Unit 7: Redox &
Electrochemistry
What’s the point ?
REDOX reactions are important in …
C3H8O + CrO3 + H2SO4 
Cr2(SO4)3 + C3H6O + H2O
• Purifying metals
(e.g. Al, Na, Li)
• Producing gases
(e.g. Cl2, O2, H2)
• Electroplating metals
• Electrical production (batteries, fuel cells)
• Protecting metals from corrosion
• Balancing complex chemical equations
• Sensors and machines (e.g. pH meter)
What is Redox?
• REDOX stands for REDuction/OXidation
• Oxidation is often thought of as a combination
of a substance with oxygen (rusting, burning)
• Just like with acid/base definitions the
definition of oxidation is expanded
• Oxidation refers to a loss of electrons
• Reduction refers to a gain of electrons
• As a mnemonic remember LEO says GER
• Loss Electrons = Oxidation
• Gain Electrons = Reduction
• Read 12.1 (pg. 443 - 445)
Testing concepts
Q- What is oxidation? What is reduction?
Represent each as a chemical equation.
A- oxidation = loss of e– … X X+ + e–
reduction = gain of e– … X + e– X–
Q- Why are 2Na + Cl2  2NaCl & 2H2 + O2  H2O
considered redox reactions?
A- Both involve the transfer of electron density
(Na has no charge, the atoms in diatomic
molecules have no partial charge. After reaction
the atoms have different shares of the electrons
because of different EN values)
Q- Is it possible to oxidize a material without
reducing something else?
A- No. A lost e– is taken up by something else.
Testing concepts
Q- Define oxidizing agent, reducing agent.
A- An oxidizing agent causes oxidation by being
reduced itself
A reducing agent causes reduction by being
oxidized itself
Q- PE 1
A- CaCl2 is an ionic compound with a positive
calcium ion and negative chlorine ions
Ca + Cl2  CaCl2
Ca  Ca2+ + 2e–, Cl2 + 2e– 2Cl–.
Thus Ca is losing electrons (oxidation) and
Cl is gaining electrons (reduction).
These are called “half reactions”
Oxidation numbers
• We will see that there is a simple way to keep
track of oxidation and reduction
• This is done via “oxidation numbers”
• An oxidation number is the charge an atom
would have if electrons in its bonds belonged
completely to the more electronegative atom
• E.g. in HCl, Cl has a higher EN (pg. 255).
Thus, oxidation numbers are Cl = -1, H = +1
• Notice that oxidation numbers are written as
+1 vs. 1+ to distinguish them from charges.
• Instead of referring to EN chart, a few rules
are followed to assign oxidation numbers
• Refer also to study note
Rules (and rationale - 12.2)
1. Any element, when not combined with atoms
of a different element, has an oxidation # of
zero. (O in O2 is zero)
2. Any simple monatomic ion (one-atom ion) has
an oxidation number equal to its charge (Na+
is +1, O2– is –2)
3. The sum of the oxidation numbers of all of the
atoms in a formula must equal the charge
written for the formula. (if the oxidation
number of O is –2, then in CO32– the oxidation
number of C is +4)
Rules (and rationale - 12.2)
4. In compounds, the oxidation # of IA metals is
+1, IIA is +2, and aluminum (in IIIA) is +3
5. In ionic compounds, the oxidation # of a
nonmetal or polyatomic ion is equal to the
charge of its associated ion. (CuCl2, Cl is –1)
6. F is always –1, O is always –2 (unless
combined with F), H is usually +1
or rule 5
rule 6 3 6 4 3 6 3 6 6 3 5 6 3 6
total +1 +5 -6 +2 +12 -14 -4 +6 -2 +1 -1 +2 +5 -8
Ox.# +1 +5 -2 +1 +6 -2 -2 +1 -2 +1 -1 +1 +5 -2
HNO 3 K2Cr2O7 C2H6O AgI H2PO4–
PE 2 (450), 12.9, 12.12 (484), 12.10, 12.11, 12.13 (484)
More practice
PE 2 pg. 450: answers in back of book
12.9, 12.12 (pg. 484): answers in back of book
12.10: a) Cl=+7, b) Cr=+3, c) Sn=+4, d) Au=+3
12.11: a) Na= +1, H = +1, P= +5, O= -2
b) Ba= +2, Mn= +6, O= -2
c) Na= +1, S= +2.5, O= -2
d) Cl= +3, F= -1
12.13: a) +2, b) +4, c) +3, d) +5, e) -2, f) 0,
g) –1, h) -3 I) –1/3
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