1
OXIDATION NUMBERS (Section 4.4)
Oxidation number is the charge of an atom in a molecule if all the bonding is
considered ionic.
Oxidation number is different from formal charge.
Using oxidation numbers gives us another way to account for electrons in
chemical changes.
Rules of assigning oxidation numbers
1. The oxidation number of an atom as an element is always zero.
Examples
Fe (s)
F–F
Ox. # = 0
Ox. # = 0
 for each fluorine atom
2. The oxidation number of an atom as an ion is the same as the charge of the ion.
Examples
Fe3+ (aq)
F- (aq)
Ox. # = 3
Ox. # = -1
3. In polyatomic compounds, consider the oxidation number of the most
electronegative element to be its “normal” ionic charge.
Example: Water molecule
O
H
Ox. # (O) = -2
H
Example: Carbon tetrafluoride
F
Ox. # (F) = -1
F
C
F
F
Rules of thumb
Ox. # (O) = -2
Ox. # (F) = -1
Ox. # (H) = +1
- Careful! These rules of thumb can be broken.
2
4. The sum of all the oxidation numbers of the atoms equals the charge of the
compound.
Example: What is the oxidation number of the phosphorus atom in phosphorus
trichloride?
Cl
Cl
P
Cl
Chlorine is most electronegative atom and its “normal” ionic charge is –1.
Ox. # (Cl) = -1
The sum of oxidation numbers must be zero.
Example: What are the oxidation numbers of the atoms in the chlorate ion.
O
Cl
O
O
DEFINITIONS
Oxidation – process of losing electrons
Reduction – process of gaining electrons
Oxidizing agent – substance that causes oxidation in another substance
Reducing agent – substance that causes reduction in another substance
**Source of confusion**
Oxidizing agent is reduced.
Reducing agent is oxidized.
0
+1 -1
+2 -1
0
Consider Zn (s) + 2 HCl (aq)  ZnCl2 (aq) + H2 (g)
Zn metal is losing e-. (oxidation)
H+ ions are gaining e-. (reduction)
Zn (s)  Zn2+ (aq) + 2 e2 H+ (aq) + 2 e-  H2 (g)
Zn is the reducing agent. (It causes H+ to reduce.)
H+ is the oxidizing agent. (It causes Zn to oxidize)
3
HALF-REACTIONS
Reduction-Oxidation (Redox) reactions can be broken into two parts: an
oxidation reaction and a reduction reaction.
These two parts are called half-reactions.
0
+2
+2
0
Consider Zn (s) + CuSO4 (aq)  ZnSO4 (aq) + Cu (s)
Oxidation: Zn (s)  Zn2+ (aq) + 2 eReduction: Cu2+ (aq) + 2 e-  Cu (s)
Half-reactions must occur in pairs
- We can never have only oxidation or only reduction.
- e- lost in oxidation must be gained through reduction.
# of e- lost = # of e- gained
BALANCING REDOX EQUATIONS
1. Divide equation into half-reactions
2. Balance each half-reaction
a.) Balance atoms other than H and O
b.) Balance oxygen atoms by adding water to the appropriate side
c.) Balance hydrogen atoms by adding H+ to appropriate side
d.) Balance charge by adding e- to appropriate side
3. Multiply each equation by integer so that
# of e- lost = # of e- gained
4. Add equations together; cancel spectator species
a.) If reaction occurs in basic conditions, “neutralize” H+ by adding OH- to
both sides and canceling extra water molecules.
Example: Zn (s) + VO2+ (aq)  Zn2+ (aq) + V3+ (aq) in acidic conditions
Balance: Zn (s)  Zn2+ (aq)
a) OK
b) OK
c) OK
d) Zn (s)  Zn2+ (aq) + 2 eBalance: VO2+ (aq)  V3+ (aq)
a) OK
b) VO2+ (aq)  V3+ (aq) + H2O (l)
c) VO2+ (aq) + 2 H+ (aq)  V3+ (aq) + H2O (l)
d) VO2+ (aq) + 2 H+ (aq) + e-  V3+ (aq) + H2O (l)
Zn (s)  Zn2+ (aq) + 2 e2{VO (aq) + 2 H+ (aq) + e-  V3+ (aq) + H2O (l)}
Zn (s) + 2 VO2+ (aq) + 4 H+ (aq) + 2 e-  Zn2+ (aq) + 2 e- + 2 V3+ (aq) + 2 H2O (l)
Zn (s) + 2 VO2+ (aq) + 4 H+ (aq)  Zn2+ (aq) + 2 V3+ (aq) + 2 H2O (l)
Ta Da!! The balanced redox reaction in acidic sol.
2+
4
What about same reaction is basic conditions?
Add OH- to each side to neutralize H+.
Zn (s) + 2 VO2+ (aq) + 4 H+ (aq) + 4 OH- (aq)
 Zn2+ (aq) + 2 V3+ (aq) + 2 H2O (l) + 4 OH- (aq)
Zn (s) + 2 VO2+ (aq) + 2 H2O (l)  Zn2+ (aq) + 2 V3+ (aq) + 4 OH- (aq)
Ta Da Again!! The balanced redox rxn. in basic sol.
Example: Balance the following rxn in basic solution:
C3H8O3 (aq) + MnO4- (aq)  CO32- (aq) + MnO2 (aq)
Balance: C3H8O3 (aq)  CO32- (aq)
Balance: MnO4- (aq)  MnO2 (aq)
3  {C3H8O3 (aq) + 20 OH- (aq)  3 CO32- (aq) + 14 H2O (l) + 14 e-}
14  { MnO4- (aq) + 2 H2O (l) + 3 e-  MnO2 (s) + 4 OH- (aq)}
3 C3H8O3 (aq) + 4 OH- (aq) + 14 MnO4- (aq)
 9 CO32- (aq) + 14 H2O (l) + 14 MnO2 (s)
Example: IO3- (aq) + Cr3+ (aq)  I2 (s) + Cr2O72- (aq) in acidic conditions
Balance: IO3- (aq)  I2 (s)
Balance: Cr3+ (aq)  Cr2O72- (aq)
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3{2 IO3- (aq) + 12 H+ (aq) + 10 e-  I2 (s) + 6 H2O (l)}
5{2 Cr3+ (aq) + 7 H2O (l)  Cr2O72- (aq) + 14 H+ (aq) + 6 e-}
6 IO3 + 36 H+ + 30 e- + 10 Cr3+ + 35 H2O
 3 I2 + 18 H2O + 5 Cr2O72- + 70 H+ + 30 e6 IO3- + 10 Cr3+ + 17 H2O  3 I2 + 5 Cr2O72- + 34 H+
Double check: Can coefficients be reduced any further?
VOLTAIC CELLS
Voltaic cells are batteries.
In 1800 Alessandro Volta discovered that stacking plates of metal between layers
of wet blotter paper created electricity. (Voltaic ‘pile’)
Electrons spontaneously flow from reducing agent to oxidizing agent.
- Batteries exploit spontaneous redox reaction!
Batteries are also known as galvanic cells.
Consider reaction in a nickel/cadmium battery
Cd (s) + Ni2+ (aq)  Cd2+ (aq) + Ni (s)
Cd is being oxidized as Ni2+ is being reduced.
Schematic of a Voltaic Cell
flow of eanode
cathode
-
+
K+
Cl-
K+
Cl-
Ni
Cd
Cl-
K+
Cd
Cd
2+
salt bridge
Cd2+
Cd2+
Cd (s)  Cd2+ (aq)
Definitions
Electrode where oxidation occurs is anode.
Electrode where reduction occurs is cathode.
Note: Oxidation  Anode
(both vowels)
Reduction  Cathode
(both consonants)
At the anode, Cd (s) loses e to form Cd2+ (aq)
- The cadmium metal slowly dissolves.
At the cathode, Ni2+ (aq) gains e- to form Ni (s)
- The nickel ions plate on electrode. (It gets bigger.)
Ni2+
Ni2+
Ni
Ni2+
Ni2+ (aq)  Ni (s)
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Salt Bridges
The electrons flowing from anode to cathode can be used to do electrical work
such as lighting a light bulb or heating a toaster.
Charges must flow in a circuit so that electrical charge is conserved.
Thus a salt bridge is needed to allow charges to flow.
A salt bridge is a glass tube filled with brine (concentrated salt) solution with
semipermeable plugs at the ends of the tubes.
For a battery to work over any substantial period of time, the two metal solutions
must be separated.
The semipermeable plugs allow only ions (such as K+ and Cl-) from brine solution
to leave.
The flow of oppositely charged ions out of the salt bridge completes the electrical
circuit in a battery.
Voltaic Cell Notation
An abbreviated way of describing a voltaic cell has been developed by electrochemists
where the essential features of a voltaic cell can be written on one line.
1. All phase boundaries are indicated with a vertical line.
Cd (s) | Cd2+ (aq) (The solid is a different phase than the aqueous solution.)
2. The salt bridge is indicated by a double vertical line.
3. The anode is written first (on the left), then the cathode (on the right)
Cd (s) | Cd2+ (aq) || Ni2+ (aq) | Ni (s)
Note the two ions in the cell come in “contact” with each other via the salt bridge.
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ELECTROMOTIVE FORCE (EMF)
EMF is the strength of the reducing agent’s e- pushing combined with the strength
of the oxidizing agent’s e- pulling.
The SI unit to measure EMF is the volt.
1J
1V 
1C
1 volt of EMF gives 1 Coulomb of charge 1 Joule of energy.
The charges in a system (e. g., a copper wire) move randomly until a voltage is applied.
e-
ee
e
e-
-
e
-
-
e-
e-
Voltage causes charges to move a single direction. Voltage induces electrical
current.
e-
+
e-
e-
ee-
e-
e
-
e-
Voltage comes from a difference in potential energy.
- Electrons flow from high potential energy to low potential energy.
- Flow of electrons is like flow of water in a waterfall from a height (high
potential energy) to a depth (low potential energy).
Cell potentials and half reactions
For the Ni/Cd cell, the EMF at std. conc. is 0.12 V.
The total voltage is the sum of the voltages due to each half-reaction.
Cd(s)  Cd2+ (aq) + 2 eNi2+ (aq) + 2 e-  Ni (s)
Cd (s) + Ni2+ (aq)  Cd2+ (aq) + Ni (s)
E = 0.40 V
E = - 0.28 V
E = 0.12 V
-
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STANDARD REDUCTION POTENTIAL
We are not able to measure the potential of a reduction half-reaction by itself.
- Cell potential is due to both ox. and red.
Ecell = Ered + Eox
- To find reduction potential, we must set a standard, i.e. a reference point.
The standard that is chosen is the reduction of H+ ion.
2 H+ (aq, 1M) + 2 e-  H2 (g, 1atm) at 25 C
E = 0.00 V by definition.
Thus the standard reduction potentials are measured relative to the reduction of
hydrogen ion.
The standard reduction potentials for many half-reactions are often compiled
(as in Appendix I).
Ni2+ (aq) + 2 e-  Ni (s)
E = - 0.28 V
Cd2+ (aq) + 2 e-  Cd (s)
E = - 0.40 V
Standard concentration for all standard reduction potentials is 1M.
Standard reduction potentials are the source of the activity series from Chapter 4.
Calculating standard cell potentials
- Cell potential is due to both ox. and red.
Ecell = Ered + Eox
- When the half-reaction is reversed, the potential has the opposite sign.
- Thus Eox has the opposite sign of the standard reduction potential.
*Source of Confusion*
Electrical potentials are intensive quantities.
- They do not depend on stoichiometric coeff.
- To find potential of cell, add potentials of half-reactions.
- Remember: oxidation half-reaction has opposite sign from standard
reduction potential.
**Source of confusion**
When calculating standard cell potentials, one does not multiply standard
reduction potential by stoichiometric coefficients.
This method is different than the procedure to calculate Hrxn, Srxn and Grxn.
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Example: Calculate the standard EMF for the galvanic cell reaction:
Cr (s) + Cu2+ (aq)  Cr3+ (aq) + Cu (s)
Balanced Equation
Write each half-reaction
Find standard reduction potentials in table
Cr3+ (aq) + 3 e-  Cr (s)
E = - 0.74 V
2+
Cu (aq) + 2 e  Cu (s)
E = 0.34 V
Read Section 19.7 for information about specific voltaic cells.
Strength of Oxidizing and Reducing Agents
The more positive the reduction potential, the stronger the reduction.
- Therefore, the better the oxidizing agent.
F2 (g) + 2 e-  2 F- (aq)
H2O2 (aq) + 2 H+ (aq) + 2 e-  2 H2O (l)
E = 2.87 V
E = 1.776 V
- Fluorine gas is one of the best oxidizing agents.
- Hydrogen peroxide is also a good oxidizing agent.
The more negative the reduction potential, the more positive the oxidation
potential. Thus the stronger the oxidation.
- Therefore, the better the reducing agent.
Li+ (aq) + e-  Li (s)
Mg2+ (aq) + 2 e-  Mg (s)
- Lithium is one of the best reducing agents.
- Magnesium is also a good reducing agent.
E = - 3.07 V
E = - 2.37 V
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SPONTANEOUS ELECTROCHEMICAL RXNS
Using chemical intuition, which reaction is spontaneous?
Cl2(g) + 2 e-  2 Cl-(aq)
E = 1.36 V
+
Na (aq) + e  Na(s)
E = - 2.71 V
- Chlorine taking e- is spontaneous.
In general, reactions with E positive are spontaneous.
Reactions with E negative are not spontaneous.
Note: To decide spontaneity, one needs to consider total cell potential, not only
half-reaction.
Free Energy and Electric Potential
For a general electrochemical reaction,
G = -n F E
n – number of moles of electrons transferred in reaction, i. e., stoichiometric
coefficient of e- in balanced redox equation.
F – Faraday’s constant = charge of 1 mole of electrons
F = 96,485 C/mol = 96,485 J/Vmol
E – cell potential
Equation relates intensive property (EMF) to extensive property (G) via moles
of e- transferred.
Since equation is general
G = -n F E

G = -n F E
- Thus we can calculate G of a reaction by measuring E.
- Also note that equation shows us that cells with positive EMF spontaneously
react.
- Spontaneous electrochemical reactions create electrical current.
Example: Calculate G for the reaction Zn (s) + Cl2 (g)  ZnCl2 (aq) from the
standard reduction potentials of the half-reactions.
Zn (s) + Cl2 (g)  Zn2+ (aq) + 2 Cl- (aq)
Zn (s)  Zn2+ (aq) + 2 eCl2 (g) + 2 e-  2 Cl- (aq)
E = 0.76 V
E = 1.36 V
E = 2.12 V
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Example: For the reaction Ni(NO3)2 (aq) + Cd (s)  Ni (s) + Cd(NO3)2 (aq)
what is the expected voltage in an voltaic cell where [Ni2+] = [Cd2+] = 1 M?
Grxn = -23.2 kJ/mol
Nernst Equation
Equation relates EMF with concentration of electrolyte
G = G + RT ln Q
-n F E = -n F E + RT ln Q
Recall:

E  E0 
RT
ln Q
nF
At 25 C, E  E 0 
0.0257 V
ln Q
n
Example: For the reaction, Cr (s) + Ni2+ (aq)  Cr3+ (aq) + Ni (s), calculate the
EMF when
a) [Ni2+] = [Cr3+] = 1 M
Cr (s)  Cr3+ (aq) + 3 eNi2+ (aq) + 2 e-  Ni (s)
Total cell potential
E = 0.74 V
E = - 0.23 V
E = 0.51 V
b) [Ni2+] = 2.0 M; [Cr3+] = 0.010 M
Balanced equation is 2 Cr(s) + 3 Ni2+(aq)  2 Cr3+(aq) + 3 Ni(s)
High amount of reactant and low amount of product creates greater driving force
in a reaction that is already spontaneous.
Review relationships between G, E, and Q(or K)
G  G 0  RT ln Q
G = -n F E
RT
E  E0 
ln Q
nF
at eq.
at eq.
at eq.
G 0   RT ln K
G = 0, E = 0
RT
E0 
ln K
nF
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ELECTROLYSIS
The process of adding electrical energy to make a nonspontaneous process,
spontaneous.
Applications of electrochemical cells
Galvanic cells
- source of electrical energy (battery)
Electrolytic cells
- refining of ores
- produce alkali, alkaline earth metals
- produce H2 and O2 from water (fuel cells)
- metal plating (gold, chromium, etc…)
Consider the formation of KBr
Br2(l) + 2 e-  2 Br-(aq)
K(s)  K+(aq) + e-
E = 1.07 V
E = 2.92 V
E = 3.99 V
Balanced reaction is 2 K (s) + Br2 (l)  2 KBr (aq)
Calculate K (equilibrium constant)
E0 
3.992
RT
ln K  K  e 0.0257  e 311  10135
nF
This reaction is definitely spontaneous!!
Is there anyway to make K(s) and Br2(l) from KBr(s)?
Yes!, by reversing the voltage, i.e., forcing e- on K+ and ripping e- from Br-.
When an external voltage is applied that is greater than 3.99 V, the reverse
reaction occurs.
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Schematic of an Electrolytic Cell
EMF source
(battery)
-
flow of e-
+
cathode
anode
-
+
Pt
Pt
+
Br-
K+
K(l)
Br-
Br-
Br-
K+
BrK+
K
Br
K+
K+
Br-
K+
K+ (l)  K (l)
Br2(g)
-
2 Br- (l)  Br2 (g)
Liquid potassium is depositing on the cathode.
Gaseous bromine is forming on the anode.
Note: cell reaction is 2 KBr (l)  2 K (l) + Br2 (g). We must use the molten salt
rather than the aqueous solution because if we attempt the reaction
2 KBr (aq)  2 K (l) + Br2 (g) other reactions with the solvent (water) occur.
*Source of Confusion*
The following table gives the signs of the electric potential at each of the
electrodes in each type of cell.
Galvanic
Electrolytic
Cathode
+
-
Anode
+
However current always flows from anode to cathode.
The amount of K(s) or Br2 (l) can be controlled by controlling the number of
moles of electrons in an oxidation/reduction process.
- The number of moles of electrons used (i.e., total charge used) is controlled
by adjusting the time that the electrical current is allowed to flow.
amount of ch arg e
- current 
I
time
- SI unit of current is Ampere (A)
- 1A 
1C
1s
-Q=It
Q – charge
I – current
t – time
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Example: An iron spoon is placed in an electrochemical cell with AuCl3 solution
to be gold plated. How long must the spoon be in the electrolytic cell
if the spoon is to be plated with 0.293 g of Au and the current of the
cell is 1.03 A.
1st Question: How many moles of e- are needed for the reaction Au3+(aq) + 3 e-  Au(s)
2nd Question: How many coulombs is 0.00446 mole-?
3rd Question: If current is 1.03 A, how long does it take for 431 C to flow?
Example: How many grams of barium metal can be produced by the passage of
173 C of charge in an electrolytic cell of molten BaCl2?
Ba2+ (l) + 2 e-  Ba (s)