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Electrochemistry
• Topics Covered
– Oxidation-reduction
reactions
– Balancing oxidationreduction equations
– Voltaic cells
– Cell EMF
– Spontaneity of redox
reactions
– Batteries
– Electrolysis
• Reading Quiz
– Read Chapter 20
– Know definitions for all
words in bold
– Table 20.1
– Sample exercises
20.1, 20.5, 20.6
Oxidation-Reduction Review
• Oxidation refers to the loss of electrons
(LEO)
• Reduction refers to the gain of electrons
(GER)
• Thus oxidation-reduction occurs when
electrons are transferred from the atom
that is oxidized to the atom that is reduced
Oxidation-Reduction Reactions
Oxidation-Reduction Reactions
• Oxidation-reduction reactions are often called
redox reactions
• By writing the oxidation number of each element
above or below the equation we can see the
oxidation state changes that occur
• In any redox reaction both oxidation and
reduction must occur
• The substance that is being reduced is
called the oxidizing agent or oxidant and
is gaining electrons in the reaction
• The substance being oxidized is called the
reducing agent or reductant and is
losing electrons to the oxidant
Sample Problem
Sample Problem
• The nickel-cadmium (nicad) battery uses the
following redox reaction to generate electricity:
Cd(s) + NiO2(s) + 2H2O(l) → Cd(OH)2(s) + Ni(OH)2(s)
• Identify the oxidant and reductant in the
following equation:
2H2O + Al + MnO4 → Al(OH)4- + MnO2(s)
• Identify the substances that are oxidized and
reduced, and indicate which are oxidizing agents
and which are reducing agents.
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Balancing Redox Reactions
Balancing by Half-Reaction
• When balancing reactions we must obey the law
of conservation of mass
• As we balance redox reactions we must also
balance the loss and gain of electrons
• It is often easiest to do this by separating a
reaction into oxidation and reduction halfreactions
• We can use half-reactions to balance
complicated redox reactions in acidic solution by
using the following procedure:
– Sn2+ + 2Fe3+ → Sn4+ + 2Fe2+
– Oxidation: Sn2+ → Sn4+ + 2e– Reduction: 2Fe3+ + 2e- → 2Fe2+
Balancing by Half-Reaction
– Multiply each half-reaction by an integer so
that the number of electrons lost in one
reaction equals the number of electrons
gained in the other
– Add the two half-reactions and simplify by
cancelling species that occur on both sides
– Check the equation to make sure that there
are the same number of atoms of each kind
and the same total charge on each side
Balancing Redox Reactions in
Basic Solution
• If a redox reaction occurs in basic solution,
the equation must be balanced using OHand H2O instead of H+ and H2O
• The half-reactions can be balanced initially
as if they occurred in acidic solution
• The H+ ions can then be “neutralized” by
adding an equal number OH- ions to each
side and canceling the resulting water
molecules
– Divide the equation into two incomplete half reactions,
one for oxidation and one for reduction
– Balance each half reaction
•
•
•
•
First balance the elements other than H and O
Next, balance the O atoms by adding H2O
Then balance the H atoms by adding H+
Finally, balance the charge by adding e-
Sample Problems
• Balance the following reactions in acidic
solution:
– MnO4- + C2O42- → Mn2+ + CO2
– Cr2O72- + Cl- → Cr3+ + Cl2
– Cu + NO3- → Cu2+ + NO2
Sample Problems
• Complete and balance the following reactions in
basic solution:
– CN- + MnO4– NO2- + Al
CNO- + MnO2
NH3 + Al(OH)4-
– Cr(OH)3 + ClO-
CrO42- + Cl2
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Voltaic Cells
Spontaneous Redox Reaction
• The energy released in a spontaneous redox
reaction can be used to perform electrical work
• This task is accomplished through a voltaic (or
galvanic) cell
• This device transfers electrons in a redox
reaction through an external pathway rather than
directly between reactants
• A strip of zinc metal is
immersed in an aqueous
copper sulfate solution
• The redox reaction takes
place at the metal-solution
interface and involves direct
transfer of two electrons from
Zn atoms to Cu2+ ions
• As time passes, a darkcolored deposit of copper
metal appears on the zinc,
and the blue color due to
Cu2+ fades from the solution
Voltaic Cell
Voltaic Cell
• In this set-up, the Zn metal and the Cu2+ ions are not
in direct contact
• Instead, the Zn is in direct contact with Zn 2+ and the
Cu is in contact with Cu2+ in another compartment
• The reduction of the Cu2+ can occur only through an
external circuit
Voltaic Cell
• The two solid metals that are connected by the external circuit
are called electrodes
• The electrode at which oxidation occurs is called the anode
• The electrode at which reduction occurs is called the cathode
• Each of the two compartments is called a half-cell
Voltaic Cell
• Anode (oxidation half-reaction):
– Zn(s)
Zn2+(aq) + 2e-
• Cathode (reduction half-reaction):
– Cu2+(aq) + 2e-
Cu(s)
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Voltaic Cell
• Electrons become available as zinc metal is oxidized
at the anode
• They flow through the external circuit to the cathode,
where they are consumed as Cu 2+ is reduced
Voltaic Cell
• For the cell to work, the solutions must remain neutral
• As Zn2+ ions enter solution, anions must flow from the
cathode to keep the solution electrically neutral
• As Cu2+ ions leave solution, cations must flow from
the anode
Voltaic Cell
• The zinc electrode loses mass, the zinc solution
becomes more concentrated
• The copper electrode gains mass and the copper
solution becomes more dilute and loses color
Voltaic Cell
• This migration of ions can be through a porous
material or (as in this case) through a salt bridge
• A salt bridge consists of a U-shaped tube that
contains an electrolyte solution whose ions will not
react with other ions in the cell
• The solution is often a gel so that it stays in the tube
Voltaic Cell
Voltaic Cell
• Whatever means is used to allow ions to migrate
between half-cells, anions always migrate towards
the anode and cations toward the cathode
• In any voltaic cell, the electrons flow from the anode
(labeled negative) to the cathode (labeled positive)
• The shorthand notation shows the anode half-cell on
the left of the salt bridge and the cathode half-cell on
the right
• A single vertical line indicates the phase boundary
between the metal and the metal ion solution
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Cell EMF
• The difference in potential energy per electric charge
between two electrodes in measured in volts (V)
– 1 V = 1 J/C
• The potential difference (voltage) between two
electrodes provides the driving force that pushes the
electrons through the external circuit
• We call this potential difference the electromotive force
or emf
• The emf of a cell, denoted Ecell, is also called the cell
potential or cell voltage
Standard Reduction Potentials
• The emf for a given volatic cell depend on the
particular cathode and anode used
• Because there are so many different types of
cells, it is easiest to calculate the emf from the
cell by knowing the standard reduction
potential, E°red, for each half-cell
• The standard cell potential can then be
calculated as follows:
– E°cell = E°red(cathode) - E°red(anode)
Cell EMF
• The emf of a given cell changes with
concentration and temperature
• Standard conditions for a cell are 25 C and 1 M
solution concentrations (1 atm pressure for
gases)
• Under these conditions, the emf is called the
standard emf or the standard cell potential
• Standard emf is denoted Eºcell
Standard Reduction Potentials
• Because it is impossible to measure E°red directly, a
reference point is needed for comparison
• The reference point used is the reduction of H +
– 2H+(aq, 1 M) + 2eH2(g, 1 atm)
– E°red = 0 V
• The electrode designed to produce this half-reaction is
called a standard hydrogen electrode (SHE)
• E°cell is the voltage of any cell produced when a SHE is
the anode
Sample Problem
• Calculate the standard cell potential of the
galvanic cell shown above
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Sample Problem
• Calculate the standard cell potential of the
galvanic cell shown above
Sample Problems
• Rank the following ions in order of
increasing strength as oxidizing agents:
NO3-, Ag+, Cr2O72-
Oxidizing and Reducing Agents
• The more positive the E°red value for a
half-reaction, the greater the tendency for
reactant of the half-reaction to be reduced
and, therefore, to oxidize another species
• The half reaction with the smallest
reduction potential is most easily reversed
as an oxidation
Spontaneity of Redox Reactions
• Any reaction that can occur in a voltaic cell to
produce a positive emf must be spontaneous
• In general, we can write:
– E = E red(reduction process) - E
process)
• Rank the following species from strongest
to weakest reducing agent: I-, Fe, Al
Sample Problem
• Determine whether or not the following
reactions are spontaneous under standard
conditions
– Cu(s) + 2H+(aq) → Cu2+(aq) + H2(g)
– Cl2(g) + 2I-(aq) → 2Cl-(aq) + I2(s)
red(oxidation
• A positive value of E indicates a spontaneous
process and a negative value of E indicates a
nonspontaneous one
EMF and Free Energy Change
• Gibbs free energy and emf of a redox reaction
are related in the following way:
– ΔG = -nFE or ΔG = -nFE (at standard conditions)
– n is a positive number that represents the number of
electrons transferred in the reaction
– F is called Faraday’s constant and is the quantity of
charge on 1 mol of electrons
• 1 F = 96,500 C/mol = 96,500 J/V-mol
– A positive value of E and a negative value of ΔG both
indicate that a reaction is spontaneous
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Sample Problems
• Calculate ΔG for the following reactions
using standard reduction potentials.
– 4Ag(s) + O2(g) + 4H+(aq) → 4Ag+(aq) + 2H2O(l)
– 3Ni2+ + 2Cr(OH)3 + 10OH- → 3Ni + 2CrO42- + 8H2O
The Nernst Equation
• To find the emf of a cell under
nonstandard conditions we can use the
Nernst equation:
Ecell
Ecell
RT
ln Q or
nF
Ecell
Ecell
0.0592
log Q @ 298 K
n
Effect of Concentration on Cell
EMF
• As a voltaic cell discharges, reactants are
consumed and products are generated, so
the concentrations of these substances
change
• The emf progressively drops until E = 0, at
which point the cell is “dead”
• At that point the concentrations of
reactants and products are in equilibrium
Sample Problem
• Given the following reaction:
Cr2O72-(aq) + 14H+(aq) + 6I-(aq) → 2Cr3+(aq) + 3I2(s) + 7H2O(l)
• Calculate the emf at 298 K when [Cr2O72-] = 2.0
M, [H+] = 1.0 M, [I-] = 1.0 M, and [Cr3+] = 1.0 x
10-5 M.
• Q is the reaction quotient
Sample Problem
• If the voltage of a Zn-H+ cell is 0.45 V at
25 C when [Zn2+] = 1.0 M and PH2 = 1.0
atm, what is the concentration of H+?
Sample Problem
• Using standard reduction potentials,
calculate the equilibrium constant for the
oxidation of Fe2+ by O2 in acidic solution
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Sample Problem
• Calculate the equilibrium constant for the
following reaction:
Batteries
Br2(l) + 2Cl-(aq) → Cl2(g) + 2Br-(aq)
• A battery is a portable, self-contained
electrochemical power source that consists of
one or more voltaic cells
• Smaller batteries, like AAs or AAAs, contain a
single voltaic cell and produce around 1.5 V
• Larger batteries consist of a series of voltaic
cells and can produce greater voltages, such as
a 12 V car battery
Lead-Acid Battery
Lead-Acid Battery
• A 12 V lead-acid battery consists of six voltaic cells in series
– Cathode: PbO2(s) + HSO4-(aq) + 3H+(aq) + 2ePbSO4(s) +
2H2O(l)
+
– Anode: Pb(s) + HSO4 (aq)
PbSO4(s) + H (aq) + 2e– Complete: PbO2(s) + Pb(s) + 2HSO4-(aq) + 2H+(aq)
2PbSO4(s) + 2H2O(l)
– Eºcell = 2.041 V
• Because the reactants are solids, there is no need to separate the
anode and cathode into separate compartments
• The reverse reaction plus supplied electrical energy recharges the
battery
Alkaline Battery
Dry Cell Batteries
• The most common primary
(non-rechargeable) battery is
the alkaline battery
• The name comes from the
fact that anode consists of
powdered metal suspended
in a concentrated KOH gel
• The emf of an alkaline
battery is about 1.55 V at
room temperature
• Alkaline batteries provide far
superior performance over
older “dry cell” batteries
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Fuel Cells
Fuel Cells
• In a hydrogen-oxygen fuel
cell, gaseous H2 is oxidized
to water at the anode, and
gaseous O2 is reduced to
hydroxide ion at the cathode
• The net reaction is the
conversion of H2 and O2 to
water
• Fuel cells are not technically
batteries because they are
not self-contained
Corrosion
Corrosion
• Corrosion reactions are spontaneous redox
reactions in which metals are converted into
an unwanted compound by some substance
in the environment
• For nearly all metals, oxidation is a
thermodynamically favorable process and
can be very destructive
• The rusting of iron is one of the most common
and expensive forms of corrosion
• 20% of all iron used each year is to replace
rust-damaged items
•
•
•
•
The metal and a surface water droplet constitute a tiny galvanic cell in
which iron is oxidized to Fe2+ in a region of the surface (anode region)
remote from atmospheric O2
O2 is reduced near the edge of the droplet at another region of the
surface (cathode region)
Electrons flow from anode to cathode through the metal, while ions flow
through the water droplet
Dissolved O2 oxidizes Fe2+ further to Fe3+ before it is deposited as rust
(Fe2O3)
Corrosion
Corrosion
• A layer of zinc protects iron from oxidation, even when the zinc
layer becomes scratched
• The zinc (anode), iron (cathode), and water droplet (electrolyte)
constitute a tiny galvanic cell
• Oxygen is reduced at the cathode, and zinc is oxidized at the
anode, thus protecting the iron from oxidation
• Protecting a metal from
corrosion by making it the
cathode in an
electrochemical cell is called
cathodic protection
• The metal that is oxidized
while protecting the cathode
is called the sacrificial
anode
• Underground pipes, ship
hulls, and even coffins are
often protected by sacrificial
anodes
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Electrolysis
Electrolysis
• It is possible to use electrical energy to cause
nonspontaneous redox reactions to occur
• Such processes that are driven by an outside
energy source are called electrolysis reactions
and take place in electrolytic cells
• An electrolytic cell consists of two electrodes in a
molten salt or aqueous solution
• Just as in voltaic cells, oxidation occurs at the
anode and reduction at the cathode
Electrolysis
• Electrolysis of
molten sodium
chloride
• Chloride ions are
oxidized to Cl2 gas
at the anode, and
Na+ ions are
reduced to sodium
metal at the cathode
Electroplating
• Electroplating uses
electrolysis to
deposit a thin layer
of metal on another
metal
• This is used for
beautification or
protection from
corrosion
Sample Problem
• Electrolysis of AgF(aq) in an acidic solution
leads to the formation of silver metal and oxygen
gas. Write the half-reaction that occurs at each
electrode and calculate the minimum emf
needed for this process to occur under standard
conditions.
Quantitative Aspects of Electrolysis
• For any half reaction, the amount of a substance that is
reduced or oxidized is directly proportional to the number
of electrons passed into the cell
• The charge of one mol of electrons is a faraday (F)
• The quantity of charge passing through an electrical
circuit is measured in coulombs (C)
– Remember 1 F = 96,500 C/mol e-
• The number of coulombs passing through a cell can be
calculated as follows:
– Coulombs = amperes x seconds
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Sample Problem
Sample Problems
• Calculate the number of grams of
aluminum produced in 1.00 hour by the
electrolysis of molten AlCl3 if the current in
10.0 A.
• The half-reaction for the formation of
magnesium metal upon electrolysis of molten
MgCl2 is Mg2+ + 2e- → Mg. Calculate the mass
of magnesium formed upon passage of a current
of 60.0 A for a period of 4.00 x 103 s.
• How many seconds would be required to
produce 50.0 g of Mg if the current is 100.0 A?
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