Chemistry 12
Unit 5 – Oxidation Reduction
Chemistry 12 – Unit 5: Electrochemistry
Definitions: (species means atom, ion or molecule)
Electrochemistry – ________________________________________________
__________________________________________________________(p. 189)
Oxidation – a species undergoing oxidation loses electrons
(charge becomes more positive)
Reduction – a species undergoing reduction gains electrons
(charge becomes more negative)
Oxidizing agent – The species being reduced
(gains electrons, causes the other one to be oxidized)
LEO says GER
Losing
Electrons is
Oxidization
Gaining
Electrons is
Reduction
Reducing agent – The species being oxidized
(loses electrons, causes the other one to be reduced)
Redox Reaction – ___________________________________________
(p. 190)
Redox – Short for Oxidation – Reduction
2 e-
OAR
Cu2+ (aq) + Zn (s)
E.g.)
Oxidizing
agent
Æ Cu(s) + Zn2+(aq)
Reducing
agent
The
Oxidizing
Agent is
Reduced
Redox identification
Reduction – charge decreases gets more –ve (gains electrons)
E.g.) Pb2+(aq) + Mg0(s) Æ Pb0(s) + Mg2+(aq)
Oxidation - Charge increases gets more +ve (loses electrons)
Question
In the reaction:
2Fe2+ + Cl2 Æ 2Fe3+ + 2Cl-
Identify the following:
a)
b)
c)
d)
The Oxidizing Agent: ___________
The species being oxidized:______
The reducing agent:____________
The species being reduced:______
Unit 5-Oxidation-Reduction NOTES
e)
f)
g)
h)
The species gaining electrons:____
The species losing electrons:_____
The product of oxidation_________
The product of reduction_________
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Chemistry 12
Unit 5 – Oxidation Reduction
Half-Reactions
-Redox reactions can be broken up into oxidation half reactions & reduction half reactions.
e.g.) Redox rx: Pb2+(aq) + Zn(s) Æ Pb(s) + Zn2+(aq)
Electrons on the LEFT side (or GER)
Means REDUCTION
Reduction Half-rxn: Pb2+(aq) + 2e- Æ Pb(s)
Oxidation Half-rxn:
Write the oxidation half reactions & reduction half reactions for
the following redox reactions.
E.g. 1
Pb2+(aq) + Zn(s) Æ Pb(s) + Zn2+(aq)
Oxidation half rx: ___________________________
Reduction half rx: ___________________________
E.g. 2
F2(g) + Sn2+(aq) Æ 2F-(aq) + Sn4+(aq)
Note: Half-rx’s always have e-‘s, redox
(oxidation-reduction) reactions never
show e-‘s!
Oxidation half rx: ___________________________
Reduction half rx: ___________________________
Read Hebden p.189-191
Do #1-2 p. 192
5.2 Oxidation numbers
Oxidation numbers (p. 193) -________________________________________
__________________________________________________ (even if it’s not!)
Rules to find oxidation number of an atom
1) In elemental form: Oxidation number of atoms = 0
(Single atoms of monatomic elements) or (diatomic molecules of diatomic
elements)
Eg) Mn, Cr, N2, F2, Sn, O2, etc.
The oxidation # of each atom = 0
2) In monatomic ions: oxidation # = charge
Eg)
In Cr3+ -oxidation # of Cr = +3
S2- -oxidation # of S = -2
3) In ionic compounds
a) the oxidation # of Alkali Metals is always +1
eg) NaCl
Ox # of Na & K = +1
K2CrO4
Unit 5-Oxidation-Reduction NOTES
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Chemistry 12
Unit 5 – Oxidation Reduction
b) the oxidation # of Halogens when at the end (right side) of the formula
is always –1
Ox # of Cl, Br and F = -1
eg) CaCl2 AlBr3
KF
Note: Halogens are not always –1! (Only when it is written last in formula.)
4) In molecules or polyatomic ions:
a) Ox. # of oxygen is almost always –2
e.g.) KOH CrO42- Li3PO4
Ox # of
O is –2
(Remember, “O” in O2 has
an Ox. # of ______)
b) An exception is Peroxides in which ox. # of O = -1
Hydrogen Peroxide: H2O2
Ox # of O’s = -1
Alkali Peroxides: Na2O2
5) In molecules or ions:
a) Hydrogen is almost always +1
e.g.)
HNO3
H2SO4 HPO42-
Every “H” has an ox # of +1
b) An exception is metallic hydrides, which have an ox # of -1
e.g.) NaH
CaH2 (In each one of these Ox. # of H = -1)
(ox # of “H” in NH3? _______)
(ox # of “H” in H2 = ______)
Finding oxidation numbers of each atom in a molecule or PAI
In a neutral molecule the total charge = 0
e.g.) NH3 Å Total charge = 0 (no charge)
In a polyatomic ion – the total ionic charge is written on the top right
e.g.) CrO42-
Total ionic charge (TIC) = -2
Oxidation numbers of all atoms add up to total ionic charge (TIC)
Unit 5-Oxidation-Reduction NOTES
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Chemistry 12
Unit 5 – Oxidation Reduction
e.g.) Find the oxidation # of Cr in CrO42oxidation # of O? ___
sum of O4? ___
TIC? ___
X – 8 = -2
X = +6 So ox # of Cr here = +6
e.g.) Find ox # of Cl in HClO4
HClO4
+1 + x + 4 (-2) = 0
e.g.) Find Ox # of Cr in Cr2O72-
Cr2O722x + 7(-2) = -2
e.g.) Find ox # of P in Li3PO4
Li3 P O4
3(+1) + x + 4 (-2) = 0
Changes in oxidation numbers
When an atom’s oxidation # is increased, it is _________________.
e.g.)
Half-rx: Fe2+ Æ Fe3+ + e-
More complex:
-When Mn3+ changes to MnO4-, is Mn oxidized or reduced?
Mn3+ Æ MnO4- What is the ox # of Mn before & after the reaction? Before ____ After ____
- The ox # of Mn is _________________..
- In this process, Mn is _________________.
Unit 5-Oxidation-Reduction NOTES
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Chemistry 12
Unit 5 – Oxidation Reduction
e.g.) oxidation or reduction
NO2- Æ NO3-
e.g.) oxidation or reduction
H2O2 Æ
H2O
E.g.: Is the following reaction a redox reaction and if so what is the ixidized species
and the reduced species
3SO2 + 3H2O + ClO3- Æ 3SO42- + 6H+ + Cl-
Oxygen - There are no O2 molecules or peroxides, so “O” in all these has an ox # = -2
Hydrogen - H is (+1) in both of these so it doesn’t change
Coefficients are
just for balancing
Sulphur
3SO2 Æ 3SO42Ox # of S is +4 ÆOx # of S is +6
Ox # of S increases so S is being oxidized
Note:
¾
¾
¾
R.A.O., the reducing agent is oxidized
The species SO2 is acting as the reducing agent.
The element S is being oxidized so S is losing electrons.
Look at the species with Cl:
ClO3-
Æ
Cl-
Ox. # of Cl = +5 Æ Ox. # of Cl = -1
Decrease in ox # so Cl is being reduced
Therefore, because S is ____________ and Cl is ______________ this is a redox reaction.
Which specific atom acting as the oxidizing agent? ________________
Given the redox reaction:
2MnO4- + 3C2O42- + 4H2O Æ 2MnO2 + 6CO2 + 8OHFind:
The species being reduced: _____________.
a)
b)
c)
d)
e)
The species undergoing oxidation: _____________.
The oxidizing agent: ________________.
The reducing agent: ________________.
The species gaining electrons: ______________.
f)
The species losing electrons: ______________.
Read p. 193-194 Do Exercises3 - 6 on p. 194-195 of SW.
Unit 5-Oxidation-Reduction NOTES
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Chemistry 12
Unit 5 – Oxidation Reduction
Chemistry 12 – Unit 5: Electrochemistry
Definitions: (species means atom, ion or molecule)
Electrochemistry – the branch of chemistry which is concerned with the
conversion of chemical energy to electrical energy and vice versa (p. 189)
Oxidation – a species undergoing oxidation loses electrons
(charge becomes more positive)
Reduction – a species undergoing reduction gains electrons
(charge becomes more negative)
Oxidizing agent – The species being reduced
(gains electrons, causes the other one to be oxidized)
LEO says GER
Losing
Electrons is
Oxidization
Gaining
Electrons is
Reduction
Reducing agent – The species being oxidized
(loses electrons, causes the other one to be reduced)
Redox Reaction – a reaction involving the loss or gain of electrons
(p. 190)
Redox – Short for Oxidation – Reduction
2 e-
OAR
Cu2+ (aq) + Zn (s)
E.g.)
Oxidizing
agent
Æ Cu(s) + Zn2+(aq)
Reducing
agent
The
Oxidizing
Agent is
Reduced
Redox identification
Reduction – charge decreases gets more –ve (gains electrons)
E.g.) Pb2+(aq) + Mg0(s) Æ Pb0(s) + Mg2+(aq)
Oxidation - Charge increases gets more +ve (loses electrons)
Question
In the reaction:
2Fe2+ + Cl2 Æ 2Fe3+ + 2Cl-
Identify the following:
i)
j)
k)
l)
The Oxidizing Agent: ___________Cl2
The species being oxidized:______Fe2+
The reducing agent:____________ Fe2+
The species being reduced:______ Cl2
Unit 5-Oxidation-Reduction NOTES
m)
n)
o)
p)
The species gaining electrons:____ Cl2
The species losing electrons:_____ Fe2+
The product of oxidation_________Fe3+
The product of reduction_________Cl-
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Chemistry 12
Unit 5 – Oxidation Reduction
Half-Reactions
-Redox reactions can be broken up into oxidation half reactions & reduction half reactions.
e.g.) Redox rx: Pb2+(aq) + Zn(s) Æ Pb(s) + Zn2+(aq)
Electrons on the LEFT side (or GER)
Means REDUCTION
Reduction Half-rxn: Pb2+(aq) + 2e- Æ Pb(s)
Oxidation Half-rxn: Zn(s) Æ 2e- + Zn2+(aq)
Write the oxidation half reactions & reduction half reactions for
the following redox reactions.
E.g. 1
Pb2+(aq) + Zn(s) Æ Pb(s) + Zn2+(aq)
Oxidation half rx: ___________________________
Reduction half rx: ___________________________
E.g. 2
F2(g) + Sn2+(aq) Æ 2F-(aq) + Sn4+(aq)
Note: Half-rx’s always have e-‘s, redox
(oxidation-reduction) reactions never
show e-‘s!
Oxidation half rx: ___________________________
Reduction half rx: ___________________________
Read Hebden p.189-191
Do #1-2 p. 192
5.2 Oxidation numbers
Oxidation numbers (p. 193) -the charge that an atom would possess if the
species containing the atom was made up of ions (even if it’s not!)
Rules to find oxidation number of an atom
4) In elemental form: Oxidation number of atoms = 0
(Single atoms of monatomic elements) or (diatomic molecules of diatomic
elements)
Eg) Mn, Cr, N2, F2, Sn, O2, etc.
The oxidation # of each atom = 0
5) In monatomic ions: oxidation # = charge
Eg)
In Cr3+ -oxidation # of Cr = +3
S2- -oxidation # of S = -2
6) In ionic compounds
b) the oxidation # of Alkali Metals is always +1
eg) NaCl
Ox # of Na & K = +1
K2CrO4
Unit 5-Oxidation-Reduction NOTES
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Chemistry 12
Unit 5 – Oxidation Reduction
b) the oxidation # of Halogens when at the end (right side) of the formula
is always –1
Ox # of Cl, Br and F = -1
eg) CaCl2 AlBr3
KF
Note: Halogens are not always –1! (Only when it is written last in formula.)
4) In molecules or polyatomic ions:
c) Ox. # of oxygen is almost always –2
e.g.) KOH CrO42- Li3PO4
Ox # of
O is –2
(Remember, “O” in O2 has
an Ox. # of ______)
d) An exception is Peroxides in which ox. # of O = -1
Hydrogen Peroxide: H2O2
Ox # of O’s = -1
Alkali Peroxides: Na2O2
5) In molecules or ions:
a) Hydrogen is almost always +1
e.g.)
HNO3
H2SO4 HPO42-
Every “H” has an ox # of +1
b) An exception is metallic hydrides, which have an ox # of -1
e.g.) NaH
CaH2 (In each one of these Ox. # of H = -1)
(ox # of “H” in NH3? _______)
(ox # of “H” in H2 = ______)
Finding oxidation numbers of each atom in a molecule or PAI
In a neutral molecule the total charge = 0
e.g.) NH3 Å Total charge = 0 (no charge)
In a polyatomic ion – the total ionic charge is written on the top right
e.g.) CrO42-
Total ionic charge (TIC) = -2
Oxidation numbers of all atoms add up to total ionic charge (TIC)
Unit 5-Oxidation-Reduction NOTES
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Chemistry 12
Unit 5 – Oxidation Reduction
e.g.) Find the oxidation # of Cr in CrO42oxidation # of O? -2
sum of O4? -8
TIC? -2
X – 8 = -2
X = +6 So ox # of Cr here = +6
e.g.) Find ox # of Cl in HClO4
HClO4
+1 + x + 4 (-2) = 0
1+x–8=0
x–7=0
x = +7
2x – 14 = -2
2x = +12
x = +6
3+x–8=0
x–5=0
x = +5
e.g.) Find Ox # of Cr in Cr2O72-
Cr2O722x + 7(-2) = -2
e.g.) Find ox # of P in Li3PO4
Li3 P O4
3(+1) + x + 4 (-2) = 0
Changes in oxidation numbers
When an atom’s oxidation # is increased, it is oxidized.
e.g.)
Half-rx: Fe2+ Æ Fe3+ + e-
More complex:
-When Mn3+ changes to MnO4-, is Mn oxidized or reduced?
Mn3+ Æ MnO4- What is the ox # of Mn before & after the reaction? Before 3+
After 7+
- The ox # of Mn is increased.
- In this process, Mn is oxidized
Unit 5-Oxidation-Reduction NOTES
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Chemistry 12
Unit 5 – Oxidation Reduction
e.g.) oxidation or reduction
NO2- Æ NO3-
e.g.) oxidation or reduction
H2O2 Æ
H2O
E.g.: Is the following reaction a redox reaction and if so what is the ixidized species
and the reduced species
3SO2 + 3H2O + ClO3- Æ 3SO42- + 6H+ + Cl-
Oxygen - There are no O2 molecules or peroxides, so “O” in all these has an ox # = -2
Hydrogen - H is (+1) in both of these so it doesn’t change
Coefficients are
just for balancing
Sulphur
3SO2 Æ 3SO42Ox # of S is +4 ÆOx # of S is +6
Ox # of S increases so S is being oxidized
Note:
¾
¾
¾
R.A.O., the reducing agent is oxidized
The species SO2 is acting as the reducing agent.
The element S is being oxidized so S is losing electrons.
Look at the species with Cl:
ClO3-
Æ
Cl-
Ox. # of Cl = +5 Æ Ox. # of Cl = -1
Decrease in ox # so Cl is being reduced
Therefore, because S is oxidized and Cl is reduced this is a redox reaction.
Which specific atom acting as the oxidizing agent? Cl in ClO3Given the redox reaction:
2MnO4- + 3C2O42- + 4H2O Æ 2MnO2 + 6CO2 + 8OHFind:
The species being reduced: _____________.
g)
h)
i)
j)
k)
The species undergoing oxidation: _____________.
The oxidizing agent: ________________.
The reducing agent: ________________.
The species gaining electrons: ______________.
l)
The species losing electrons: ______________.
Read p. 193-194 Do Exercises3 - 6 on p. 194-195 of SW.
Unit 5-Oxidation-Reduction NOTES
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