Redox
Principles and application
Course outline
Oxidation and reduction
Redox reactions
explain oxidation and
reduction as an electron
transfer process
calculate oxidation numbers
identify and name oxidants
and reductants in equations
identify oxidation-reduction
reactions using oxidation
numbers
Oxidation and reduction
Redox reactions
describe, write equations for
and interpret observations for:
metal displacement reactions
halogen displacement
reactions
write balanced simple redox
equations (metal/metal ion,
metal/hydrogen ion and
halogen/halide ion)
Course outline
Oxidation and reduction
cathode processes
Redox Applications – Electrolytic
Cells
role of the electrolyte
describe and explain how an
electric current is conducted in an
electrolytic cell
describe and explain the following
during the operation of an
electrolytic cell:
anode processes
direction of ion migration
direction of electron flow in
external circuit
electrode product prediction for
molten metal halides only
predict and name the electrode
products for the electrolysis of
molten metal halides only.
OXIDATION
Used to be viewed as
Gaining oxygen
2 Mg + O2 → 2 MgO
Magnesium gains oxygen
Losing hydrogen
CH4 + 2 S → C + 2 H2S
Carbon loses hydrogen
REDUCTION
Used to be viewed as
Losing oxygen
FeO + C → Fe + CO
Iron loses oxygen
(Note: carbon also gains oxygen = oxidation)
Gaining hydrogen
CH4 + 2 O2 → CO2 + 2 H2O
Oxygen gains hydrogen
(Note: carbon also loses hydrogen and
gains oxygen
=
oxidation)
It was observed that very
often when one
substance was being
oxidised another was
being reduced at the
same time.
Modern view of
oxidation and reduction
Oxidation is defined as
losing of electrons LEO
Reduction is defined as
gaining of electrons GER
Modern view of
oxidation and reduction
Oxidation Is Loss
of electrons
Reduction Is Gain
of electrons
OIL RIG
A redox reaction is an
electron transfer
process
REDOX = electron transfer
- 4e- 4e4e
-
Reducing agent
4e Loses electrons
`
2 Fe (ON=0)
4e
2+O2 –
2
Fe
+
4e
4e- Gains electrons O2 (ON=0)
4e- 4e- 4e- 4e- Oxidising
agent
REDOX = electron transfer
- 2e- 2e2e
-
Reducing agent
2e Loses electrons
`
2 Na (ON=0)
2e
+Cl –
2
Na
+
2e
2e- Gains electrons Cl2 (ON=0)
2e- 2e- 2e- 2e- Oxidising
agent
An oxidising agent
Oxidises another substance by
taking electrons from it
Is also called the oxidant
Gains electrons, and therefore
Becomes reduced
A reducing agent
Reduces another substance by
donating electrons to it
Is also called the reductant
Loses electrons, and therefore
Becomes oxidized
A redox equation analysed
Electron
S gains electrons
transfer
Is reduced
Is the oxidant
2+ +
Mg
+
S
→
Mg
0
0
+2
Mg loses electrons
Is oxidized
Is the reductant
S–22–
What is the ON of P in PH3
ONP + 3 ONH = 0 (neutral)
ONP + 3 (+1) = 0
ONP + 3
= 0
ONP =
–3
ON of P in H3PO4
3H
+ P + 4O
= 0
3(+1) + P + 4(–2)
= 0
(+3)
= 0
+ P + (–8)
P + (–5)
P = +5
= 0
ON of P in H3PO3
3H + P + 3O
3(+1) + P + 3(–2)
(+3) + P + (–6)
P + (–3)
P = +3
= 0
= 0
= 0
= 0
ON of Cr in CrO3
Cr
+
3O
Cr
+
3(– 2) = 0
Cr
+
(– 6)
Cr
= 0
= 0
= +6
ON of Cr in Cr2O722Cr
+
7O
=–2
2Cr
+
7(– 2) = – 2
2Cr
= 14 – 2
2Cr
= + 12
Cr
= +6
ON of Cr in CrO42-
Cr
+
4O
Cr
+
4(– 2) = – 2
Cr
=–2
= 8–2
= +6
ON of Cr in (NH4)2CrO4
2NH4 + Cr + 4O
= 0
2(+1) + Cr + 4(–2) = 0
Cr +2 – 8
Cr
= 0
= +6
ON of C in CH4
C
+ 4H
= 0
C
+ 4(+1)
= 0
C
= –4
ON of C in CH2O
C + 2H
+ O
= 0
C + 2(+1) +(– 2)
= 0
C + 2 + (– 2)
= 0
C = 0
How can combined C have an ON of zero?
O is more electronegative
than C
H is less electronegative
than C
C has ‘lost’ these 2
electrons
e
e
O
e
e
e
e
e
e
C
C has
‘gained’
these 2
electrons
Carbon has ‘lost’ 2 electrons to Oxygen
but ‘gained’ 2 from Hydrogen
ON of C in CH3OH
C + 4H
+ O
= 0
C + 4(+1) + (– 2) = 0
C + 4 + (– 2)
C + 2
C
= 0
= 0
=–2
ON of C in C6H6
6C
+
6C
+ 6(+1) = 0
6C +
6H
=
0
6
=
0
6C
= –6
C =–1
ON of Cl in HClO4
H
+ Cl + 4O
= 0
(+1) + Cl + 4(– 2) = 0
(+1) + Cl + (– 8)
= 0
Cl + (– 7) = 0
Cl = +7
ON of Se in H2SeO3
2H
+ Se + 3O
= 0
2(+1) + Se + 3(– 2) = 0
(+2) + Se + (– 6)
Se + (– 4)
Se
= 0
= 0
= +4
ON of Se in SeO3
Se + 3O
= 0
Se + 3(– 2)
= 0
Se + (– 6)
= 0
Se
= +6
ON of Mn in MnO4
Mn + 4O
=–1
Mn + 4(– 2) = – 1
Mn + (– 8)
=–1
Mn
= +7
ON of Mn in MnO4
Mn + 4O
=–2
Mn + 4(– 2)
=–2
Mn + (– 8)
=–2
Mn
= +6
2–
ON of Mn in MnO2
Mn + 2O
= 0
Mn + 2(– 2)
= 0
Mn + (– 4)
= 0
Mn
= +4
ON of Mn in Mn2O3
2Mn + 3O
= 0
2Mn + 3(– 2) = 0
2Mn + (– 6)
= 0
2Mn
= +6
Mn
= +3
In HClO4 the
•ON of Cl is +7
•ON of O is – 2
H + Cl + 4O = 0
+1 + Cl – 8 = 0
Cl = 8 – 1 = +7
•ON of H is +1
With the aid of an electron-dot diagram,
show this is consistent with the
electronegativities of the atoms in each bond.
EN: O = 3.4 Cl = 2.8 H = 2.2
And also these
2 electrons
e
e
ee
O
e
e
ee
ee
ee
e
e
eOe
ee
As well these 2
electrons
Cl has ‘lost’ 7
electrons
ON = +7
Cl has ‘lost’ these
2 electrons
e
e
eO e
ee
Cl
e
e
ee
O
ee
e
e
H
H has ‘lost’
And this 1 1 electron
ON = +1
electron
Each O has
‘gained’ 2
electrons
ON = – 2
In H2SO4 the
2H + S + 4O = 0
•ON of S is +6
+2 + S
•ON of O is – 2
S = 8 – 2 = +6
–8
=0
•ON of H is +1
With the aid of an electron-dot diagram,
show this is consistent with the
electronegativities of the atoms in each bond.
EN: O = 3.4 S = 2.6 H = 2.2
And also these
2 electrons
e
e
ee
O
e
e
ee
ee
ee
e
e
eOe
ee
As well as
this 1
electron
S has ‘lost’ 6
electrons
ON = +6
S has ‘lost’ these
2 electrons
e
e
eOe
ee
S
e
e
ee
O
ee
H
e
e
And this 1
electron
Each O has
‘gained’ 2
electrons
ON = – 2
H
Each H has
‘lost’ 1
electron
ON = +1
Write the half equation for the
conversion of IO3– to I2
1. Write the ‘skeleton’ equation including the reactant and
product species. Determine the oxidation state of each
species, to decide which element is being
oxidized/reduced.
IO3– →
(+5)(–2)
I2
(0)
I is reduced
I + 3(–2) = –1
so I = (+5)
Write the half equation for the
conversion of IO3– to I2
2. Balance the numbers of atoms of the element that is
being oxidized/reduced
IO3–
→
I2
2 IO3– →
I2
Write the half equation for the
conversion of IO3– to I2
3. Balance the number of oxygen atoms by adding H2O to
the side that needs more O
IO3–
→
I2
2 IO3–
→
I2
2 IO3–
→
I2 + 6 H2O
Write the half equation for the
conversion of IO3– to I2
4. Balance the number of hydrogen atoms by adding H+ to the
side that needs more H
IO3–
2 IO3–
2 IO3–
→
→
→
2 IO3– + 12 H+ →
I2
I2
I2 + 6 H2O
I2 + 6 H2O
Write the half equation for the
conversion of IO3– to I2
5. Balance the charge by adding electrons to the side that has
more positive charge
IO3–
2 IO3–
2 IO3–
→
→
→
2 IO3– + 12 H+ →
I2
I2
I2 + 6 H2O
I2 + 6 H2O
2 IO3– + 12 H+ + 10e – → I2 + 6 H2O
Write the half equation for the
conversion of IO3– to I2
Now amend the equation for this reaction in BASIC solution
Add 12 OH– to both sides of the half
equation.
2 IO3– + 12 H+ + 10e – → I2 + 6 H2O
12 OH–
12 OH–
Write the half equation for the
conversion of IO3– to I2
Now amend the equation for this reaction in BASIC solution
The OH– and H+ neutralise
2IO3–+ 12H2O + 10e –→ I2+6H2O +12 OH–
Write the half equation for the
conversion of IO3– to I2
Now amend the equation for this reaction in BASIC solution
Cancel out excess H2O
6
2IO3–+ 12H2O + 10e –→ I2+6H2O +12 OH–
2IO3–+ 6H2O + 10e –→ I2 + 12 OH–
Write the half equation for the
conversion of Mn2+ to MnO4–
1. Write the ‘skeleton’ equation including the reactant and
product species. Determine the oxidation state of each
species, to decide which element is being
oxidized/reduced.
Mn2+ → MnO4–
(+2)
(+7)(–2)
Mn is oxidized
Mn + 4(–2) = –1
So Mn = +7
Write the half equation for the
conversion of Mn2+ to MnO4–
2. Balance the numbers of atoms of the element that is
being oxidized/reduced
Mn2+ → MnO4–
Mn is already balanced
Write the half equation for the
conversion of Mn2+ to MnO4–
3. Balance the number of oxygen atoms by adding H2O to
the side that needs more O
Mn2+ → MnO4–
Mn2+ + 4 H2O → MnO4–
Write the half equation for the
conversion of Mn2+ to MnO4–
4. Balance the number of hydrogen atoms by adding H+ to the
side that needs more H
Mn2+ → MnO4–
Mn2+ + 4 H2O → MnO4–
Mn2+ + 4 H2O → MnO4– + 8 H+
Write the half equation for the
conversion of Mn2+ to MnO4–
5. Balance the charge by adding electrons to the side that has
more positive charge
Mn2+ → MnO4–
Mn2+ + 4 H2O → MnO4–
Mn2+ + 4 H2O → MnO4– + 8 H+
Mn2+ + 4 H2O → MnO4– + 8 H+ + 5e –
Write the half equation for the
conversion of Mn2+ to MnO4–
Now amend the equation for this reaction in BASIC solution
Add 8 OH– to both sides of the half
equation.
Mn2+ + 4 H2O → MnO4– + 8 H+ + 5e –
8 OH
–
8 OH–
Write the half equation for the
conversion of Mn2+ to MnO4–
Now amend the equation for this reaction in BASIC solution
The OH– and H+ neutralise
Mn2++ 4 H2O + 8 OH– → MnO4– + 8 H2O + 5e –
Write the half equation for the
conversion of Mn2+ to MnO4–
Now amend the equation for this reaction in BASIC solution
Cancel out excess H2O
4
Mn2++ 4 H2O + 8 OH– → MnO4– + 8 H2O + 5e –
Mn2+
+
–
8 OH → MnO4– + 4 H2O + 5e –
Write the two half equations
and the net redox equation for the
disproportionation (self-redox) of
Cu+ to Cu2+ and Cu
The two half reactions are:
Cu+ → Cu2+
and
Cu+ → Cu
Disproportionation (self-redox) of
Cu+ to Cu2+ and Cu
The two half reactions are:
Cu+ → Cu2+
and
Cu+ → Cu
These equations need only to be
balanced for charge
Cu+ → Cu2+ + e–
Cu+ + e– → Cu
Disproportionation (self-redox) of
Cu+ to Cu2+ and Cu
Now combine the two half equations
Each equation has one electron,
so simply add all reactants
and all products
Cu+
→ Cu2+ + e–
Cu+ + e– → Cu
2 Cu+
→ Cu2+ + Cu
Write the two half equations
and the net redox equation
for the disproportionation (self-redox)
of ClO2 to ClO3– and ClO2– in basic
solution
Some ClO2 molecules are converted to ClO3– and
other ClO2 molecules are converted to ClO2–
Use the rules to write the two half equations for acidic solution,
and then for basic solution.
Disproportionation (self-redox) of
ClO2 to ClO3– and ClO2–
The two half reactions are:
ClO2 → ClO3– and ClO2 → ClO2–
(+4)
(+5)
(+4)
(+3)
Use the rules to complete each half equation
Disproportionation (self-redox) of
ClO2 to ClO3– and ClO2–
Balance each for O by adding H2O
ClO2 + H2O → ClO3–
ClO2
→ ClO2–
Disproportionation (self-redox) of
ClO2 to ClO3– and ClO2–
Balance each for H by adding H+
ClO2 + H2O → ClO3– + 2H+
ClO2
→ ClO2–
Disproportionation (self-redox) of
ClO2 to ClO3– and ClO2–
Balance charges by adding electrons
ClO2 + H2O → ClO3– + 2H+ + e –
ClO2 + e – → ClO2–
Disproportionation (self-redox) of
ClO2 to ClO3– and ClO2–
Now add the two half equations
ClO2 + H2O → ClO3– + 2H+ + e –
ClO2 + e – → ClO2–
2ClO2 + H2O → ClO3– + ClO2– + 2H+
Disproportionation (self-redox) of
ClO2 to ClO3– and ClO2–
Add 2 OH– to each side
2ClO2 + H2O → ClO3– + ClO2– + 2H+
2 OH–
2 OH–
Disproportionation (self-redox) of
ClO2 to ClO3– and ClO2–
Add 2 OH– to each side
Then cancel excess H2O
1
2ClO2 + H2O → ClO3– + ClO2– + 2H2O
2 OH–
2ClO2 + 2OH– → ClO3– + ClO2– + H2O
Disproportionation (self-redox) of
MnO42– to MnO4– and MnO2
The two half reactions are:
MnO42–
→ MnO4–
(+6)
MnO42–
(+6)
(+7)
→ MnO2
(+4)
Use the rules to complete each half equation
Disproportionation (self-redox) of
MnO42– to MnO4– and MnO2
Balance O by adding H2O
MnO42–
→ MnO4–
MnO42–
→ MnO2 + 2H2O
Disproportionation (self-redox) of
MnO42– to MnO4– and MnO2
Balance H by adding H+
MnO42–
→ MnO4–
MnO42– + 4H+
→ MnO2 + 2H2O
Disproportionation (self-redox) of
MnO42– to MnO4– and MnO2
Balance charges by adding electrons
MnO42–
→ MnO4– + e –
MnO42– + 4H+ + 2e –→ MnO2 + 2H2O
Disproportionation (self-redox) of
MnO42– to MnO4– and MnO2
Electrons lost =
2MnO42–
Electrons gained
→ 2MnO4– + 2e –
MnO42– + 4H+ + 2e –→ MnO2 + 2H2O
Disproportionation (self-redox) of
MnO42– to MnO4– and MnO2
Add the two half equations
2MnO42–
→ 2MnO4– + 2e –
MnO42– + 4H+ + 2e – → MnO2 + 2H2O
3MnO42– + 4H+ → 2MnO4– +MnO2 +2H2O
Chemistry Project
Redox Titration
Content
Definition
Experiment
Discussion
Definition
Redox reaction
• A redox reaction is involving transfer of
electrons.
• Oxidation is a process in which a substance
loses electrons.
• Reduction is a process in which a substance
gains electrons.
• They cannot take place without each other.
Definition
Redox reaction
•
An oxidizing agent is a substance that oxidizes others
by accepting electrons.
•
A reducing agent is a substance that reduces others by
donating electrons.
Definition
Titration
•
Titration is a process of volumetric analysis that is used
to determine the amount of solute in a solution.
•
In simple titration, indicator is used to detect the
equivalence point of reaction.
Definition
Example
•
Iodometric titration
•
Titration involving oxidation of iron (Ⅲ)
•
Titration involving reduction of iron (Ⅱ)
( In some case,indicators need not to add.
e.g. Titration with potassium manganate (Ⅶ)
or potassium dichromate (Ⅵ)
)
Experiment of iodimetric titrition
1)
Thoery
A standard iodine solution is titrated with a thiosulphate
solution with unknown molarity. The standardized
thiosulphate solution can be used to titrate another
iodine solution of unknown concentration.
Experiment of iodimetric titrition
2)• Since iodine is volatile, we cannot prepare
standard solution of iodine directly by accurately
weighing a certain amount of iodine solid &
dissolving the iodine in water .
• Moreover , iodine is only very slightly soluble in
water
Experiment of iodimetric titrition
• so a standard solution of iodine is first prepare by dissolving a
known of pure potassium iodate(v) solid into an acidic medium
(diluted H2SO4)containing excess iodide.
IO3- (aq) +5I- (aq) + 6H+ (aq) ------> 3 I2 (aq) +3H2O (l)
or [IO3- (aq) +8I- (aq) + 6H+ (aq) ------> 3 I3- (aq) + 3H2O (l) ]
Experiment of iodimetric titrition
3) 1ST titration
• The standard iodine solution is then used to
titrate with a thiosulphate solution of unknown
concentration .
• Since the colour of iodine in iodide cannot be used
to accurately detect the end point starch is used
as the indicator.
(the change in colour ‘brown->yellow->colourless’ is very
difficult to observe)
Starch + iodine  blue ‘complex’
Experiment of iodimetric titrition
• Since starch irreversibly combines with iodine at
a high concentration of I2(aq), so that the I2 will
not be released from starch at the end point .
• The starch solution should be added at the later stage of the
titration (when the solution just turns from brown to pale yellow ).
After the addition of starch ,the mixture turns deep blue .
• The end point is shown by the complete
decolourisation of the blue colour .
Experiment of iodimetric titrition
4) 2nd titration
• After standardization ,the thiosulphate solution
can be used to titrate another solution of
unknown concentration of iodine.
This method of titration (using iodine as an oxidizing
agent in the titration ) is known as the iodimetric titration
or iodimetry.
Experiment of iodimetric titrition
IO3- (aq) +5I- (aq) + 6H+ (aq) ------> 3 I2 (aq) +3H2O (l)
or
…..(1)
[IO3- (aq) +8I- (aq) + 6H+ (aq) ------> 3 I3- (aq) + 3H2O (l)]
I2 (aq) + 2S2O32- (aq) -------> S4O62- (aq) + 2I- (aq) ……(2)
or
[I3- (aq) + 2S2O32- (aq) -------> S4O62- (aq) +3I- (aq)]
Experiment of iodimetric titrition
The concentration of thiosulphate solution can be determined as
follow:
no. of moles of I2 = 3 x no. of moles of IO3no. of moles of I2 = 1/2 x no. of moles of S2O32-
no. of moles of S2O32- = 6 x no. of moles of IO3-
Determination of iron(Ⅲ)
1) Excess NaI (aq) is added to an oxidizing agent
(e.g. Fe3+) with unknown molarity.
2) The iodine liberated is then determined by the titration with
standard thiosulphate solution .
2Fe3+ (aq) + 2I- (aq) ----> 2Fe2+ (aq) + I2 (aq)
This method of titration (using iodide as reducing agent in the
titration )is known as the iodometric titration or iodometery.
Determination of iron (Ⅱ)
1)Standard acidified KMnO4 (aq) is added to an reducing
agent (e.g. Fe2+)
MnO4- (aq) +8H+ (aq) +5Fe2+ (aq)  Mn2+ (aq) +4H2O (l) +5Fe3+ (aq)
2) An indicator is not required as acidified potassium
permanganate(Ⅶ) itself serves as an indicator.
Determination of iron (Ⅱ)
3) When manganate (Ⅶ) is added to iron (Ⅱ) in acidic
solution , manganate(Ⅶ) is reduced to manganate(Ⅱ)
while iron (Ⅱ) is oxidized to iron (Ⅲ).
4) The colour of the reaction mixture at the equivalence
point is yellow (due to the presence of FeFe3+ (aq) )
5) When an extra drop of acidified potassium manganate
(Ⅶ)and thus become light purple .
Conclusion
Reducing agent can be determined by titration with
potassium permanganate
Oxidizing agent can be determined by titration with
sodium iodide (iodometric titration)
Discussion
Error in the idiometric titration
A standard solution of iodine must be used
immediately because its molarity changes
with time because:
•
iodine is volatile =>I2 can escape from the solution , causing the
decrease of [I2] with time
•
iodine can be oxidized by air (promoted by acids, heat & light
)=>[I2]increase with time
Discussion
Error in the titration with KMnO4
• An aqueous solution of KMnO4 is unstable .
4MnO4- + light -------> MnO2 (s) +2 O2 (g) + MnO22- (aq)
• So the KMnO4 (aq) should be stored in a
brown bottle & should be standardized before
use.
Making Galvanic Cells
No bridge
Cu – Al
0V
Cu – Zn
0V
Al – Zn
0V
Voltage Points to Deposit
2.0 V
Cu
Cu
1.1 V
Cu
Cu
0.9 V
Zn
Zn
3) The salt bridge allows the cell to continue to function. K+ or
NO3– can move into one of the half cells to balance the
charge created by + metal ions forming or leaving solution.
4) No. We did not get the same values in the lab
Cu – Zn cell
Electron flow
Cu (+)
Cu2+
Salt bridge
Zn (–)
Zn2+
Zn  Zn2+ + 2e– (oxidation - LEO)
Cu2+ + 2e–  Cu (reduction - GER)
Cu2+ + Zn  Cu + Zn2+
Al – Zn cell
Electron flow
Al (–)
Al3+
Salt bridge
Zn (+)
Zn2+
Zn2+ + 2e–  Zn (reduction - GER)
Al  Al3+ + 3e– (oxidation - LEO)
2Al + 3Zn2+  2Al3+ + 3Zn
Making Galvanic Cells
Cu-Zn
voltage
Al-Zn
voltage
No bridge
0
No bridge
0
0 sec
 0.9
0 sec
 0.5
30 sec
 0.9
30 sec
 0.5
Deposit
Cu
Deposit
Al
Points to
Cu
Points to
Al
3) The salt bridge allows the cell to continue to function. K+ or
NO3– can move into one of the half cells to balance the
charge created by + metal ions forming or leaving solution.
4) No. We did not get the same values in the lab
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Electron flow e-
Fe (+)
Salt bridge
Fe2+
Mg (–)
Mg2+
Mg  Mg2+ + 2e– (oxidation - LEO)
Fe2+ + 2e–  Fe (reduction - GER)
Mg + Fe2+  Mg2+ + Fe
Cu and Al
Electron flow e-
Cu (+)
Salt bridge
Cu2+
Al (–)
Al3+
Al  Al3+ + 3e– (oxidation - LEO)
Cu2+ + 2e–  Cu (reduction - GER)
3Cu2+ + 2Al  3Cu + 2Al3+
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