Chapter 21- Electrochemistry
Reduction-Oxidation or
REDOX chemistry
21-1: Introduction to
Electrochemistry

Background: In Chemistry 11 we saw many
Single Replacement reactions, such as:


Mg(s) + 2HCl(aq) ssd MgCl2(aq) + H2(g)
In fact, this is a two part “REDOX” reaction:
1)
Mg(s) ssd Mg2+ (aq) + 2 e—
Mg (s) has LOST two Electrons during an
OXIDATION reaction (* LEO = loss of
electrons is oxidation)
2)
2H+ (aq) + 2 e— ssd H2 (g)
The H+ ions have GAINED two Electrons during a
REDUCTION reaction (*GER = gain of
electrons is reduction)
Note: The Cl— ions in the original reaction did not
participate and are SPECTATOR ions
Background…. Cont’d

Oxidation and Reduction reactions occur
simultaneously… every electron lost in
the oxidation is gained in the reduction.

As each of these reactions is one-half of
the overall reaction, we call them… HALFREACTIONS!

Back to our example:
1)
2)
Mg(s) sd Mg2+ (aq) + 2 e—
2H+ (aq) + 2 e— sd H2 (g)
OXIDATION
REDUCTION
Mg(s) + 2H+(aq) sd Mg2+(aq) + H2 (g)
These two reactions can be combined to yield the
overall NET reaction.
Mg gets oxidized, but CAUSES the reduction of H+,
and Mg is called the REDUCING AGENT
H+ gets reduced, but CAUSES the oxidation of Mg
and H+ is called the OXIDIZING AGENT

The Reducing Agent gets oxidized and the
Oxidizing Agent gets reduced
2+
2+
Another Example

3CuCl2(aq) +2Al(s) ssd 2AlCl3(aq) + 3Cu(s)

Two half reactions:

1)
Cu2+(aq) + 2e— ssd Cu(s)
Cu2+ ion GAINS 2 electrons and is
REDUCED
 2)
Al(s) ssd Al3+ (aq) + 3e—


Aluminum atom LOSES 3 electrons and is
OXIDIZED

Cu2+ ion is the OXIDIZING AGENT and gets
REDUCED

Al atom is the REDUCING AGENT and gets
OXIDIZED

REMEMBER: LEO the Lion says GER

-- LEO= Loss of Electrons is Oxidation

-- GER= Gain of Electrons is Reduction
21-2 Oxidation Numbers
Oxidation and reduction can be explained
in terms of the loss and gain of electrons.
Oxidation numbers are used to help keep
track of electrons in chemical reactions
 OXIDATION NUMBER = the theoretical
charge an atom would have in a molecule
if every molecule was IONIC. This is as if
the electrons were completely transferred.

For example, in the compound H2O:

O has an electronegativity of 3.16

H has an electronegativity of 2.1
 Since O is more electronegative than H,
then we assume that the O removes the
electron from each H and the oxidation
numbers are: H = +1
O = -2

Redox reactions can be defined in terms
of oxidation number:
 An element is said to be oxidized if its
oxidation number increases in a reaction.
An element is said to be reduced if its
oxidation number decreases in a reaction.

Rules for Assigning Oxidation #’s
Rule #1. The oxidation number of an atom
of any element in its elemental state
(uncombined form) is ZERO, no matter
how complex the molecule.

H2 = 0, F2 = 0. Be = 0. Li = 0, Na = 0,
O2 = 0, P4 = 0, S8 = 0

RULE #2. For an ion composed of only one
atom, the oxidation number is equal to the
charge on the ion.
 K+ = +1, Mg2+ = +2, Al3+ = +3, F- = -1, O2- = -2

All alkali metal ions (Li+, Na+, K+, Rb+, Cs+,
Fr+) have an oxidation number of +1.

All alkaline earth metal ions (Be+2, Mg+2,
Ca+2, Sr+2, Ba+2, Ra+2) have an oxidation
number of +2.

Rule #3: The oxidation number of oxygen
in most compounds is –2 (e.g. H2O, CaO)
 There are three exceptions to this rule:

In OF2 O=+2 since F is more
electronegative than oxygen

In H2O2 , or HO2- , or O22- (the peroxide
family)
O= -1
 O2 : the oxidation number is 0 (see rule
#1)


Rule # 4. Fluorine has an oxidation
number of –1 in all of its compounds.
Rule #5. The oxidation number of
hydrogen is mostly +1, except when it is
bonded to elements less electronegative
than itself or if it is in its elemental form.

NaH
Na = +1, H = -1

CH4
C = -4, H = +1



H2
H=0

Rule #6. In a NEUTRAL molecule, the
sum of the oxidation numbers of all the
atoms must be ZERO. In a POLYATOMIC
ION, the sum of the oxidation numbers of
all the atoms in the ion must be equal to
the NET CHARGE of the ion.
Examples
Assign Oxidation Numbers to each
element in the following:
 a) NH4+
 b) Rb2O
 c) H2Cr2O7

Homework



Optional: Read HEBDEN WORKBOOK
Section 5-1 "Introduction". Do Exercises p.192
#1-2.
3. Optional: Read HEBDEN WORKBOOK
Section 5-2 "Oxidation Numbers". Do Exercise
p.194 #3-5.
4. Do HANDOUT "Section 21•1/21•2
Provincial Exam Questions".