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Redox Reactions
 Cu(s) + 2Ag+(aq) +2NO3-(aq)  Cu2+(aq) + 2NO3- (aq) + 2Ag(s)
 Removing the spectator ions we get our net ionic equation:
Cu(s) + 2Ag+ (aq)  Cu2+(aq) + 2Ag (s)
Copper began as a neutral atom with no charge but changes
into an ion with a 2+ charge. This happens when it loses 2
electrons. Cu (s)  Cu2+ (aq) + 2 eCopper was oxidized because it lost electrons. Silver went
from an ion Ag+ to a neutral atom Ag. The only way this can
happen is to gain electrons. It has been reduced.
LEO the
Lion Says
Loss of
Electrons is
Gain of
Electrons is
 Oxidation numbers are our system for keeping track of
what gains and what loses electrons.
 An oxidation number is a positive or negative number
assigned to an atom in a molecule or ion that reflects a
partial gain or loss of electrons.
 Main Rules:
1. The oxidation number of a pure element (not an ion)
is zero (0).
2. The oxidation number of a monatomic ion (by itself or
in an ionic compound) is equal to its charge.
3. The oxidation number of hydrogen is almost always +1
when it is in a compound.
4. The oxidation number of oxygen is almost always -2 when
it is in a compound.
Two exceptions: peroxides O is -1
oxygen with fluorine O is +2.
5. The sum of the oxidation number in a compound is
Example: Mn2O7
O Is -2 (rule 4)
-2 x 7 = -14 total
The sum of the oxidation numbers must be zero.
The total of the oxidation numbers of Mn must be +14.
+14/2 = +7. The oxidation number of Mn must be +7.
6. The sum of the oxidation number of a polyatomic ion
is equal to the charge on that ion.
Example: Cr2O7 2Oxygen - -2 x 7 = -14
The sum of the oxidation numbers must be -2 (instead
of zero) as that is the overall charge on the ion.
-14 + 12 = -2
+12/2 = +6 Cr
The oxidation number always refers to each individual
atom in the compound not the total for that element.
 Oxidation cannot occur without reduction
 Reducing Agent: substance that is oxidized
 Oxidizing Agent: substance that is reduced
 The oxidizing agent is usually the entire compound
that the element that is being reduced is in, not just
the element that is being reduced. Agents are always
on the reactants side.
 Sodium metal reacts with chlorine gas:
Na + Cl2  2 NaCl
Na 0 to +1
Cl 0 to -1
+1 -1
lost 1 e
gained 1 e
This is a redox reaction.
oxidized (reducing agent)
reduced (oxidizing agent)
Zn + HNO3  Zn(NO3)2 + NO2 + H2O
Assign oxidation #’s
Is this a redox reaction?
What is the oxidizing agent?
Remember that an increase in the oxidation number
means oxidation
Decrease in the oxidation number means reduction.
4 HCl + O 2  2 H2O + 2 Cl2
In Ox no
Final Ox No
e’ lost or Oxidized or
Oxidized (lost)
reducing agent
oxidizing agent
 Observations:
Cu(s) + Ag+ (aq)  Cu 2+(aq) + Ag (s)
 Looking at the number of atoms the net ionic equation
appears balanced. The charges however ARE NOT.
in ox no
change in e’
change in e
balance for electrons
Coefficient Total
x 1 =
gained 1
x 2 =
Cu (s) + 2 Ag + 
Cu 2+ (aq) + 2 Ag (s)
SnCl2 + HgCl2  SnCl4 + HgCl
Final Change Coeff
SnCl2 + 2 HgCl2  SnCl4 + 2HgCl
Total e
MnO41- + Fe2+ + H1+  Mn2+ + Fe3+ + H2O
element in ox final ox change
x 1 = 5
x 5 = 5
MnO41- + 5 Fe2+ + H1+  Mn2+ + 5 Fe3+ + H2O
But the H and O are not balanced:
MnO41- + 5 Fe2+ + 8H1+  Mn2+ + 5Fe3+ + 4H2O
NH3 + O2 
in. ox final
NO2 + H2O
change no. atoms no. e’ balance
x 4 = 28
x 7 = 28
Which O does the coefficients go in front of??????
Do the N first, O in diatomic second, H last.
4 NH3 + 7 O2  4 NO2 + 6 H2O
 Using this method we will break an equation into the
oxidation reaction and the reduction reaction.
These separate equations are referred to as halfreactions because the two halves cannot occur with out
the other. (or two halves make a whole)
The spectator ions are removed from the equations.
Each half reaction is balanced separately. Electrons are
added to balance the charges.
The electrons lost must equal the electrons gained.
Everything is put back together including the
spectator ions.
Mg (s) + Cl2(g)  MgCl2(s)
What is oxidized? What is reduced?
Mg is oxidized 0 to +2
Cl is reduced 0 to -1
Mg  Mg+2
Mg  Mg2+ + 2 e’
Cl2  2 ClCl2 + 2 e’  2 Cl-
electrons are equal so it is already balanced.
Mg (s) + Cl2(g)  MgCl2(s)
Cu (s) + AgNO3(aq)  Cu(NO3)2 + Ag (s)
Cu  Cu2+ + 2e’
Ag+ + 1 e’  Ag
The electrons do not balance.
Cu  Cu2+ + 2e’
2Ag+ + 2 e’  2Ag
Cu + 2Ag+  Cu2+ + 2Ag
Return spectator ions:
Cu (s) + 2 AgNO3(aq)  Cu(NO3)2 + 2Ag (s)
MnO4- + Fe2+ + H+  Mn2+ + Fe3+ + H2O
Fe oxidized Fe +2 to +3 Mn reduced + 7 to +2.
Fe2+  Fe3+ + 1 e’
Mn7+ + 5 e’  Mn2+
5 Fe2+  5 Fe3+ + 5 e’
MnO4- + 5 Fe + H+  Mn2+ + 5 Fe3+ + H2O
(The hydrogen and the oxygen must be included in the
half reaction and balanced.).
MnO4- + 5 Fe2+ + 8 H+  Mn2+ + 5 Fe3+ + 4 H2O
 During redox reactions, electrons pass (flow) from one
substance to another.
 Electrochemistry is the branch of chemistry that deals with
the conversion of chemical energy to electrical energy.
1. Electrochemical Cells – spontaneous chemical reactions
convert chemical energy into electrical energy. Batteries
are an example.
2. Electrolytic Cells – electrical energy is used to cause nonspontaneous chemical reactions to occur. Rechargeable
batteries and electroplating are examples.
 The electrons that are released by the oxidation half
reaction are passed along to the reduction reaction.
 An external circuit needs to be created. Reactions will
occur without one but electricity will not be created.
Zn(s) + Cu2+ (aq)  Zn2+ (aq) + Cu(s)
zinc is oxidized
(reducing agent)
copper is reduced (oxidizing agent)
An electrolytic solution (conducts electricity due to
the presence of ions) is needed.
Each beaker is a half cell. The metal strips become the electrodes.
The electrodes are connected by a wire. This becomes the external circuit.
A salt bridge contains an electrolytic solution. This allows the electrons to flow
And becomes the internal circuit.
Anode = Oxidation
- post of the cell
source of electrons
An Ox
Reduction = Cathode
+ post of the cell
Consumes electrons
Red Cat
 Electrons travel from the Zn anode to the Cu cathode
through the wire of the external circuit.
 At the anode Zn2+ ions go into solution.
 The excess positive charges attract the negative NO3ions from the salt bridge.
Electrochemical Cell Simulation
 External Circuit – Electrons flow from anode to
 Internal Circuit- anions to anode; cations to cathode
 An entire complete cell is comprised of:
2 half-cells (electrodes in their solution)
internal circuit (salt bridge and the half cells)
external circuit (the wire connecting the two
 When the cell reaches equilibrium, the voltage will be
 What determines which element is oxidized and which
is reduced?
Metals like to lose electrons so they tend to be
oxidized. Metals are arranged according to the activity
In our cell example Zn is the anode (oxidation) while
Cu is the cathode (reduction).
Where are these elements on the activity series?
Zinc is quite a ways higher on the series and therefore
more easily oxidized.
 This table allows us to determine the voltage of the
electrochemical cells.
 All values on the table are determined based on a halfreaction with a hydrogen half-cell.
2H+(aq) + 2e’  H2(g)
Eo = 0.00 v
• Means: 25 oC, 100 kPa, 1 mol.L-1
Positive values on this table mean that they are better
at competing for electrons and will be reduced. The
hydrogen will be oxidized.
→ Cu(s)
H2 (g) → 2H (aq) + 2e___________________________
Cu2+(aq) + H2 (g) → 2H+(aq) + Cu(s)
0.34 V
0.00 V
0.34 V
In the Table of Standard Reduction Potentials that zinc has a negative E° indicating that it is not as
good at competing for electrons as hydrogen.
Zn2+(aq) + 2e- → Zn(s) E° = -0.76 V
Therefore if zinc and hydrogen are paired together in an electrochemical cell, the hydrogen would be
reduced (gain the electrons) and zinc would be oxidized (losing electrons). To determine the net
redox reaction as well as the voltage of the electrochemical cell we reverse the zinc equation (write it
in oxidation form), and also reverse it's sign before adding the equations and E° together:
Zn(s) → Zn2+(aq) + 2e2H+(aq) + 2e- → H2 (g)
Zn(s)+ 2H+(aq) → Zn2+(aq) + H2 (g)
0.76 V
0.00 V
0.76 V
 We can now use the table to calculate the voltage of
our zinc-copper cell as well as use the table to explain
why zinc is the anode (oxidized) and copper is the
cathode (reduced).
 Locate the Cu/Cu2+ half-reaction.
Cu: Cu2+ (aq) + 2e’  Cu(s) Eo = 0.34 V
Zn: Zn2+ (aq) + 2e’  Zn(s) Eo = -0.76 V
The copper is larger than zinc so it will be reduced. Zinc
will be oxidized.
Zn  Zn2+ + 2 e’
0.076 V
Cu2+ + 2 e’  Cu
0.034 V
Cu2+ (aq) + Zn(s)  Zn2+ (aq) + Cu (s)
1.10 V
 Always reverse the half reaction that will result in a
positive value for Eo when the equations are added
 A positive Eo value means that the reaction is
spontaneous and electrochemical cells always involve a
spontaneous chemical reaction.
 Batteries are electrochemical cells used to generate
Types of Batteries:
1. Dry Cells (Primary batteries)
non rechargeable; electrolytes are paste not liquid;
used for flashlights, radios; toys etc.
Consists of a zinc case (anode) , a graphite rod (cathode)
and an electrolytic paste
2. Secondary Batteries
An example is a car battery (lead-acid).