)
Electrolysis of Molten Salts
A molten salt is the simplest case.
Na+(l) + Cl-(l)
NaCl(s) melting
2 Na+(l) + 2 Cl-(l)
Anode:
Cathode:
electrolysis
2 Cl -(l)
2 Na(l) + Cl2(g)
Cl 2(g) + 2 e -
2 mol
2 Faradays
1 mol
Na+(l) + e-
Na(l)
1 mol
1 mol
1 Faraday
1 mole electrons = 1 Faraday Charge = 96,500 Coulombs .
1 ampere = 1 Coulomb/second
The amount of material produced depends directly
on the amount of current passed through the cell.
Some Salts can be Electrolysed in Water
CuBr 2(aq) :
Electrolysis Products
Cu 2+ (aq) + 2 Br - (aq)
Anode Rxn: 2 Br-(aq)
Cu (s) + Br
Br2(l) + 2 e-
Cathode Rxn: Cu2+(aq) + 2 e-
2(l)
- ( Ered = 1.07 V)
Cu(s)
Ered = 0.34 V
Ecell = - 0.73 V
An applied voltage > 0.73 V will cause:
the reduction of Copper ion to Copper metal
.
and the oxidation of Bromide ion to Bromine. .
8
How much copper metal can be plated by
a current of 5.0 amperes in 1 hour?
You will need to calculate the number of
Faradays delivered by this current and relate
that to the numbers of moles of copper.
9