Transcript Chapter 18 - Richsingiser.com

Daniel L. Reger
Scott R. Goode
David W. Ball
www.cengage.com/chemistry/reger
Chapter 18
Electrochemistry
Oxidation and Reduction
• Oxidation is the loss of electrons by a
chemical process.
• When sodium forms a compound, Na+ is
formed. Sodium is oxidized.
• Reduction is the gain of electrons by a
chemical process.
• When Cl- ions are formed from elemental
chlorine, chlorine is reduced.
Oxidation-Reduction (“Redox”)
• An oxidation-reduction reaction, or
redox reaction, is one in which electrons
are transferred from one species to
another.
• In every redox reaction, at least one species
is oxidized and at least one species is
reduced.
• 2Na(s) + Cl2(g) → 2NaCl(s) is a redox
reaction because Na is oxidized and Cl is
reduced.
Oxidizing and Reducing Agents
• An oxidizing agent is the reactant that
accepts electrons, causing an oxidation to
occur.
• The oxidizing agent is reduced.
• A reducing agent is the reactant that
supplies electrons, causing a reduction to
occur.
• The reducing agent is oxidized.
• In the reaction of sodium with chlorine, Na is
the reducing agent and Cl2 is the oxidizing
agent.
Half-reactions
• In a half-reaction, either the oxidation or
reduction part of a redox reaction is
given, showing the electrons explicitly.
• Half-reactions emphasize the transfer of
electrons in a redox reaction.
• For 2Na(s) + Cl2(g) → 2NaCl(s) :
• Na → Na+ + 1e• Cl2 + 2e- → 2Cl-
oxidation half-reaction
reduction half-reaction
Oxidation States
• The oxidation state is the charge on the
monatomic ion, or the charge on an atom
when the shared electrons are assigned
to the more electronegative atom.
• Electron pairs shared by atoms of the same
element are divided equally.
• In CaCl2, an ionic compound:
• calcium has an oxidation state of +2.
• chlorine has an oxidation state of -1.
Review
• Assign oxidation numbers to each atom
in the following substances.
(a) PF3
(b) CO (c) NH4Cl
Review
• Balance the following equation in acid
solution:
• Cr2O72- + C2H5OH → Cr3+ + CO2
Review
• Balance the following equation in basic
solution:
• Zn + ClO- → Zn(OH)42- + Cl-
Voltaic Cells
• A voltaic cell (also known as a
galvanic cell) is an apparatus that
produces electrical energy directly from
a redox reaction.
• All voltaic cells depend on redox
reactions.
Voltaic Cell - Diagram
Voltaic Cells
• The oxidation half-cell contains the
reaction Zn(s) → Zn2+(aq) + 2e-.
• The reduction half-cell contains the
reaction Cu2+(aq) + 2e- → Cu(s).
• Since the half-cells are physically
separated, electrons must travel from one
side to another through a connecting
wire.
Voltaic Cell Conventions
• Assume that oxidation occurs in the
left half-cell and reduction occurs in
the right half-cell.
• Zn(s) → Zn2+(aq) + 2e- occurs in left.
• Cu2+(aq) + 2e- → Cu(s) occurs in right.
• A voltmeter is connected to the wire
coming out of each metal electrode.
• If the measured voltage is positive, the
chemical reaction is spontaneous as
written.
Electrodes and Half-Cells
• Redox reactions not involving metals can
be used in half-cells by using an inert
electrode, like gold, platinum, or carbon
to provide electrical contact. Examples of
half-reactions include:
• two soluble ions
• Fe3+(aq) + e- → Fe2+(aq)
• gas-ion reactions
• Cl2(g) + e- → 2Cl-(aq)
• insoluble salts
• AgCl(s) + e- → Ag(s) + Cl-(aq)
Example of Inert-Electrode Half-Cell
The Hydrogen Electrode
Electrical Potential
• Electromotive force or emf (E) is the
electrical driving force that “pushes”
electrons from the oxidation half-cell to
the reduction half-cell.
• Cell potential is the potential energy per
unit charge that is characteristic of each
half-cell reaction. It is measured in volts
(V).
• 1 volt = 1 joule/coulomb (1 V = 1 J/C)
Standard Potentials
• The overall potential of a cell depends on
the concentrations of the species in the
reaction.
• Standard potential, Ecell, is the cell
potential when each species in the
reaction is present in its standard state.
• Solids, liquids, and gases are in their pure
state at 1 atm pressure.
• Solutes have 1 M concentration.
Potentials of Voltaic Cells
• For the reaction
Zn(s) + Cu2+(aq) → Cu(s) + Zn2+(aq)
the Ecell is +1.10 V.
• A positive cell potential means that the
reaction is spontaneous in the forward
direction.
Additivity of Cell Potentials
• Cell potentials are additive:
Cu(s) + 2Ag+(1 M) → 2Ag(s) + Cu2+(1 M)
Zn(s) + Cu2+(1 M) → Zn2+(1 M) + Cu(s)
+0.46 V
+1.10 V
Zn(s) + 2Ag+(1 M) → 2Ag(s) + Zn2+(1 M)
E°cell =+1.10 V + 0.46 V = +1.56 V
Standard Reduction Potential
• The standard reduction potential of a
half-reaction is the potential of the
reduction reaction relative to the standard
hydrogen electrode as the oxidation.
Standard Reduction Potentials Table
• Half-reactions written as reductions,
where electrons are reactants.
• Standard reduction potential for
2H+(aq,1 M) + 2e- → H2(g, 1 atm)
is set to 0.000 V.
• If a reduction reaction is reversed to
make oxidation, change sign on
reduction potential.
Ag+(aq, 1 M) + e- → Ag(s) E = 0.80 V
Ag(s) → Ag+(aq, 1 M) + e- E =-0.80 V
Standard Reduction Potentials Table
Reduction Half-Reaction
F2(g) + 2e- → 2F-(aq)
Ag+(aq) + e- → Ag(s)
Fe3+(aq) + e- → Fe2+(aq)
Sn4+(aq) + 2e- → Sn2+(aq)
2H+(aq) + 2e- → H2(g)
Co2+(aq) + 2e- → Co(s)
Fe2+(aq) + 2e- → Fe(s)
Zn2+(aq) + 2e- → Zn(s)
Mg2+(aq) + 2e- → Mg(s)
E (V)
2.87
0.80
0.77
0.15
0.00
-0.28
-0.44
-0.76
-2.37
Calculating Cell Potential
• The standard potential of a cell
reaction is given by
Ecell = Ered + Eox
• Both half-reactions must transfer the
same number of electrons.
• Multiplying the coefficients of a halfreaction to balance the electrons does
NOT change the potential.
Fe3+(aq) + e- → Fe2+(aq)
E = 0.77 V
2Fe3+(aq) + 2e- → 2Fe2+(aq) E = 0.77 V
Example: Using Standard Potentials
• Calculate the potential of the cell
reaction
2Fe3+ + 2I- → 2Fe2+ + I2
from the potentials in Table 18.1.
Test Your Skill
• Write the spontaneous cell reaction and
calculate Ecell for the voltaic cell made
up from the two half-reactions below.
Sn4+(aq) + 2e- → Sn2+(aq)
Mg2+(aq) + 2e- → Mg(s)
E = 0.15 V
E = -2.37 V
Test Your Skill
• Write the spontaneous cell reaction and
calculate Ecell for the voltaic cell made
up from the two half-reactions below.
Sn4+(aq) + 2e- → Sn2+(aq)
Mg2+(aq) + 2e- → Mg(s)
E = 0.15 V
E = -2.37 V
• Answer:
Mg(s) + Sn4+(aq) → Mg2+(aq) + Sn2+(aq)
Ecell = 2.52 V
Activity Series
• An activity series arranges halfreactions in order of decreasing potential.
Use of Activity Series
• Species with large positive reduction
potentials are oxidizing agents – they
oxidize species below them in the activity
series.
• Species with large negative reduction
potentials are reducing agents – they
reduce species above them in the
activity series.
Activity Series
• What happens when Fe is added to 1 M
solutions of Zn(NO3)2 and Co(NO3)2?
Co2+(aq) + 2e- → Co(s)
Fe2+(aq) + 2e- → Fe(s)
Zn2+(aq) + 2e- → Zn(s)
E = -0.28 V
E = -0.44 V
E = -0.76 V
• Fe reacts with Co2+ but not with Zn2+.
Test Your Skill
• From the data in Table 18.1, select a
metal that reduces Ag+(aq) but does
not reduce AgCl(s). Write the reaction
and calculate its standard potential.
Cell Potentials and DG
• DG = -nFE
where DG = free energy change;
n = number of electrons transferred;
F = Faraday constant, 96,485 C/mol e-;
E = cell potential.
Cell Potentials and DG
• Under standard conditions, DG = -nFE.
• This means that spontaneous reactions
have positive cell potentials.
Calculating DG
• Calculate DG for the reaction below.
AgCl(s) → Ag+(aq) + Cl-(aq)
using the following information:
AgCl(s) + e  Ag(s) + Cl (aq)
+
Ag(s)  Ag (aq) + e
-
-
E° = 0.222 V
E° = -0.80 V
Test Your Skill
• Use the data in Appendix H to calculate
DG for the following reaction.
2Na(s) + 2H2O(l) → 2Na+(aq) + 2OH-(aq) + H2(g)
Relation of E to Keq
• DG° = -RT ln Keq
• DG° = -nFE°
• -nFE° = -RT ln Keq
• E° =
RT
nF
ln Keq =
• At 298 K, E° =
2 .3 0 3 R T
nF
0 .0 5 9 1
n
log Keq
log Keq
Calculating Keq from E
• Determine Keq for the following reaction.
Fe(s) + Pb2+(aq) → Pb(s) + Fe2+(aq)
Test Your Skill
• Determine Keq for the reaction below at
25 C.
2Ag(s) + Ni2+(aq) → Ni(s) + 2Ag+(aq)
The Nernst Equation
• The Nernst equation is used to calculate
cell potentials under non-standard
conditions:
E ce ll  E ce ll 
2 .3 0 3 R T
nF
lo g Q  E ce ll 
0 .0 5 9 1
lo g Q
n
where Q is the reaction quotient. The
second expression is for 25 C only.
The Nernst Equation
• What is the cell potential for a cell
composed of Zn, 0.500 M Zn(NO3)2, Cu,
and 1.500 M Cu(NO3)2?
Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
E = 1.10 V
Test Your Skill
• What is the cell potential for a cell
composed of Cu, 1.250 M Cu(NO3)2, Fe,
and 0.100 M FeSO4?
Applications of Voltaic Cells
• Measurement of species concentration:
the pH meter.
• The half-cell potential depends on the
concentration of hydrogen ions in
solution.
• A pH electrode is an electrochemical
cell with a known reference cell as a
part.
• A pH meter is simply a volt meter
calibrated to display pH instead of volts.
Applications: Batteries
• All batteries are voltaic cells.
• In a dry cell, a Zn case is the anode,
which is in contact with a moist paste of
MnO2, NH4Cl, and a carbon electrode.
• Cell reaction:
Zn(s) + 2NH4+(aq) + MnO2(s) →
Zn2+(aq) + Mn2O3(s) + 2NH3(aq) + H2O(l)
Dry Cell
Alkaline Dry Cell
•
•
•
•
NH4Cl is replaced with KOH or NaOH.
Zn anode corrodes less.
Voltage is more constant.
Overall cell reaction:
Zn(s) + MnO2(s) → ZnO(s) + Mn2O3(s)
Alkaline Dry Cell
Lead Storage Battery
• Used for high current applications.
• Overall reaction:
Pb(s) + PbO2(s) + 4H+ + 2SO42- →
2PbSO4(s) + 2H2O(l)
• No salt bridge needed, because
oxidizing and reducing agents do not
come into contact with each other.
• Rechargeable as long as the PbSO4
product adheres to electrodes (not
limitless).
Lead Storage Battery
Fuel Cells
• A fuel cell is a voltaic cell in which
reactants are supplied continuously,
products are removed continuously, and
electrical energy is produced from the
chemical reaction.
• The H2/O2 fuel cell uses the reaction
2H2(g) + O2(g) → 2H2O(l)
and is used to produce electricity and
water on space missions.