Redox Reactions
The corrosion of metals, especially the
corrosion of iron (rusting), has a great
economic impact.
Rust is one of the common causes of
bridge accidents.
The Kinzua Bridge in Pennsylvania
was blown down by a tornado in
2003 largely because the central
base bolts holding the structure to
the ground had rusted away,
leaving the bridge resting by
gravity alone.
Redox Reactions
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Redox Reactions
Explain why this reaction is a redox reaction!
Give more than one answer.
Gain/loss of oxygen/hydrogen, gain/loss of electrons (transfer), increase/decrease in Ox.number
Remember:
• A redox reaction is any reaction involving a
transfer of electrons.
• In all redox reactions, oxidation and reduction
happen at the same time.
• Oxidation is loss of electrons/ increase in oxidation
number.
• Reduction is gain of electrons/decrease in
oxidation number.
• Oxidising agents (oxidants) are themselves
reduced.
• Reducing agents (reductants) are themselves
oxidised.
Redox Reactions
Examples:
Write the oxidation number for each of the
underlined atoms:
PbO2
+4
BrO3-
+5
[FeSCN]2+ +3
Redox Reactions
Complete worksheet #1
Example:
Use oxidation numbers to determine whether the
following reaction is a redox reaction, and if so,
which element has been oxidised, and which has
been reduced: It is a redox reaction. The nitrogen in ammonia
has been oxidised (ON increased).
The oxygen has been reduced (ON decreased).
0
-3
0
-2
4NH3(g) + 3O2(g) → 2N2(g) + 6H2O(l)
Colours of species in redox reactions
The colour of any species depends on the oxidation number.
You are expected to know a number of species and their colours.
Reduced form
brown
Cu
SO2
Mn2+
H2O2
H2O
Cr3+ blue/green
Fe2+ pale green
ClBrH2
solid
gas
aq
liquid
liquid
aq
aq
aq
aq
gas
Oxidised form
blue
Cu2+
SO42H+/MnO4- purple
O2
H2O2
orange
Cr2O72orange
Fe3+
pale green
Cl2
red/orange
Br2
H+
aq
aq
aq
gas
liquid
aq
aq
gas
liquid
aq
Reduced form
MnO2 brown
MnO42- green
II2 in I- = I3- brown
H2S
Pb2+
brown
NO2
C2O42S2O32red/orange
Br2
Oxidised form
solid
H2O/MnO4- purple
aq
aq
OH-/MnO4- purple
aq
aq
I2 in I- = I3IO3S
PbO2
NO3CO2
S4O62BrO3-
brown
aq
aq
gas
aq
gas
aq
aq
liquid
aq
yellow/white solid
brown
solid
aq
gas
aq
aq
C2O42S2O32S4O62BrO3IO3-
Names of unfamiliar ions:
Oxalate ion (from oxalic acid)
Thiosulfate ion
Tetrathionate
Bromate
Iodate
Rhubarb contains
oxalic acid. This makes
your teeth feel furry.
Citrobacter live in the intestines of warm
blooded animals, without oxygen. They
can respire by converting tetrathionate
into thiosulfate.
Recognising species (observations):
Some species can be recognised by their colour.
For example:
The purple permanganate ion (MnO4-) can be
reduced in alkaline conditions (presence of OH-).
The product is the green manganate ion (MnO42-).
Observation: the purple solution turns green.
Recognising species (observations):
Complete worksheet #2
Some species can not be recognised by their
colour as they are colourless or do not have a
distinct colour.
Therefore you need to memorise some tests to
detect these species:
I2:
turns blue/black with starch solution
SO42- : forms a white precipitate with H+/BaCl2
Fe3+ : forms a blood red solution with KSCN
Cl2 : turns damp starch-iodide paper blue/black
Balancing complicated half equations
Cr2O72- → Cr3+
1. Balance the atoms that aren’t O or H.
Cr2O72- → 2Cr3+
2. Balance the oxygen by adding water.
Cr2O72- → 2Cr3+ + 7H2O
3. Balance the hydrogens by adding H+.
Cr2O72- + 14H+ → 2Cr3+ + 7H2O
4. Add electrons to the side that is more positive.
Cr2O72- + 14H+ + 6e- → 2Cr3+ + 7H2O
Balancing complicated half equations
SO2 → SO421. Balance the atoms that aren’t O or H.
SO2 → SO422. Balance the oxygen by adding water.
SO2 + 2H2O → SO423. Balance the hydrogens by adding H+.
SO2 + 2H2O → SO42-+ 4H+
4. Add electrons to the side that is more positive.
SO2 + 2H2O → SO42-+ 4H+ + 2e-
Combining the half equations
The number of electrons in the two
half equations needs to be the same.
Therefore the half equations have to
be multiplied by a factor.
Combining the half equations
Cr2O72- + 14H+ + 6e- → 2Cr3+ + 7H2O
SO2 + 2H2O → SO42-+ 4H+ + 2eMultiply whole
equation by 3
3 SO2 + 6H2O → 3 SO42-+ 12H+ + 6e-
Combining the half equations
Cr2O72- + 14H+ + 6e- → 2Cr3+ + 7H2O
3 SO2 + 6H2O → 3 SO42-+ 12H+ + 6eCr2O72- + 14H+ + 3 SO2 + 6H2O→2Cr3+ + 7H2O + 3 SO42-+ 12H+
5. Cancel out H+ and H2O.
Cr2O72- + 2H+ + 3 SO2 → 2Cr3+ + H2O + 3 SO42-
Balancing redox equations in
acidic/neutral conditions
Last year you learned the steps to balance
redox reactions.
One of the steps included adding H+ ions.
H+ ions are present in small amounts in
neutral conditions (same as OH- ions).
In acidic conditions H+ ions are present in a
higher concentration than OH- ions.
It is therefore ok if your redox equations
contain H+ ions as they are present.
Balancing redox equations in
acidic/neutral conditions
The permanganate ion can be reduced in
neutral, acidic and in alkaline conditions.
Each of these conditions leads to a
different product.
Write down the balanced half equations
for the reduction or permanganate in
acidic conditions (producing Mn2+),
Answers
Acidic:
+
2+
MnO4 + 8H + 5e → Mn + 4H2O
Now write a balanced half equation for the
reduction of permanganate in neutral
conditions (producing MnO2).
Answers
Neutral:
+
MnO4 + 4H + 3e → MnO2+ 2H2O
H+ ions are relatively scarce. Add
OH- ions to both sides of the
equation to neutralise.
-
MnO4 + 2H2O +
3e
→ MnO2+
4OH
Balancing redox equations in alkaline
conditions
The permanganate ion can also be reduced
in alkaline conditions, producing the
manganate ion (MnO42-).
Write a balanced half equation using the
usual steps.
If adding H+ ions (which are scarce in the
alkaline solution) is necessary, add an
equal amount of OH- ions to both sides of
the equation.
Answer
Alkaline:
2MnO4 + e → MnO4
Complete worksheet #3
Key Learning Outcomes
You should now be able to:
•
•
•
Determine the oxidation number of any atom in a
compound or ion and use oxidation numbers to
identify the oxidised and reduced species in a
reaction.
Recall common oxidising and reducing agents, state
the colours of the reagents and their products, and
recall any other observations or conditions
characteristic of their use.
Write ion-electron equations for oxidation and
reduction half-reactions and combine the half
equations to give a balanced ionic equation.