Oxidation and Reduction Reactions 2
IB Topic 9.2: Redox Equations
Text Reference: Higher Level Chemistry p. 326-329
IB Assessment Statements
9.2.1
Deduce simple oxidation and reduction half-equation
given the species involved in a redox reaction.
!
9.2.2
Deduce redox equations using half-equations.
Use H+ and H2O when balancing half-equations in acidic solution.
!
9.2.3
Define the terms oxidizing agent and reducing agent.
!
9.2.4
Identify the oxidizing and reducing agents in redox equations.
The Goal of Balancing Redox Equations
1. the atoms goal
number of atoms of each element is the same on both the reactant and product side of the equation
2. the ELECTRON goal
number of electrons gained in the reduction half reaction
=
number of electrons lost in the oxidation half reaction
Simple Redox Equations
Synthesis, Decomposition and Single Displacement Reactions
Example 1:
Sodium metal reacts with bromine gas to produce sodium bromide.
skeleton equation
Na + Br2 → NaBr
Na+ + Br–
unbalanced net
ionic equation
half equations
Na + Br2 → Na+ + Br–
2 Na
→ 2 Na+
+ 21e–
Br2 + 2e– → 2Br–
final equation
2Na + Br2 → 2Na+ + 2Br–
2Na + Br2 → 2NaBr
X2
X1
Add e– to balance
charge.
Multiply to balance
electrons.
Simple Redox Equations
Synthesis, Decomposition and Single Displacement Reactions
Example 2:
Aluminum is immersed in a solution of copper (II) chloride.
Al + CuCl2 → Cu + AlCl3
skeleton equation
Cu2+ + 2Cl–
unbalanced NET
ionic equation
half equations
final equation
Al3+ + 3Cl–
Cl– ions do not change
oxidation number.
Al + Cu2+ → Cu + Al3+
2 Al
→ 2 Al3+ + 63e–
X2
3 Cu2+ + 62e–
→ 3 Cu
X3
2Al + 3Cu2+ → 3Cu + 2Al3+
Add e– to balance
charge.
Multiply to balance
electrons.
More Complex Redox Equations
Many redox reactions involving covalent molecular compounds or
polyatomic ions occur under certain acid-base conditions.
Example: Incomplete Oxidation of Ethanol
acidified Cr2O72-
CH3CH2OH
ethanol
half reactions:
[O]
CH3CHO
ethanal
C2H5OH → CH3CHO
Cr2O72– + H+ → Cr3+
Water, hydrogen ions and/or hydroxide ions will often be part of the reaction, and
will appear in the equations for the half-reactions and often the final equation.
More Complex Redox Equations
There is a methodical way to balance more complex HALF-reactions.
Each “skeleton” half reaction should first be balanced using this method.
Then balance electrons gained and lost, and add the half equations together.
http://www.fanpop.com/clubs/hawaii/images/23319255/title/aloha
A
O
H
E
Balance all ATOMS except O and H.
Balance OXYGEN by adding H2O to
the side which needs oxygen.
Balance HYDROGEN by adding H+
to the side which needs hydrogen.
Add ELECTRONS to the side with
the more positive charge in order to
balance charge.
More Complex Redox Equations
Half Reaction Example 1:
Dichromate ions react to form chromium (III) ions.
skeleton half equation
A
O
H
E
Cr2O72– + 14H+ + 6e– → 2 Cr3+ + 7H2O
+12
+6
What happens to dichromate?
reduced
What type of agent is dichromate?
oxidizing agent
More Complex Redox Equations
Half Reaction Example 2:
Iodine reacts to form iodate (V) ions.
skeleton half equation
A
O
H
E
I2 + 6H2O → 2 IO3 +
–
0
+
12H
+
–
10e
+10
What happens to iodine?
oxidized
What type of agent is iodine?
reducing agent
More Complex Redox Equations
Complete Reaction Example 1:
The reaction between copper atoms and nitrate ions in acidic solution produces nitrogen monoxide and copper (II) ions
skeleton equation
Cu + NO3– → NO + Cu2+
A
O
half equations
H balance the
E
electrons!
final
equation
→ 3Cu2+
3 Cu
6 –
+2e
x3
8
6
4
2 NO3– + 4H+ +3e– →2NO + 2H2O x2
3Cu + 2NO3 +
–
+
8H
OA = NO3–
→
2+
3Cu +
RA = Cu
2NO + 4 H2O
More Complex Redox Equations
Complete Reaction Example 2:
Permanganate (manganate VII) ions react with sulfur dioxide to form manganese (II) ions and sulfate ions.
skeleton equation
A
O half equations
H balance the
E
electrons!
final
equation
MnO4– + SO2 → Mn2+ + SO42–
8
16
10
2 MnO4– + 8H+ +5e– → 2Mn2+ + 4H2O x2
4 20
2 10
10
5 SO2 + 2H2O →5 SO42– + 4H+ +2e– x5
2MnO4– + 5SO2 + 2H2O → 2Mn2+ + 5SO42– + 4H+
OA = MnO4–
RA = SO2
More Complex Redox Equations
Complete Reaction Example 3: Disproportionation Reactions
The same reactant is both oxidized and reduced.
Chlorine reacts with water to form hydrochloric acid and hypochlorous acid (HClO)
skeleton equation
Cl2 + H2O → HCl + HClO
A
O
half equations
H balance the
E
electrons!
final
equation
+
–
+
2H
Cl2
+2e →2 HCl
Cl2 + 2H2O →2 HClO + 2H+ +2e–
2Cl2 + 2H2O → 2HCl + 2HClO
OA = Cl2
RA = Cl2
Stoichiometry and Redox Reactions
mass
A
mass
B
use molar mass
use molar mass
moles
A
use C = n/V
solution
A
(volume or
concentration)
moles
B
use C = n/V
solution
B
(volume or
concentration)
Stoichiometry and Redox Reactions
A 0.0484M standard solution of potassium permanganate was titrated against 25.00mL of an iron (II) sulfate solution. !
The equivalence point, as indicated by a faint pink colour, was reached when 15.50mL of
potassium permanganate solution had been added. !
Calculate the concentration of the iron (II) sulfate solution. !
Balanced equation:
!
MnO4- + 8 H+ + 5 Fe2+ → Mn2+ + 4 H2O + 5 Fe3+
moles
A
MnO4
–
C = 0.0484 M
V = 15.50 mL
Step 1
C = n/V
solution A
(volume or
concentration)
Step 2
moles
B
C = n/V
Step 3
solution B
(volume or
concentration)
Fe2+
C=?
V = 25.00 mL
Stoichiometry and Redox Reactions
KMnO4
A 0.0484M standard solution of potassium permanganate was titrated against 25.00mL of an iron (II) sulfate solution. !
The equivalence point, as indicated by a faint pink colour, was reached when 15.50mL of
potassium permanganate solution had been added. !
Calculate the concentration of the iron (II) sulfate solution. FeSO4
!
Balanced equation:
!
MnO4- + 8 H+ + 5 Fe2+ → Mn2+ + 4 H2O + 5 Fe3+
Step 1:
moles MnO 4– = C x V
= 0.0484 M x 0.01550 L
= 0.0007502 mol
2+
moles
Fe
Step 2:
Step 3: [Fe
2+
2+
5
mol
Fe
= 0.0007502 mol MnO 4– x
= 0.003751 mol
–
1 mol MnO 4
n
0.003751 mol
−1
]=
=
= 0.150 mol L
V
0.02500 L
∴ [FeSO4] = 0.150 mol L-1