Transcript No Slide Title

REDOX
Before you start it would be helpful to…
• Recall the layout of the periodic table
• Be able to balance simple equations
REDOX
CONTENTS
• Definitions of oxidation and reduction
• Calculating oxidation state
• Use of H, O and F in calculating oxidation state
• Naming compounds
• Redox reactions
• Balancing ionic half equations
• Combining half equations to form a redox equation
• Revision check list
OXIDATION & REDUCTION - Definitions
OXIDATION
GAIN OF OXYGEN
2Mg + O2
——> 2MgO
magnesium has been oxidised as it has gained oxygen
REMOVAL (LOSS) OF HYDROGEN
C2H5OH
——> CH3CHO + H2
ethanol has been oxidised as it has ‘lost’ hydrogen
OXIDATION & REDUCTION - Definitions
REDUCTION
GAIN OF HYDROGEN
C2H4 + H2
——> C2H6
ethene has been reduced as it has gained hydrogen
REMOVAL (LOSS) OF OXYGEN
CuO + H2 ——> Cu + H2O
copper(II) oxide has been reduced as it has ‘lost’ oxygen
However as chemistry became more sophisticated, it
was realised that another definition was required
OXIDATION & REDUCTION - Definitions
OXIDATION AND REDUCTION IN TERMS OF ELECTRONS
Oxidation and reduction are not only defined as changes in O and H
...
OXIDATION
Removal (loss) of electrons
species will get less negative or more positive
REDUCTION
Gain of electrons
species will become more negative or less positive
REDOX
When reduction and oxidation take place
OXIDATION & REDUCTION - Definitions
OXIDATION AND REDUCTION IN TERMS OF ELECTRONS
Oxidation and reduction are not only defined as changes in O and H
...
OXIDATION
Removal (loss) of electrons
species will get less negative or more positive
REDUCTION
Gain of electrons
species will become more negative or less positive
REDOX
When reduction and oxidation take place
OIL - Oxidation Is the Loss of electrons
RIG - Reduction Is the Gain of electrons
OXIDATION NUMBER
Used to...
tell if oxidation or reduction has taken place
work out what has been oxidised and/or reduced
ATOMS AND SIMPLE IONS
The number of electrons which must be added or removed to become neutral
atoms
Na in Na =
0
neutral already ... no need to add any electrons
cations
Na in Na+ =
+1
need to add 1 electron to make Na+ neutral
anions
Cl in Cl¯ =
-1
need to take 1 electron away to make Cl¯ neutral
OXIDATION NUMBER
Used to...
tell if oxidation or reduction has taken place
work out what has been oxidised and/or reduced
ATOMS AND SIMPLE IONS
The number of electrons which must be added or removed to become neutral
atoms
Na in Na =
0
neutral already ... no need to add any electrons
cations
Na in Na+ =
+1
need to add 1 electron to make Na+ neutral
anions
Cl in Cl¯ =
-1
need to take 1 electron away to make Cl¯ neutral
Q.
What are the oxidation states of the elements in the following?
a) C
b) Fe3+
c) Fe2+
d) O2-
e) He
f) Al3+
OXIDATION NUMBER
Used to...
tell if oxidation or reduction has taken place
work out what has been oxidised and/or reduced
ATOMS AND SIMPLE IONS
The number of electrons which must be added or removed to become neutral
atoms
Na in Na =
0
neutral already ... no need to add any electrons
cations
Na in Na+ =
+1
need to add 1 electron to make Na+ neutral
anions
Cl in Cl¯ =
-1
need to take 1 electron away to make Cl¯ neutral
Q.
What are the oxidation states of the elements in the following?
a) C (0)
b) Fe3+ (+3)
c) Fe2+ (+2)
d) O2- (-2)
e) He (0)
f) Al3+ (+3)
OXIDATION NUMBER
MOLECULES
The SUM of the oxidation states adds up to ZERO
ELEMENTS
H in H2
=
0
both are the same and must add up to Zero
COMPOUNDS
C in CO2 =
O in CO2 =
+4
-2
1 x +4 and 2 x -2 = Zero
OXIDATION NUMBER
MOLECULES
The SUM of the oxidation states adds up to ZERO
ELEMENTS
H in H2
=
0
both are the same and must add up to Zero
COMPOUNDS
C in CO2 =
O in CO2 =
+4
-2
1 x +4 and 2 x -2 = Zero
Explanation
• because CO2 is a neutral molecule, the sum of the oxidation states must be zero
• for this, one element must have a positive OS and the other must be negative
OXIDATION NUMBER
MOLECULES
The SUM of the oxidation states adds up to ZERO
ELEMENTS
H in H2
=
0
both are the same and must add up to Zero
COMPOUNDS
C in CO2 =
O in CO2 =
+4
-2
1 x +4 and 2 x +2 = Zero
HOW DO YOU DETERMINE WHICH IS THE POSITIVE ONE?
• the more electronegative species will have the negative value
• electronegativity increases across a period and decreases down a group
• O is further to the right than C in the periodic table so it has the negative value
OXIDATION NUMBER
MOLECULES
The SUM of the oxidation states adds up to ZERO
ELEMENTS
H in H2
=
0
both are the same and must add up to Zero
COMPOUNDS
C in CO2 =
O in CO2 =
+4
-2
1 x +4 and 2 x +2 = Zero
HOW DO YOU DETERMINE THE VALUE OF
AN ELEMENT’S OXIDATION STATE?
• from its position in the periodic table and/or
• the other element(s) present in the formula
OXIDATION NUMBER
COMPLEX IONS
The SUM of the oxidation states adds up to THE CHARGE
e.g.
NO3SO42NH4+
sum of the oxidation states
sum of the oxidation states
sum of the oxidation states
=
=
=
-1
-2
+1
Examples
in SO42-
the oxidation state of
S = +6
O = -2
+6 + 4(-2) = -2
there is ONE S
there are FOUR O’s
so the ion has a 2- charge
OXIDATION NUMBER
COMPLEX IONS
The SUM of the oxidation states adds up to THE CHARGE
e.g.
NO3SO42NH4+
sum of the oxidation states
sum of the oxidation states
sum of the oxidation states
=
=
=
-1
-2
+1
Examples
What is the oxidation state (OS) of Mn in MnO4¯ ?
•
•
•
•
•
•
the oxidation state of oxygen in most compounds is
there are 4 O’s so the sum of its oxidation states
overall charge on the ion is
therefore the sum of all the oxidation states must add up to
the oxidation states of Mn four O’s must therefore equal
therefore the oxidation state of Mn in MnO4¯is
-2
-8
-1
-1
-1
+7
+7 + 4(-2) = - 1
OXIDATION NUMBER
CALCULATING OXIDATION NUMBER - 1
Many elements can exist in more than one oxidation state
In compounds, certain elements are used as benchmarks to work out other values
HYDROGEN
+1
except
0
-1
atom (H) and molecule (H2)
hydride ion, H¯ in sodium hydride NaH
OXYGEN
-2
except
0
-1
+2
atom (O) and molecule (O2)
in hydrogen peroxide, H2O2
in F2O
FLUORINE
-1
except
0
atom (F) and molecule (F2)
OXIDATION STATES
CALCULATING OXIDATION NUMBER - 1
Many elements can exist in more than one oxidation state
In compounds, certain elements are used as benchmarks to work out other values
HYDROGEN
+1
except
0
-1
atom (H) and molecule (H2)
hydride ion, H¯ in sodium hydride NaH
OXYGEN
-2
except
0
-1
+2
atom (O) and molecule (O2)
in hydrogen peroxide, H2O2
in F2O
FLUORINE
-1
except
0
atom (F) and molecule (F2)
Q.
Give the oxidation state of the element other than O, H or F in...
SO2
NH3
NO2
NH4+
IF7
Cl2O7
NO3¯
NO2¯
SO32-
S2O32-
S4O62-
MnO42-
What is odd about the value of the oxidation state of S in S4O62- ?
OXIDATION NUMBER
A.
The oxidation states of the elements other than O, H or F are
SO2
O = -2
2 x -2 = - 4
overall neutral
S = +4
NH3
H = +1
3 x +1 = +3
overall neutral
N=-3
NO2
O = -2
2 x -2 = - 4
overall neutral
N = +4
NH4+
H = +1
4 x +1 = +4
overall +1
N=-3
IF7
F = -1
7 x -1 = - 7
overall neutral
I = +7
Cl2O7
O = -2
7 x -2 = -14
overall neutral
Cl = +7
NO3¯
O = -2
3 x -2 = - 6
overall -1
N = +5
NO2¯
O = -2
2 x -2 = - 4
overall -1
N = +3
SO32-
O = -2
3 x -2 = - 6
overall -2
S = +4
S2O32-
O = -2
3 x -2 = - 6
overall -2
S = +2
S4O62-
O = -2
6 x -2 = -12
overall -2
S = +2½ ! (10/4)
MnO42-
O = -2
4 x -2 = - 8
overall -2
Mn = +6
What is odd about the value of the oxidation state of S in S4O62- ?
An oxidation state must be a whole number (+2½ is the average value)
(14/2)
(4/2)
OXIDATION NUMBER
CALCULATING OXIDATION STATE - 2
The position of an element in the periodic table can act as a guide
METALS
NON-METALS
• have positive values in compounds
• value is usually that of the Group Number
Al is +3
• mostly negative based on their usual ion
Cl
usually -1
OXIDATION STATES
CALCULATING OXIDATION STATE - 2
Q.
What is the oxidation state of each element in the following compounds/ions ?
CH4
PCl3
NCl3
CS2
ICl5
BrF3
PCl4+
H3PO4
NH4Cl
H2SO4
MgCO3
SOCl2
OXIDATION STATES
CALCULATING OXIDATION STATE - 2
Q.
What is the oxidation state of each element in the following compounds/ions ?
CH4
C=-4
H = +1
PCl3
P = +3
Cl = -1
NCl3
N = +3
Cl = -1
CS2
C = +4
S = -2
ICl5
I = +5
Cl = -1
BrF3
Br = +3
F = -1
PCl4+
P = +4
Cl = -1
H3PO4
P = +5
H = +1
O = -2
NH4Cl
N = -3
H = +1
Cl = -1
H2SO4
S = +6
H = +1
O = -2
MgCO3
Mg = +2 H = +4
O = -2
SOCl2
S = +4
O = -2
Cl = -1
OXIDATION STATES
THE ROLE OF OXIDATION STATE IN NAMING SPECIES
To avoid ambiguity, the oxidation state is often included in the name of a species
manganese(IV) oxide shows that
Mn is in the +4 oxidation state in MnO2
sulphur(VI) oxide for SO3
S is in the +6 oxidation state
dichromate(VI) for Cr2O72-
Cr is in the +6 oxidation state
phosphorus(V) chloride for PCl5
P is in the +5 oxidation state
phosphorus(III) chloride for PCl3
P is in the +3 oxidation state
Q.
Name the following... PbO2
SnCl2
SbCl3
TiCl4
BrF5
OXIDATION STATES
THE ROLE OF OXIDATION STATE IN NAMING SPECIES
To avoid ambiguity, the oxidation state is often included in the name of a species
manganese(IV) oxide shows that
Mn is in the +4 oxidation state in MnO2
sulphur(VI) oxide for SO3
S is in the +6 oxidation state
dichromate(VI) for Cr2O72-
Cr is in the +6 oxidation state
phosphorus(V) chloride for PCl5
P is in the +5 oxidation state
phosphorus(III) chloride for PCl3
P is in the +3 oxidation state
Q.
Name the following... PbO2
lead(IV) oxide
SnCl2
tin(II) chloride
SbCl3
antimony(III) chloride
TiCl4
titanium(IV) chloride
BrF5
bromine(V) fluoride
REDOX REACTIONS
OXIDATION AND REDUCTION IN TERMS OF ELECTRONS
Oxidation and reduction are not only defined as changes in O and H
REDOX
When reduction and oxidation take place
OXIDATION
Removal (loss) of electrons ‘OIL’
species will get less negative or more positive
REDUCTION
Gain of electrons ‘RIG’
species will become more negative or less positive
O
X
I
D
A
T
I
O
N
+7
+6
+5
+4
+3
+2
+1
0
-1
-2
-3
-4
R
E
D
U
C
T
I
O
N
REDOX REACTIONS
OXIDATION AND REDUCTION IN TERMS OF ELECTRONS
Oxidation and reduction are not only defined as changes in O and H
REDOX
When reduction and oxidation take place
OXIDATION
Removal (loss) of electrons ‘OIL’
species will get less negative or more positive
REDUCTION
Gain of electrons ‘RIG’
species will become more negative or less positive
REDUCTION in O.S.
Species has been REDUCED
e.g. Cl is reduced to Cl¯ (0 to -1)
INCREASE in O.S.
Species has been OXIDISED
e.g. Na is oxidised to Na+ (0 to +1)
O
X
I
D
A
T
I
O
N
+7
+6
+5
+4
+3
+2
+1
0
-1
-2
-3
-4
R
E
D
U
C
T
I
O
N
REDOX REACTIONS
OXIDATION AND REDUCTION IN TERMS OF ELECTRONS
REDUCTION in O.S.
Species has been REDUCED
Q.
INCREASE in O.S.
Species has been OXIDISED
State if the changes involve oxidation (O) or reduction (R) or neither (N)
Fe2+
—>
Fe3+
I2
F2
—>
—>
I¯
F2O
REDOX REACTIONS
OXIDATION AND REDUCTION IN TERMS OF ELECTRONS
REDUCTION in O.S.
Species has been REDUCED
Q.
INCREASE in O.S.
Species has been OXIDISED
State if the changes involve oxidation (O) or reduction (R) or neither (N)
Fe2+
—>
Fe3+
O
+2 to +3
I2
F2
—>
—>
I¯
F2O
R
R
0 to -1
0 to -1
Redox reactions of metals with acids
Mg (s) + 2HCl (aq)  MgCl2 (aq) + H2 (g)
Redox reactions of metals with acids
Mg (s) + 2HCl (aq)  MgCl2 (aq) + H2 (g)
We can assign oxidaton numbers to each atom in any equation in
order to –
Identify whether a redox reaction has taken place
Work out what has been oxidised and what has been reduced.
Redox reactions of metals with acids
Mg (s) + 2HCl (aq)  MgCl2 (aq) + H2 (g)
We can assign oxidaton numbers to each atom in any equation in
order to –
Identify whether a redox reaction has taken place
Work out what has been oxidised and what has been reduced.
The metal is oxidised, forming positive metal ions
The hydrogen ion in the acid is reduced, forming the element
hydrogen, as a gas
We can write the above equation to show the role of the hydrogen
ion, H+
Mg (s) + 2H+ (aq)  Mg2+ +H 2 (g)