Copyright Sautter 2003
ELECTROCHEMISTRY
• All electrochemical reactions involve oxidation and reduction.
• Oxidation means the loss of electrons (it does not always involve
oxygen).
• Reduction means the gain of electrons (gaining of negatives, that is
electrons, reduces the oxidation number of an atom. We will discuss
oxidation number latter in this program).
• Whenever electrons are lost by one substance they must be gained by
another.
• The substance that loses electrons is referred to as a reducing agent
(it lets another substance be reduced, that is, gain the electrons).
• The substance that gains electrons is referred to as an oxidizing agent
(it lets another substance be oxidized, that is, lose the electrons).
• These terms are important in electrochemistry!!
OXIDATION & REDUCTION
• OXIDATION:
Zn(s)  Zn+2(aq) + 2 eMetallic zinc is oxidized to zinc ion. Metallic zinc is
serving as a reducing agent.(electron loser)
• REDUCTION:
Cu+2(aq) + 2 e- Cu(s)
Copper ion is reduced to copper metal. Copper ion is
serving as an oxidizing agent (electron gainer)
• In the overall reaction two electrons are transferred
from the zinc metal to the copper ion.
Zn(s) + Cu+2(aq) Zn+2(aq) + Cu(s)
OXIDATION NUMBERS
• In the previous reactions the loss and gain of
electrons was obvious. Zn(s) must lose two
negative electrons in order to form a Zn+2(aq) ion.
And Cu+2(aq) must gain two negative electrons to
form a Cu(s) atom.
• What about more complicated reactions? How do
we tell which substance loses and which gains
electrons and how many?
• For example:
H+ + MnO4- + Cl-  Mn+2 + Cl2 + H2O
Here we need oxidation numbers!
OXIDATION NUMBERS
• Before discussing the mechanics of oxidation numbers it
is important to realize that:
• (1) oxidation numbers have no physical significance.
They are merely a way to tell which substances gain or
lose electrons and how many.
• (2) oxidation numbers are assigned to atoms never
molecules. Molecules contain atoms with oxidation
numbers but they themselves cannot be assigned
oxidation numbers.
• (3) there a several rules used to assign oxidation
numbers. They must be observed carefully.
OXIDATION NUMBERS RULES
• (1) Elements have oxidation numbers of zero. For
example Cu has an oxidation number of zero. In
Cl2 each atom of chlorine has an oxidation number
of zero.
• (2) The oxidation number of a monatomic ion
(consisting of one atom) is the charge on the ion.
In Cu+2 the oxidation number is +2. In Cl- the
oxidation number is –1.
• (3) The oxidation number of combined oxygen is
always –2 except in peroxides such as hydrogen
peroxide, H2O2. In CO2 each oxygen atom has an
oxidation number of –2. In SO3 each oxygen is –2.
OXIDATION NUMBERS RULES
• (4) Combined hydrogen is always +1 except in hydrides (metal
atoms and hydrogen like NaH). In water H2O each hydrogen has
an oxidation number of +1. In methane, CH4, each hydrogen is
+1.
• (5) The sum of all of the oxidation numbers of the atoms in a
molecule equal its charge. If no charge is shown the sum must
equal zero. This allows us to determine the oxidation numbers of
elements not discussed in the first four oxidation number rules!
• Oxidation numbers of other elemental groups:
Elements of Column I in a compound = +1
Elements of Column II in a compound = +2
It is important to realize that the oxidation number of an atom
can change when it combines differently during reaction. This
is the basis for deciding how electrons are transferred during a
chemical reaction!
FINDING UNKNOWN OXIDATION NUMBERS
• Problem: What is the oxidation number of Mn in MnO4-, the
permanganate ion?
• Solution: Mn is unknown
Each O is –2 (rule 3)
All of the oxidation numbers must add up to –1 (rule 5)
Mn + 4(-2) = -1, Mn = +7 in permanganate
• Problem: What is the oxidation number of Cr in K2Cr2O7, potassium
dichromate?
• Solution: Cr is unknown
Each K from Column I is +1
Each O is –2 (rule 3)
All the oxidation numbers add up to 0 (no charge is shown for K2Cr2O7)
2(+1) + 2 Cr + 7(-2) = 0,
2 Cr = 12,
Cr = 12 / 2 = +6
Cr = +6 in potassium dichromate
USING OXIDATION NUMBERS TO
DETERMINE ELECTRON TRANSFERS
• Using oxidation numbers we will find the oxidizing agent, the
reducing agent, the number of electrons lost and gained, the
oxidation half-reaction, the reduction half-reaction and the final
balanced equation for:
H+ + MnO4- + Cl-  Mn+2 + Cl2 + H2O
• Step I – Find the oxidation number of each atom.
H+ + MnO4- + Cl-  Mn+2 + Cl2 + H2O
+1 (+7, -2) -1
+2
0 (+1, -2)
• Step II – Determine which oxidation numbers change
MnO4-  Mn+2
2 Cl-  Cl2
If oxidization # s decrease,
+7
 +2
2(-1)  0
Reduction occurs
5 electrons gained
2 electrons lost If oxidization # s increase,
Oxidation occurs
(MnO4- is the oxidizing agent. It gained electrons from Cl- )
(Cl- is the reducing agent. It lost electrons to MnO4- )
USING OXIDATION NUMBERS TO
DETERMINE ELECTRON TRANSFERS
• Step III - Electrons lost by a reducing agent must always equal
electrons gained by the oxidizing agent!
• Therefore: 2 (MnO4- + 5e-  Mn+2 )
2MnO4- + 10e-  2Mn+2 (reduction half-reaction*)
• And
5(2Cl-  Cl2 + 2e-)
10 Cl-  5Cl2 + 10e- (oxidation half-reaction)
• Step IV -Completing the first half-reaction with H+ ions and H2O
molecules: 16 H+ + 2MnO4- + 10e-  2Mn+2 + 8 H2O
• Step V - Adding the oxidation half-reaction and reduction halfreaction the 10 electrons gained and lost cancel to give the overall
reaction:
16 H+ + 2MnO4- + 10 Cl-  2 Mn+2 + 5 Cl2 + 8 H2O
• * Half-reaction refers to the reaction showing either the electron gain
(reduction) or the electron loss (oxidation) step of the reaction.
ELECTRICAL POTENTIAL, CHARGE,
ELECTRON FLOW
• For electrons to transfer from one substance to another the
oxidizing agent (electron gainer) must have a greater
attraction for those electrons than the reducing agent
(electron loser). This where electrical potentials come into
play.
• Electrical potential is measured in volts. Electrical charge
is measured in coulombs. The flow of electrons from one
substance to another is measured in amperes.
• 1 coulomb is the charge on 6.25 x 1018 electrons.(The
charge on 1 electron is –1.6 x 10-19 coulombs).
1 ampere (amp) is the flow rate of 1 coulomb per second
1 volt is 1 joule of energy moving 1 coulomb of charge
between two substances.
ELECTRICAL QUANTITIES
6.25 x 10 18electrons
1 coulomb
1 second
1 joule
1 amp = 1 coul / sec
1 volt = 1 joule / coul
1 coulomb of charge
ELECTRICAL POTENTIAL
• Electrical potential is measured in volts and indicates the
tendency of electrons to move from one substance to
another. Potential depends on a variety of factors such as
the concentration of reactant materials, temperature, gas
pressures and the nature of the materials involved.
• Standard Reduction Potentials (measuring the tendency
of a substance to gain electrons) are determined at 25 0C,
1 molar concentrations, 1 atm pressures and are
compared to hydrogen ion’s tendency to be reduced
(gain electrons) to hydrogen gas (2H+(aq) + 2e-  H2(g))
which is assigned a potential of 0.00 volts.
• Substances which gain electrons better than H+ ion are
assigned positive potentials. Those which gain electrons
more poorly than H+ ion are assigned negative potentials.
ELECTRICAL POTENTIAL
Some Standard Reduction Potentials
E0 (volts)
ELECTRICAL POTENTIALS
• The very best oxidizing agent (electron gainer) available is
flourine gas, F2. Its reduction potential is +2.87 volts.
• The very best reducing agent (electron loser) is lithium
metal, Li. Its oxidization potential is +3.05 volts.
• When Li is combined with F2, the reaction is highly
spontaneous. Electrons move readily from the Li to the F2.
The standard potential of this cell (redox reaction) is +5.92
volts.
• By contrast, combining Li+ ion with F- ion results in a
nonspontaneous process.(Li+ has already lost its electrons
and F- has already gained its electrons). The standard
potential for this cell is –5.92 volts
ELECTRICAL POTENTIALS
• From the Standard Reduction Potential Table:
(1) F2(g) + 2e-  2 F-(aq)
2.87 volts
(2) Li+(aq) + 1 e-  Li(s)
-3.05 volts
•
Since Li metal serves as a reducing agent, it must be shown to
lose electrons, and the half-reaction must be reversed. This requires
that the E0 sign be changed.
(3) Li(s)  Li+(aq) + 1e+3.05 volts
•
and since 2 electrons must be transferred to F2
(4) 2 Li(s)  2 Li+(aq) + 2e+3.05 volts
(note that the E0 is not multiplied by two.)
• Combining half-reaction (1) and (4) we get
Li(s) + F2(g)  Li+(aq) + 2 F-(aq) + 5.92 volts
For the reverse reaction which is nonspontaneous, the potential is
–5.92 volts.
ELECTRICAL POTENTIALS
• From the previous frames we can conclude:
(1) Standard redox potentials for a reaction can be
obtained from standard reduction half-reactions by adding
the half-reactions.
(2) When a substance is to act as a reducing agent, the
half-reaction from the standard reduction potentials table
must be reversed and the sign of the E0 value changed.
(3) When half-reactions are added, electrons lost and
gained must be balanced. When balancing the electrons,
although the half-reactions are multiplied, the E0 values
are not.
(4) Reactions with positive E0 values are spontaneous.
Those with negative E0 values are nonspontaneous in the
forward direction
USING ELECTRICAL POTENTIALS
• Problem: What is the electrical potential (E0) for a standard cell
(redox reaction) composed of zinc metal and copper II ions? The
products are zinc ion and copper. Is the reaction spontaneous?
Identify the cathode (reduction site) and anode (oxidation site).
• Solution:
Zn(s) + Cu+2(aq)  Zn+2(aq) + Cu(s)
0
+2
+2
0
(oxidization #s)
• Zn is the reducing agent, loses electrons (0  +2)
Cu+2 is the oxidizing agent, gains electrons (+2 0)
Zn(s)  Zn+2(aq) + 2e- +0.76 volts (anode reaction)
Cu+2(aq) + 2e-  Cu(s) +0.34 volts (cathode reaction)
• Overall reaction:
Zn(s) + Cu+2(aq)  Cu(s) + Zn+2(aq) E0 = +1.10 volts
The reaction is spontaneous (E0 net is positive)
Electrons move
through the
external circuit
from anode to cathode.
Copper is deposited
at the cathode
Cu
Cu+2
Cu
Cu+2
Cu+2
Cu+2
Cu+2
Cu
Electrons move
through the
external circuit
from anode to cathode.
Zinc is dissolved
at the anode.
Zn
Zn+2
Zn
Zn+2
Zn+2
Zn
Zn
Zn
ELECTROCHEMICAL CELL
CONVENTIONS
• Electrochemical cell that release energy
spontaneously are call voltaic or galvanic cells.
Those which require energy input in order to
function are call electrolytic cells.
• Voltaic cells are often represented as follows for
the reaction: Zn(s) + Cu+2(aq)  Zn+2(aq) + Cu(s)
Solution
concentration
Solution
concentration
Zn / Zn+2 (1.0 M) // Cu+2 (1.0 M) / Cu
Anode
Compartment
components
Salt
bridge
Cathode
Compartment
components
AN ELECTROCHEMICAL CELL
EMF
1.100
VOLTS
SALT BRIDGE
ANODE
(OXIDATION
OCCURS)
CATHODE
(REDUCTION
OCCURS)
Zn(S)  Zn+2(aq) + 2 eE0 = 0.763 volts
Cu+2(aq) + 2 e-  Cu(s)
E0 = 0.337 volts
USING NONSTANDARD POTENTIALS
• Reactions do not always involve Standard Electrical Potentials (that is
25 0C, 1molar, 1atm). Calculating potential at other concentrations and
temperatures requires the use of the Nernst Equation.
• Ecell = E0 – (RT / n F) ln Q where:
Ecell = the voltage of the non standard cell
E0 = standard cell potential
R = 8.31 joules / mole K, T = Kelvin temperature
n = the number of electrons transferred in the balanced equation
F = Faraday’s constant (96,487 coulombs / mole)
Q = reaction quotient (product concentrations / reactant concentrations)
• At equilibrium Ecell = 0 and Q = Ke therefore:
E0 = (RT / n F) ln Ke
• At 25 oC (the most common temperature used), substituting 298 K,
8.31 joules / mole K, 96,487 coulombs / mole and 2.203 (the
conversion factor from base e to base 10 logs) we get:
Ecell = E0 – (0.0592 / n) log Q and Eo= (0.0592 / n) log Ke
USING THE NERNST EQUATION
• Problem: Calculate the Ke value for the zinc / copper cell
shown in the previous diagram.
• Solution: The standard potential for the cell is 1.10 volts and
2 electrons are transferred from the zinc anode to the copper
cathode as shown.
Eo= (0.0592 / n) log Ke ,
Ke = log-1( (E0 x n) / 0.0592)
Ke = log-1 (37.16) = 1.45 x 1037
(the reaction is very spontaneous)
USING THE NERNST EQUATION
• Problem: What is the potential of the zinc / copper cell if
the zinc ion concentration is 0.10 M and the copper ion
concentration is 0.00010 M? (non standard cells)
• Zn(s) + Cu+2(aq)  Zn+2(aq) + Cu(s), E0net = 1.10 volts
Ecell = E0 – (0.0592 / n) log Q
Q = [Zn+2] / [Cu+2] = 0.10 / 0.00010 = 1000
Ecell = 1.10 – (0.0592 / 2 ) log 1000 = 1.01 volts
Based on LeChatelier’s Principle, the higher
concentration of zinc ion on the right side of the
reaction should drive the system to the left and reduce
it spontaneity. The reduced voltage of 1.01 volts as
compared to the standard state voltage of 1.10 volts
confirms that observation.
RELATING FREE ENERGY AND CELL
POTENTIALS
• Since both free energy and cell potential predict the spontaneity of
a reaction they logically must be related.
• The relationship is given by:
G = - n F Ecell and G0 = - n F E0 cell where
n = number electrons transferred in the balanced equation
F = Faraday’s constant (96,487 coulombs / mole)
E = the cells potential in volts
• Problem: What is the G0 for the zinc / copper cell?
• Solution: G0 = - n F E0 cell
G0 = - (2) (96,487) (1.10) = - 2.12 x 105 joules / mole
or - 212 KJ / mole
• A large positive E0 value correlates well with a large negative
G0 value. Both indicate a very spontaneous reaction!
ELECTROLYSIS
• Electrolysis involves the input of energy into an electrochemical cell.
Electrolytic cells are used to decompose substances, deposit metals
and determine the charges on ions in solution.
• The exact number of electrons passing into an electrolytic cell can be
determined by monitoring the current flow in amperes and the time
elapsed. One ampere is equal to a flow rate of one coulomb (6.25 x
1018 electrons) per second.
• Coulombs = amps x time (in seconds)
• Since chemistry measurements are most always done in moles a
conversion factor between coulombs and moles is called Faraday’s
constant is needed. One mole (6.02 x 1023) divided by one coulomb
(6.25 x 1018) gives 96,487 coulombs per mole (Faraday’s constant we have used this number before in this program)
ELECTROLYTIC SILVER PLATING
electrons
DIRECT
CURRENT
SOURCE
anode
cathode
Ag
Ag+ + 1 e-  Ag
Ag+ solution
ELECTROLYSIS
• Problem: How many grams of copper can be deposited
from a copper sulfate solution by a current of 2 amperes
for 20 minutes?
• Solution: Cu+2(aq) + 2e-  Cu(s)
• Amperes x time in seconds = coulombs
(2.00 coul / sec) x (20 x 60) sec = 2400 coul.
• charge in coulombs / Faraday’s constant = moles of e2400 coul x (1 moles / 96,487coul) = 0.0249 moles of e• 1 mole of copper atom requires 2 moles of electrons
0.0249 moles e- / 2 = 0.0125 moles of Cu atoms
• 0.0125 moles Cu x 63.3 grams / mole = 0.791 g of Cu
ELECTROLYSIS
• Problem: What current is needed to deposit 0.500 grams of
Cr+3 ion from solution in 1.00 hour?
• Solution:
• Cr+3 + 3 e-  Cr
•
grams Cr / (grams per mole) = grams
0.500 / 52.0 = 0.00962 moles
• 1 mole of Cr+3 needs 3 moles of electrons
0.00962 moles Cr+3 x 3 = 0.0288 moles of e• current is measured in amperes (coulombs per second)
0.0288 moles of e- x 96487 coul per mole = 2780 coul
• One hour = 60 x 60 = 3600 seconds
• Amperes = 2780 coulombs / 3600 seconds = 0.773 amps
SUMMARY OF KEY IDEAS
• (1) Oxidation number rules:
(i) Elements = 0
(ii) Monatomic ion = the ionic charge
(iii) Combined oxygen = -2
(iv) Combined hydrogen = +1
(v) The sum of all oxidation numbers = the charge on
the ion or molecule
• (2) Oxidation is electron loss (oxidation number
increases. Reduction is electron gain (oxidation
decreases)
• (3) Cathode is the site of reduction. Anode is the site of
oxidation.
SUMMARY OF KEY IDEAS
• (4) Potential is measured in volts (1 volt = 1 joule / 1 coul). If the
net E0 is + the reaction is spontaneous. If net E0 is – the reaction is
non spontaneous.
• (5) Standard state is 25 C0, 1 molar and 1 atm
• (6) Faraday’s constant = 96,487 coulombs per mole
• (7) 1 ampere = 1 coulomb per second
• (8) 1 coulomb = 6.25 x 1018 electrons
• (9) Ecell = E0 – (RT / n F) ln Q
• (10) E0 = (RT / n F) ln Ke
• (11)Ecell = E0 – (0.0592 / n) log Q
• (12) Eo= (0.0592 / n) log Ke
• (13) G = - n F Ecell and G0 = - n F E0 cell
• (14) Large Ke values, + E values and - G values mean
spontaneous reactions