Oxidation
&
Reduction
IB Chemistry
Oxidation Numbers
Rules for Assigning Oxidation Numbers
1 Oxidation numbers always refer to single ions
2 The oxidation number of an atom is always 0
3 The sum of the oxidation numbers in a neutral compound is 0
4 The sum of the oxidation numbers in a polyatomic ion is equal to the
charge of the ion
5 The oxidation number of Hydrogen is usually +1 (-1 when bonded to
a metal)
6 The oxidation number of Oxygen is always -2 (except for H2O2)
7 The oxidation numbers of alkali metals are +1
8 The oxidation numbers of alkaline earth metals are +2
Examples
• NO2
O= -2
Ø = N +(-2x2)  N= +4
• N2O5
O= -2
Ø = 2N +(-2x5)  N= +5
• HClO3
O= -2, H= +1
Ø = Cl +(-2x3) +1  Cl = +5
• Ca(NO3)2
O= -2, Ca= +2
• Fe(OH)3
O= -2, H= +1
• CO3-2
O= -2
-2 = C+(-2x3)  C =+4
• SO4-2
O= -2
-2 = S+(-2x4)  S =+6
Ø = 2N+(-2x6)+2 N=+5
Ø = Fe+(-2x3)+(1x3) Fe=+3
Determining What’s Happened…
• Careful examination of the oxidation
numbers of atoms in an equation
allows us to determine what is
oxidized and what is reduced in an
oxidation-reduction reaction
Example
• An increase in the oxidation number indicates that
an atom has lost electrons and therefore oxidized.
• A decrease in the oxidation number indicates that
an atom has gained electrons and therefore
reduced
• Example
Zn +
CuSO4  ZnSO4 + Cu
0
+2 +6 -2 +2 +6 -2
0
Zn: 0  + 2  Oxidized (lost electrons)
Cu: +2  0  Reduced (gained electrons)
Exercise
For each of the following reactions find
the element oxidized and the element
reduced
Cl2
Cu
+
KBr

KCl +
Br2
+ HNO3  Cu(NO3)2 + NO2 + H2O
HNO3 +
I2
 HIO3 +
NO2
Exercise
For each of the following reactions find
the element oxidized and the element
reduced
Cl2
0
+
KBr
+1-1

KCl +
+1-1
Br increases from –1 to 0 -- oxidized
Cl decreases from 0 to –1 -- Reduced
K remains unchanged at +1
Br2
0
Exercise
For each of the following reactions find
the element oxidized and the element
reduced
Cu + HNO3  Cu(NO3)2 + NO2 + H2O
0
+1+5-2
+2 +5 -2
+4 –2 +1-2
• Cu increases from 0 to +2. It is oxidized
• Only part of the N in nitric acid changes from +5
to +4. It is reduced
• The nitrogen that ends up in copper nitrate
remains unchanged
Exercise
For each of the following reactions
find the element oxidized and the
element reduced
HNO3 +
I2
 HIO3 +
NO2
1 +5 -2
0
+1+5-2
+4-2
•
N is reduced from +5 to +4. It is reduced
•
I is increased from 0 to +5 It is oxidized
•
The hydrogen and oxygen remain unchanged.
Oxidation-Reduction Reactions
• All oxidation reduction reactions have
one element oxidized and one element
reduced
• Occasionally the same element may
undergo both oxidation and reduction.
This is known as an auto-oxidation
reduction
Agents
• The oxidizing agent takes the electron(s) and is
itself reduced
• The reducing agent loses the electron(s) and is
itself oxidized.
Half Reactions
• Show what is happening to the oxidized
species or the reduced species (tells ½ the story)
+2 -2
0
+2 -2
0
•Zinc is oxidized & is the reducing agent
•Copper is reduced & is the oxidizing agent
•Sulfate acts as a spectator ion – doesn’t do anything
Movement of e• Write a half reaction for the reduced species
• Write a half reaction for the oxidized species
• Or…
Ionic Equations
• Add the half reactions together
• The e- must cancel out
What if e-’s don’t cancel?
• Half equation 2 needs to be multiplied by 2 to
achieve equal amounts of e- on both sides
• Now they can be added together:
Activity Series – reducing agents
• Activity series allow you to
predict whether a redox
reaction will happen or not.
– Elements higher on the chart
will displace a metal ion of an
element lower on the chart.
– Mg + Zn2+ will react to form Zn
and Mg2+
– More reactive elements are
stronger reducing agents
Redox reactions in Acidified Solutions
For each half equation:
1. If a metal is present, add coefficients to balance
2. Add water to balance oxygens
3. Add hydrogen ions to balance H
4. Add electrons to balance the charge
5. Balance half equations
6. Add half equations
7. Simplify (now an algebraic expression)
Wednesday 3/26/14
• Objective: SWBAT determine the flow of
electrons in a voltaic cell compared to an
electrolytic cell
Quiz Monday
• HW: Finish packet
• Warm up: Determine the oxidation
states of each element
HNO3 +
I2
 HIO3 +
NO2
Voltaic Cells
• Two half cells connected
together – allows for
electrons to be transferred
during the redox reaction
• Produces energy in the form
of electricity
• Half cells: a metal in contact
with an aqueous solution of
its own ions
• Zn is higher in the activity series
• Spontaneous redox reaction
• Electrons will flow towards Cu
taking place
Voltaic Cells
Electron movement
• Oxidation occurs at
the anode
• Reduction occurs at
the cathode
• Electrons flow
towards the cathode
Voltaic Cells: Purpose of Salt Bridge
Once even one
electron flows from
the anode to the
cathode, the charges
in each beaker would
not be balanced and
the flow of electrons
would not be able to
continue.
Voltaic Cells: Ions, not electrons, move through
salt bridge
• Therefore, we use a
salt bridge, usually a
U-shaped tube that
contains a salt
solution, to keep the
charges balanced.
– Cations move toward
the cathode.
– Anions move toward
the anode.
How can you remember?
a RED CAT and AN OX
REDuction = CAThode
ANode = OXidation
Electrolysis
• Used to make non-spontaneous redox
reactions occur
– Provide energy in the form of electricity from an
external source
• Electricity is passed through an electrolyte
– Electrical energy is converted to chemical energy
• Reverse of Voltaic cells
• Electrolytes: conduct energy in solution
Anodes & Cathodes
Voltaic Cells
Anode
Oxidation occurs here
Cathode
Reduction occurs here
Electrolytic Cells
+
Oxidation occurs here
Reduction occurs here
Electrolytic cells are “pumping” electrons
– not spontaneous
Voltaic cells are spontaneously occurring
+
-
Electrolysis of Molten NaCl
Na reduced, Cl oxidized