Chapter 19:
Electrochemistry
Chapter Outline
Section 19.1 Redox Reactions
Section 19.2 Balancing Redox Equations
Section 19.4 Voltaic Cells
Section 19.5 Cell Potential
Section 19.6 Free Energy and Cell Potential
Section 19.7 The Nernst Equation and Concentration Cells
Section 19.8 Voltaic Cell Applications: Batteries, Fuel Cells, and Corrosion
Section 19.9 Electrolytic Cells and Applications of Electrolysis
You are not responsible for knowing section 19.3 Redox Titrations
19.1 Redox Reactions
• Recall that a redox reaction is a chemical reaction that involves the transfer of electrons from
one reactant to another
• The loss of electrons is oxidation (LEO) and the gain of electrons is reduction (GER)
• Oxidation corresponds to an increase in oxidation number whereas reduction is a decrease in
oxidation number. Refer to chapter 4 for rules for assigning oxidation numbers.
0
0
4 Fe(s) + 3 O2(g)
-2
+3
2 Fe2O3(s)
19.2 Balancing Redox Reactions
• A redox reaction can be separated into two ____________________:
2 Al + 3 Cl2 → 2 AlCl3
Oxidation half-reaction: ____________________
Reduction half-reaction: ___________________
• Balancing a redox reaction involves balancing the mass and charge in each half-reaction.
• pH is important to consider when balancing redox reactions
Acidic solution
Basic solution
Balance for hydrogen by
adding ________
Add _________ ions to
balance ______ ions
H2O can be added in both
acidic and basic solutions
to balance hydrogen and
oxygen atoms
19.2 Balancing Redox Reactions
Example - Balance this reaction below in acidic solution.
Cr2O72−(aq) + Sn2+(aq) → Cr3+(aq) + SnO2(s)
(+6)(-2)
(+2)
(+3)
(+4)(-2)
Step 1: Assign oxidation numbers to see which atoms are oxidized and which are reduced. Separate the
redox reaction into the two half reactions.
Reduction: Cr2O72−(aq) → Cr3+(aq)
Oxidation: Sn2+(aq) → SnO2(s)
Step 2: Add water molecules to balance oxygen in both reactions
Cr2O72−(aq) → Cr3+(aq) + ________
_________ + Sn2+(aq) → SnO2(s)
Step 3: Add H+ to balance hydrogen
_________+ Cr2O72−(aq) → 2 Cr3+(aq) + 7 H2O(l)
2 H2O(l) + Sn2+(aq) → SnO2(s) + _________
Step 4: Add electrons to balance the overall charge
______ + 14 H+(aq) + Cr2O72−(aq) → 2 Cr3+(aq) + 7 H2O(l)
2 H2O(l) + Sn2+(aq) → SnO2(s) + 4 H+(aq) + ______
Step 5: Multiply one or both half-reactions by a small integer to balance the electrons.
[2 H2O(l) + Sn2+(aq) → SnO2(s) + 4 H+(aq) + 2 e−] ×
_____
____H2O(l) + ____ Sn2+(aq) → _____ SnO2(s) + ____ H+(aq) + ____ e−
Step 6: Add the half-reactions, cancelling any species that appear on both sides
2
1
6 e− + 14 H+(aq) +Cr2O72− (aq) → 2 Cr3+(aq) + 7 H2O(l)
6 H2O(l) + 3 Sn2+(aq) → 3 SnO2(s) + 12 H+(aq ) + 6 e−
2 H+(aq) +Cr2O72− (aq) + 3 Sn2+(aq) → 2 Cr3+(aq) +H2O(l) + 3 SnO2(s)
Step 7: Verify that all atoms and charges are balanced
Reactant side:
Product side:
H: _____
Cr: _____
O: ______
Sn: ______
H: _____
Cr: _____
O: ______
Sn: _____
Charge: ______
Charge: ______
19.4 Voltaic Cells
The Daniell cell is a common type of voltaic cell and consist of a
redox reaction between copper and zinc:
Reduction: Cu2+(aq) + 2 e− → Cu(s) (cathode)
• There are two main types of
electrochemical cells:
1. Voltaic Cells - consists of ____________
chemical reactions to generate
electricity
2. Electrolytic Cells – use electricity to
drive a ______________ chemical reaction
Oxidation: Zn(s) → Zn2+(aq) + 2 e− (anode)
Electrons flow from the
_________ to the __________
A ___________ is used to
maintain charge balance
19.4 Voltaic Cells – Standard Cell Notation
Anode half-reaction:
Zn(s)
Zn2+(aq) + 2 e−
Cathode half-reaction:
Cu2+(aq) + 2 e−
Overall cell reaction:
Zn(s) + Cu2+(aq)
Cu(s)
Zn2+(aq) + Cu(s)
Salt bridge
Anode half-cell
Cathode half-cell
Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s)
Electron flow
Phase boundary
Phase boundary
19.5 Cell Potential
• The electromotive force associated with the flow
of electrons is measured as the ___________________
• Cell potentials for half-reactions are measured
relative to a reference cell known as the standard
hydrogen electrode, SHE, set to 0.00 V with
respect to the half-reaction.
• Standard reduction potentials, Eored of various
half-cell reactions are measure with the SHE are
the other half-cell
• Standard conditions: _____________________________
SHE
Table 19.1
Selected Standard
Reduction Potentials
(25℃)
A more ____________
value Eo red means
that the species is
more likely to be
______________
Half-Reaction
o
E red
(V )
F2(g) + 2 e−→ 2 F−(aq)
2.87
Co3+(aq) + e−→ Co2+(aq)
1.92
MnO4−(aq) + 8 H+(aq) + 5 e−→ Mn2+(aq) + 4 H2O(l)
1.51
Pb4+(aq) + 2 e−→ Pb2+(aq)
1.46
Cl2(g) + 2 e−→2 Cl−(aq)
1.36
Cr2O72−(aq) + 14 H+(aq) + 6 e−→ 2 Cr3+(aq) + 7 H2O(l)
1.23
Ag+(aq) + e− → Ag(s)
0.80
Fe3+(aq) + e−→ Fe2+(aq)
0.77
Cu+(aq) + e− → Cu(s)
0.52
Cu2+(aq) + 2 e− → Cu(s)
0.34
Cu2+(aq) + e−→ Cu+(aq)
0.15
Sn4+(aq) + 2 e−→ Sn2+(aq)
0.15
2 H+(aq) + 2 e−→H2(g)
0.00 (SHE)
Pb2+(aq) + 2 e− → Pb(s)
−0.13
Sn2+(aq) + 2 e− → Sn(s)
−0.14
Ni2+(aq) + 2 e− → Ni(s)
−0.26
Cr3+(aq) + e−→ Cr2+(aq)
−0.41
Fe2+(aq) + 2 e− → Fe(s)
−0.44
Zn2+(aq) + 2 e− → Zn(s)
−0.76
2 H2O(l) + 2 e−→H2(g) + 2 OH−(aq)
−0.83
Al3+(aq) + 3 e− → Al(s)
−1.66
Mg2+(aq) + 2 e−→ Mg(s)
−2.38
Na+(aq) + e− → Na(s)
−2.71
19.5 Cell Potentials
• The standard cell potential, Eocell is calculated using the formula:
E cell = E cathode − E anode
Eo cathode: standard reduction
potential for the cathode halfreaction (reduction)
Eo anode: standard reduction
potential for the anode halfreaction (oxidation)
Example – Calculate the cell potential for the voltaic cells comprised of the sets of
half-cells.
A) Sn4+/Sn2+ and Cr3+/Cr2+
B) Al/Al3+ and Pb/Pb2+
A) Sn4+/Sn2+ and Cr3+/Cr2+
Look up the half-reactions in table 19.1
Sn4+(aq) + 2 e− → Sn2+(aq)
E = −0.15 V
Cr3+(aq) + e− → Cr2+(aq)
E = −0.41 V
Since the half-reaction with Sn has a more positive value of Eo , this reaction occurs at the
cathode
E cell = E cathode − E anode
E cell = 0.15 V − ( −0.41 V)= 0.56 V
B) Al/Al+3 and Pb/Pb2+
Look up the half-reactions in table 19.1
Al3+(aq) + 3 e− → Al(s)
E = −1.66 V
Pb2+(aq) + 2e− → Pb(s)
E = −0.13 V
Since the half-reaction with ________ has a more positive value of Eo , this reaction occurs at the
cathode
E cell = E cathode − E anode
∘
E𝑐𝑒𝑙𝑙
= ______________________
19.6 Free Energy and Cell Potential
• The cell potential (Ecell ) is related to the Gibbs free energy (∆G) through the formula:
G = −nFEcell
• In this equation, n is the number of ________________ transferred, and F is the Faraday
constant, F, which is the charge, in ____________, of a mole of electrons.
F = 96,485 C/mol e−
• Under standard conditions,
Recall
that
a
process
is
spontaneous when _________.
When ________, ∆G < 0. and the
reaction is _______________.
G = −nFEcell
19.7 The Nernst Equation and the Concentration of Cells
• Recall the relationship between ∆G and the reaction quotient, Q:
G = G + RT ln Q
• Combining this equation with G = −nFEcell yields the Nernst equation:
𝐸𝑐𝑒𝑙𝑙
RT
ln Q
𝑐𝑒𝑙𝑙 −
nF
= 𝐸𝑜
• The Nernst equation can be used to calculate the cell potential when solute concentrations are
some value other than ________.
19.7 The Nernst Equation and the Concentration of Cells
Example - Calculate the potential of a zinc/silver cell in which the zinc electrode is immersed in
0.100 M zinc nitrate and the silver electrode is in 0.500 M silver nitrate.
Step 1: Write the half-reactions, and determine Eo
∘
∘
∘
E𝑐𝑒𝑙𝑙
= E𝑐𝑎𝑡ℎ𝑜𝑑𝑒
− E𝑎𝑛𝑜𝑑𝑒
= ____________________________________
Step 1: Write the half-reactions, and determine n
Zn2+(aq) + 2 e− → Zn(s)
+
E = −0.76V
( Ag+(aq) + e− → Ag(s)
Zn + 2 Ag+ → Zn2+ + 2 Ag
E = +0.76V ) x 2
Since there are two moles of
electrons that are transferred in the
reaction, _________
Step 3: Calculate Q using the concentration of Zn+2
Step 4: Calculate Q using the concentration of Zn+2

E =E −
RT
nF
lnQ
19.8 Voltaic Cell Applications: Batteries and Fuel Cells
Lead storage car battery
•
Anode:
Pb(s) + SO42−(aq) → PbSO4(s) + 2 e−
•
Cathode:
MnO2(s) + H2O(l) + e−
→ MnO(OH)(s) + 2 OH−(aq)
Alkaline Dry Cell
• Anode:
Zn(s) + 2 OH−(aq)
→ Zn (OH)2(s) + 2 e−
• Cathode:
MnO2(s) + H2O(l) + e−
→ MnO(OH)(s) + 2 OH−(aq)
Hydrogen Fuel Cells
• Anode: H2(g) → 2 H+(aq) + 2 e−
• Cathode:
O2(g) + 4 H+(aq) + 4 e−
→ 2 H2O(l)
19.8 Voltaic Cell Applications: Corrosion
• The oxidation of metal structures is
known as _____________
• Corrosion can be prevented by adding a more
__________ to the iron surface. The more active
metal becomes the anode via _________________.
Anode (oxidation): Fe(s) → Fe2+(aq) + 2 e−
Cathode (reduction): O2(g) + 4 H3O+(aq) + 4 e−→ 6 H2O(l)
2 Fe2+(aq) + ½ O2(g) + 6 H2O+(l) → Fe2O3(s) + 4 H3O+(aq)
• Zinc is more active than iron (recall activity series)
and is used as a ______________________.
Oxidation: Zn(s) → Zn2+(aq) + 2 e−
Reduction: O2(g) + 4 H3O+(aq) + 4 e−→ 6 H2O(l)
19.9 Electrolytic Cells and Applications of Electrolysis
Electrolytic cells apply electrical currents to cause a nonspontaneous reaction to occur.
19.9 Electrolytic Cells and Applications of Electrolysis
• Electrolysis calculations are stoichiometric calculations based on the number of moles of electrons
passing through the electrolytic cell.
• The number of electrons available is determined by the current, which is measured in ____________, defined
as the amount of charge in coulombs per second, C/s.
1 A=
1C
and 1 C = 1 A  s
s
• 1 mol of electrons has a total charge of 96,485 C, the Faraday constant.
1 mol e− = 96,485 C
• The total charge (q) is calculated by multiplying current (I) by time (t).
q=It
19.9 Electrolytic Cells and Applications of Electrolysis
Example - Calculate the mass of copper metal that will be deposited by passing a 10.0 A current
through a solution of copper(II) sulfate for 3.00 h.
Step 1: Calculate the total charge, q
t = 3.00 h x 3600 s/1 h = 1.08 x 104 s
q=It
q = ____________________________
Step 2: Convert charge to moles of electrons using Faraday’s constant
Step 3: Convert moles of electrons to moles of copper
Cu2+(aq) + 2 e− → Cu(s)
Step 4: Convert moles of copper to mass of copper