Download Topic 9 - Anderson High School

Topic 9
Oxidation and Reduction
9.1.1 Define oxidation and reduction in terms of
electron loss and gain.
Oxidation: The loss of electrons
Fe2+(aq) → Fe3+(aq) + eReduction: The gain of electrons
2H+(aq) + 2e- → H2(g)
9.1.1 Define oxidation and reduction in terms of
electron loss and gain.
Helpful Mnemonic
LEO goes GER
Loss of Electrons is
Oxidation
Gain of Electrons is
Reduction
9.1.2 Deduce the oxidation number of an
element in a compound.
A species is oxidized when it loses electrons.
– Here, zinc loses two electrons to go from neutral
zinc metal to the Zn2+ ion.
9.1.2 Deduce the oxidation number of an
element in a compound.
A species is reduced when it gains electrons.
– Here, each of the H+ gains an electron and they
combine to form H2.
9.1.2 Deduce the oxidation number of an
element in a compound.
• It may be easier to find what is being reduced
and oxidized by splitting the equation into
“half equations”.
• For example, with Zn(s) + 2H+(aq) → Zn2+(aq) + H2(g)
It can be split up as:
Zn(s) → Zn2+(aq) + 2eand 2H+(aq) + 2e- → H2(g)
9.1.2 Deduce the oxidation number of an
element in a compound.
• It is not always easy to split equations into half
equations.
• Consider the following reaction:
N2(g) + 3H2(g)  2NH3(g)
Can you tell which is being oxidized?
If not, then we need to use oxidation numbers.
9.1.2 Deduce the oxidation number of an
element in a compound.
Oxidation Number
The charge that an atom would have if all
covalent bonds were broken so that the more
electronegative element kept all the electrons.
9.1.2 Deduce the oxidation number of an
element in a compound.
Oxidation Number Rules
–
–
–
–
–
Elements in elemental state = 0
F = -1 (always)
O = -2 (except in H2O2 where its +1)
H = +1 (except in hydrides H-)
Halides = -1 except when bonded to oxygen or other
halides higher in the group (more reactive one will
be -1)
The sum of the oxidation numbers in a neutral
compound is 0.
The sum of the oxidation numbers in a polyatomic ion is
the charge on the ion.
9.1.2 Deduce the oxidation number of an
element in a compound.
Find the oxidation number for the
following:
Nitrogen in N2
=
Carbon in CH4
=
Sulfur in H2SO4
=
Phosphorous in PCl4+ =
Iodine in IO4=
Answers: 0, -4, +6, +5, +7
– Elements in elemental
state = 0
– F = -1 (always)
– O = -2 (except in H2O2
where its +1)
– H = +1 (except in hydrides
H-)
– Halides = -1 except when
bonded to oxygen or
other halides higher in the
group (more reactive one
will be -1)
9.1.3 State the names of compounds using oxidation
numbers.
For elements that have a variable oxidation number,
the oxidation state is signified by Roman
numerals.
Example: Fe+3 would be written as Iron(III)
How would you write the following?
FeCl2
FeCl3
MnO4- Cr2O3
Answers: iron(II) chloride, iron(III) chloride, permanganate (VII),
chromium(III) oxide
Challenge: How would you write the formula for
ammonium dichromate?
Answer: (NH4)2Cr2O7
9.1.4 Deduce whether an element undergoes oxidation
or reduction in reactions using oxidation numbers.
Let’s go back to the equation:
N2(g) + 3H2(g)  2NH3(g)
What is the oxidation number for nitrogen on both sides?
Has it been oxidized or reduced?
Answer: Oxidation number goes from 0 to -3. It has
gained electrons, therefore it has been reduced.
9.1.4 Deduce whether an element undergoes oxidation or
reduction in reactions using oxidation numbers.
Consider the reaction between MnO4− and C2O42− :
MnO4−(aq) + C2O42−(aq)  Mn2+(aq) + CO2(aq)
9.1.4 Deduce whether an element undergoes oxidation or
reduction in reactions using oxidation numbers.
+7
+3
+2
+4
MnO4− + C2O42-  Mn2+ + CO2
First, assign oxidation numbers.
Next, find out if carbon and manganese are being
oxidized or reduced.
Since the manganese goes from +7 to +2, it is reduced.
Since the carbon goes from +3 to +4, it is oxidized.
9.2.1 Deduce simple oxidation and reduction halfequations given the species involved in a redox reaction.
• Let’s look at an equation that we worked with
before….
MnO4− + C2O42-  Mn2+ + CO2
• What is wrong with this equation?
• Answer: It is not balanced!
• We have worked with half equations before
(zinc and hydrogen). Now we’ll dig deeper.
9.2.1 Deduce simple oxidation and reduction half-equations
given the species involved in a redox reaction.
General rules for balancing half equations
– 1)
– 2)
– 3)
– 4)
Balance atoms being oxidized or reduced
Add H20 to balance Oxygen atoms
Add H+(aq) to balance Hydrogen atoms
Add e- to balance charge
9.2.1 Deduce simple oxidation and reduction half-equations
given the species involved in a redox reaction.
Oxidation Half-Reaction
C2O42−  CO2
To balance the carbon, we add a coefficient of 2:
C2O42−  2 CO2
9.2.1 Deduce simple oxidation and reduction half-equations
given the species involved in a redox reaction.
Oxidation Half-Reaction
C2O42−  2 CO2
The oxygen is now balanced as well. To
balance the charge, we must add 2 electrons
to the right side.
C2O42−  2 CO2 + 2 e−
9.2.1 Deduce simple oxidation and reduction half-equations
given the species involved in a redox reaction.
Reduction Half-Reaction
MnO4−  Mn2+
The manganese is balanced; to balance the
oxygen, we must add 4 waters to the right
side.
MnO4−  Mn2+ + 4 H2O
9.2.1 Deduce simple oxidation and reduction half-equations
given the species involved in a redox reaction.
Reduction Half-Reaction
MnO4−  Mn2+ + 4 H2O
To balance the hydrogen, we add 8 H+ to the
left side.
8 H+ + MnO4−  Mn2+ + 4 H2O
9.2.1 Deduce simple oxidation and reduction half-equations
given the species involved in a redox reaction.
Reduction Half-Reaction
8 H+ + MnO4−  Mn2+ + 4 H2O
To balance the charge, we add 5 e− to the left
side.
5 e− + 8 H+ + MnO4−  Mn2+ + 4 H2O
9.2.2 Deduce redox equations using half-equations.
Combining the Half-Reactions
Now we evaluate the two half-reactions
together:
C2O42−  2 CO2 + 2 e−
5 e− + 8 H+ + MnO4−  Mn2+ + 4 H2O
To attain the same number of electrons on each
side, we will multiply the first reaction by 5
and the second by 2.
9.2.2 Deduce redox equations using half-equations.
Combining the Half-Reactions
5 C2O42−  10 CO2 + 10 e−
10 e− + 16 H+ + 2 MnO4−  2 Mn2+ + 8 H2O
When we add these together, we get:
10 e− + 16 H+ + 2 MnO4− + 5 C2O42− 
2 Mn2+ + 8 H2O + 10 CO2 +10 e−
9.2.2 Deduce redox equations using half-equations.
Combining the Half-Reactions
10 e− + 16 H+ + 2 MnO4− + 5 C2O42− 
2 Mn2+ + 8 H2O + 10 CO2 +10 e−
The only thing that appears on both sides are the
electrons. Subtracting them, we are left with:
16 H+ + 2 MnO4− + 5 C2O42− 
2 Mn2+ + 8 H2O + 10 CO2
9.2.1 Deduce simple oxidation and reduction half-equations
given the species involved in a redox reaction.
9.2.2 Deduce redox equations using half-equations.
Practice
Given two half-equations:
Cr2O72-(aq) → Cr3+(aq)
Fe2+ → Fe3+
Deduce the half-equations for each, then
deduce the redox equation.
Answer
Cr2O72-(aq) + 14H+(aq) + 6Fe2+(aq) →
2Cr3+(aq) + 7H2O(l) + 6Fe3+(aq)
9.2.3 Define the terms oxidizing agent and reducing agent.
Oxidizing agent: Substance that is reduced and
causes the oxidation of another substance in a
redox reaction.
Reducing agent: Substance that is oxidized and
causes the reduction of another substance in
a redox reaction.
9.2.4 Identify the oxidizing and reducing agents
in redox equations.
Identify the oxidizing and reducing agents in
the following equations:
Sn2+(aq) + 2Fe3+(aq) → Sn4+(aq) Fe2+(aq)
Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)
9.2.4 Identify the oxidizing and reducing agents
in redox equations.
Deduce the following half equations, deduce
the redox equation, and identify the oxidizing
agent and the reducing agent.
• MnO4-(aq) → Mn-2(aq)
• SO2(aq) → SO42-(aq)
9.3.1 Deduce a reactivity series based upon the
chemical behaviour of a group of oxidizing and
reducing agents.
• Recall in acids and bases that a strong acid
had a weak conjugate base.
• Same in redox reactions. The conjugate of a
powerful oxidizing agent is a weak reducing
agent.
Strong oxidizing
agent
F2 + 2e- ↔ 2F-
Weak reducing
agent
9.3.1 Deduce a reactivity series based upon the chemical
behaviour of a group of oxidizing and reducing agents.
• Mr. F can really attract the
electrons (more
electronegative).
• When Mr. F has the electrons,
he doesn’t want to let them
go.
• So although he is a good
oxidizing agent, he is a poor
reducing agent. (He doesn’t
like to reduce the number of
his electrons!)
9.3.1 Deduce a reactivity series based upon the
chemical behaviour of a group of oxidizing and
reducing agents.
• Think back to Topic 3 on Periodicity.
• What are the trends in electronegativity?
9.3.1 Deduce a reactivity series based upon the chemical
behaviour of a group of oxidizing and reducing agents.
Compare
What exception
do you see?
Hydrogen (Lithium is another
exception)
9.3.2 Deduce the feasibility of a redox reaction from a
given reactivity series.
Cl2(aq) + 2I-(aq) → I2(aq) + 2Cl-(aq)
Feasible?
A: Yes
I2(aq) + 2Cl-(aq) → Cl2(aq) + 2I-(aq)
Feasible?
A: No
Chlorine attracts electrons more strongly than iodine,
so chlorine is a better oxidizing agent.
http://www.youtube.com/watch?v=SQit4Yxl4HU&list=EC5C
3F606F1DFA4133
9.3.2 Deduce the feasibility of a redox reaction from a
given reactivity series.
Zn(s) + Cu2+(aq) → Cu(s) + Zn2+(aq)
Feasible?
A: Yes
Cu(s) + Zn2+(aq) → Zn(s) + Cu2+(aq)
Feasible?
A: No
These examples are all displacement reactions,
because they involve a more reactive metal or
non-metal displacing the reactive one from its salt.
9.4.1 Explain how a redox reaction is used to produce
electricity in a Voltaic cell.
• Anode(-): Oxidation, forms a negative charge
• Cathode(+): Reduction, forms
a positive
charge
e
e
-
ee e2e+
2+
e
e
9.4.1 Explain how a redox reaction is used to produce electricity in a
Voltaic cell.
• Lets harness some Energy!!
Problem, the highly negative charge on
electrode causes (+) ions to be attracted
back
Zn(s)
- e
e
e
e
2+
2+
2+
2+
2+
Solution
Balance (-) charge
by replacing it
with some more
negative ions
Cu(s)
9.4.1 Explain how a redox reaction is used to produce electricity in a
Voltaic cell.
• Lets harness some Energy!!
+
+
-
-
+
-
Zn(s)
Cu(s)
+
+
-
-
e
e
2+
e
2+
2+
e
2+
2+
http://www.dynamicsci
ence.com.au/tester/solu
tions/chemistry/redox/g
alvan5.swf
9.4.2 State that oxidation occurs at the negative
electrode (anode) and reduction occurs at the positive
electrode (cathode).
• ANOX and RED CAT
• The oxidation occurs
at the anode.
• The reduction occurs
at the cathode.
• Which of the metals is
being reduced?
• So which is the
cathode?
9.4.2 State that oxidation occurs at the negative electrode
(anode) and reduction occurs at the positive electrode (cathode).
• Lead and zinc are set up in a voltaic cell.
• Which one would be oxidized? Which one is
being reduced?
• A: Zinc is being oxidized. Lead is being
reduced.
• Which one would be the cathode and which
would be the anode?
• Zinc would be the anode, lead is the cathode.
9.5.1 Describe, using a diagram, the essential
components of an electrolytic cell.
Need to have a liquid containing ions, which is called an electrolyte.
(http://www.youtube.com/watch?v=Pu9XZyQPsy0)
9.5.2 State that oxidation occurs at the positive electrode
(anode) and reduction occurs at the negative electrode
(cathode).
• The anode attracts anions.
• When the anions reach they anode, they lose
electrons.
• So are they oxidized or reduced?
• A: oxidized
• When cations reach the cathode they gain
electrons and they are reduced.
9.5.3 Describe how current is conducted in an
electrolytic cell.
• Electricity is supplied from an external source
and used to create a non-spontaneous
reaction.
• Electrolyte solution can conduct electricity
because the ions move towards oppositely
charged electrodes.
• Electrons flow From Anode To CAThode
(FATCAT)
9.5.4 Deduce the products of the electrolysis of a
molten salt.
Sodium chloride
• Negative chloride ions are attracted to the
positive ions. There they lose electrons and
are oxidized to chlorine gas:
2Cl-(l) → Cl2(g) + 2e• Positive sodium ions are attracted to the
negative cathode. They gain electrons and are
reduced to sodium metal:
Na+(l) + e- → Na(l)
9.5.4 Deduce the products of the electrolysis of
a molten salt.
Question
For every 2 mol of electrons that flow through
the circuit, how many mol of chlorine gas and
sodium metal will be produced?
A: 1 mol of chlorine gas and 2 mol of sodium.