Oxidation and Reduction and Electrochemistry Notes
Oxidation Numbers
Charge -
the net excess of electrons or protons in an atom or molecule
Oxidation Number -
oxidation number is the apparent charge of an atom within a molecule. If the
atom is a single atom and not part of a molecule, then the oxidation number
equals the charge
Relationship of Molecular Charge and Oxidation Number
Charge of the molecule = sum of the oxidation number’s of each atom
(see examples below the table)
Assigning Oxidation Numbers Within Molecules
Element
Oxidation Number
Group I metals
+1
Group II metals
+2
Hydrogen
+1: also -1 in hydrides
Oxygen
-2
Individual atoms
Charge = Oxidation #
*All other oxidation numbers must be calculated using the axiom that the total charge on a
molecule is the sum of the oxidation numbers of all the atoms that constitute the molecule
ex.
SO42-
O's oxidation number is -2 by definition.
S's oxidation number is calc'd via S + 4O's = -2
S + 4(-2) = -2
S = +6
ex
Fe(NO3)3
From the dissociation we know that there is the following
Fe(NO3)3  Fe3+ + NO31Fe's oxidation number is +3 since the oxidation number of single atoms
equals the charge.
O's oxidation number is -2 by definition.
N's oxidation number is calc'd via N + 3O's = -1
N + 3(-2) = -1
N = +5
Oxidation – Reduction Reactions
A reaction that involves the oxidation of one atom with the simultaneous reduction of another
atom.
Oxidation  +3
Reduction -
Fe  Fe3+ + 3 e1- where Fe goes from 0
is the loss of electrons
ex
is the gain of electrons ex
V2+ + 2e1-  V where V goes from +2  0
An oxidation-reduction reaction (or redox for short) therefore involves two "half-reactions" - one for the
oxidation and one for the reduction.
Oxidation and Reduction and Electrochemistry Notes
Balancing Basic Redox Reactions
1st
Split the reaction into two half reactions. Identify which is oxidation and which is
reduction.
2nd
Balance the atoms in each reaction by adding coef's (if needed)
3
rd
Balance the charge on each side of the reaction by adding the correct number of electrons
to each side (e's are one negative charge each)
4th
5
Multiply each half-reaction by the number of e's in the other
th
example
Add the two half-reactions together to get the overall redox reaction. Note that the e's are
not present in the redox reaction and should've cancelled, that's why there is step 4.
Balance the following redox reaction
Fe2+ + Cu2+
 Fe3+ + Cu
1st
half rxns Fe2+
 Fe3+
oxidation - +2  +3 loss of one electron
2+
Cu
 Cu
reduction - +2  0
gain of two electrons
2nd/3rd
balance Fe2+
 Fe3+ + e1Cu2+
+ 2e1-
 Cu
4th
multiply 2Fe2+
 2Fe3+ + 2e1{result of multiplying each by 2x}
2+
1Cu
+ 2e
 Cu {result of multiplying each by 1x}
5th
add
2Fe2+ + Cu2+
 2Fe3+ + Cu {this is the final balanced equation}
Electrochemistry Vocabulary
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Electrochemistry: the study of the interchange of chemical and electrical energy
Redox reaction: involves a transfer of electrons from the reducing agent to the oxidizing agent
Oxidation: involves a loss of electrons (an increase in the oxidation number)
Reduction: involves a gain of electrons (a decrease in the oxidation number)
Half-reactions: a redox reaction broken in to two parts, one half with the oxidation and the other
half with the reduction
Salt bridge/ porous disc: involved in the connection between the two solutions
Galvanic cell: device in which chemical energy is changed to electrical energy
Anode: the electrode compartment in which oxidation occurs
Cathode: electrode compartment in which the reduction occurs
Cell potential: (Ecell) the pull of driving force on the electrons
Volt: the unit of electrical potential (J/C)
Voltmeter: measure the cell potential
Standard hydrogen electrode: a platinum electrodes in contact with 1M H + ions and bathed by
hydrogen gas at 1 atm
Standard reduction potentials: E values corresponding to reduction half-reactions with all solutes
at 1M and all gases at 1 atm
Concentration cell: cell in which both compartments have the same components but at different
concentrations
Nernst equation: give the relationship between cell potential and concentrations of cell
components
E = E -
0.0591
n
log(Q)
Oxidation and Reduction and Electrochemistry Notes
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Glass electrode: contains a reference solution of dilute hydrochloric acid in contact with a thin
glass membrane
Ion-selective electrodes: electrodes that are sensitive to the concentration of a particular ion
Lead storage battery: lead serves as the anode and lead coated with lead dioxide serves as the
cathode
Dry cell battery: in its acid version, the zinc inner case acts as the anode, while a carbon rod in
contact with a moist paste of solid MnO2, solid NH4Cl, and a carbon rod act as the cathode
Fuel cells: a galvanic cell for which the reactants are continuously supplied
Corrosion: the process of returning metals to their natural state
Cathodic protection: method most often employed to protect steel in buried fuel tanks and
pipelines
Electrolytic cell: an apparatus that uses electrical energy to produce chemical change
Electrolysis: forcing a current through a cell to produce a chemical change for which the cell
potential is negative
Ampere: measure of current (C/s)
Down’s cell: a cell used for electrolyzing molten sodium chloride
Chlor-alkali process: for producing chlorine and sodium hydroxide by electrolyzing brine in a
mercury cell
Description of Cells
Galvanic Cells:
anode
cathode
Oxidation takes place in the anode
Reduction takes place in the cathode
Note that a porous frit or disc may substitute for the salt bridge
Oxidation and Reduction and Electrochemistry Notes
Steps for finding electrochemical potential:
1) Write two ½ reactions written as reductions
2) Compare E’s––the reaction with the highest E is reduced, the other is oxidized
3) Flip the oxidized reaction, or change the sign of E
4) Balance/add the reaction
5) Erxn = Ereduced + Eoxidized
 If E > 0, the reaction is spontaneous
 If E < 0, the reaction is nonspontaneous
This is because:
 G = -nFErxn
where n = the number of moles of electrons,
F = Faraday’s constant, 96485 C/mol,
Erxn = electrochemical potential
Oxidation and Reduction and Electrochemistry Notes
The Nernst Equation:
The Nernst equation is used to relate a cell’s electrochemical potential, E, that is not at standard conditions
(1 atm. 25 ◦C, 1 M solutions) with an electrochemical potential that is at standard conditions, E.
E = E -
RT
ln(Q)
nF
where
E = cell potential under nonstandard conditions
E = cell potential under standard conditions
R = 8.314 VC/mol K
F = 96485 C/mol
Q = the reaction quotient (use initial concentrations)
T = temperature
N = number of moles of electrons
*As the concentration of the products of a redox reaction increases, the voltage decreases; and as the
concentration of the reactants in a redox reaction increases, the voltage increases.
Eo 
RT
ln K , K is the equilibrium constant
nF
o
o
*If E is positive, then K is greater than 1 and the forward reaction is favored. If E is negative, then K is
less than 1 and the reverse reaction is favored.
Example Problems
1. Consider the galvanic cell based on the reaction
Al 3  Mg  Al  Mg 2
The half reactions are
Al 3  3e   Al
E 0  1.66V
Mg 2  2e   Mg
E 0  2.37V
Give the balanced cell reaction and calculate E0 for the cell.
(See work on page 845)
Solution: 0.71 V
2. Using the data in Table 17.1, calculate the
G for the reaction
Cu 2  Fe  Cu  Fe 2
Is this reaction spontaneous?
Oxidation and Reduction and Electrochemistry Notes
(See work on page 850)
Solution: G  1.5  10 J , spontaneous
5
3. Describe the cell based on the following half reactions

VO2  2 H   e   VO 2  H 2 O
E 0  1.00V
Zn 2  2e   Zn
E 0  0.76V
T  25 o C
where

[VO2 ]  2.0 M
[ H  ]  0.50 M
[VO 2 ]  1.0  10  2 M
[ Zn 2 ]  1.0  10 1 M
(See work on page 855)
Solution: 1.89V
3.
Determine the cell potential for the rxn
Half-Reactions:
Al 3 (aq)  Mg (s)  Al (s)  Mg 2 (aq)
 
 
Al 3  3e   Al
Mg 2  2e   Mg
-1.66 V
-2.37 V
work
2
 
 
( Al 3  3e   Al )
2

-1.66 V
3 ( Mg  Mg  2e )
2.37 V
__________________________________________________________
2 Al 3 (aq)  3Mg ( s)  2 Al ( s)  3Mg 2 (aq)
4.
Write the line notation for the cell, given:
 
0.71 V
Ag   e   Ag
Fe 3  e   Fe 2
work Line Notation:
Pt ( s) | Fe 2 (aq), Fe3 (aq) || Ag  (aq) | Ag (s)
5.
Determine the free energy of the cell that ha s the rxn:
Cu 2 (aq)  Fe( s)  Cu(s)  Fe 2 (aq)
Cu 2  2e   Cu
Fe  Fe 2  2e 
   0.34 V
    0.44 V
__________________________________________________________
Cu 2  Fe  Fe 2  Cu
G  nF 
 
0.78 V
Oxidation and Reduction and Electrochemistry Notes
G  (2mol )(96,485
C
J
)(0.78 )
mol
C
G  -1.5 x 10 5 J
6.
Determine the cell potential for the galvanic cell with the following half-reactions at 25 °C with
the given concentrations: [Ag+] = 1.0 M, [H2O2] = 2.0 M, [H+] = 2.0 M.
Ag   e   Ag
H 2 O2  2 H   2e   2 H 2 O
work
    0.80 V
H 2 O2  2H  2e  2H 2 O
   1.78 V
________________________________________________________
2 Ag  H 2 O2  2H   2 Ag   2H 2 O
   0.98 V
-2 ( Ag   e   Ag )

 cell
 cell
 cell

RT
[ Ag  ]2
   cell 
ln(
)
nF [ H  ]2 [ H 2 O2 ]
J
(8.314
)( 298K )
(1.0M ) 2
K

mol
 0.98V 
ln(
)
C
(2.0M ) 2 (2.0M ) 2
(2mol)(96485
)
mol
 1.01 V