REDOX REACTIONS:
Definitions, Oxidation States and redox
equations
Revision Guide
Outcomes
11.2.3.1 recall concepts from earlier grades: oxidation as gain of
oxygen and reduction as gain of hydrogen, oxidation and
reduction in terms of electron gain and loss
11.2.3.2 know and be able to use the rules for assigning oxidation
numbers and be able to assign oxidation numbers to individual
elements in compounds
11.2.3.3 be able to write oxidizing and reducing half equations
and to combine them into an overall redox equation
HOW DO WE DEFINE THE TERMS : “OXIDATION” AND
“REDUCTION” ?
Definition could be about:
 gain and loss of oxygen
 gain and loss of hydrogen
 gain and loss of electron ( you should know how to assign
oxidation numbers to individual elements in compounds)
YOUR TASK:
Which set represents oxidation reaction? reduction reaction?
Construct a definition for oxidation and reduction reactions based on what is
shown in the given equations.
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 ‘OIL’
species will get less negative or more positive
REDUCTION Gain of electrons
‘RIG’
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
OIL - Oxidation Is the Loss of
electrons
RIG - Reduction Is the Gain of
electrons
YOUR TASK:
Which set represents oxidation reaction? reduction reaction?
Construct a definition for oxidation and reduction reactions based on what is
shown in the given equations.
OXIDATION STATES
Used to...
tell if oxidation or reduction has taken place
work out what has been oxidised and/or reduced
construct half equations and balance redox equations
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+
d) O2-
e) He
c) Fe2+
f) Al3+
OXIDATION STATES
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 STATES
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 STATES
MOLECULES
The SUM of the oxidation states adds up to ZERO
ELEMENTS
Zero
H in H2 =
COMPOUNDS C in CO2
O in CO2
0
both are the same and must add up to
=
=
+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 STATES
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 STATES
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 STATES
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 STATES
CALCULATING OXIDATION STATE - 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 STATE - 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- ?
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 STATES
CALCULATING OXIDATION STATE - 2
The position of an element in the periodic table can act as a guide
METALS
• have positive values in compounds
• value is usually that of the Group Number
• where there are several possibilities the
values go no higher than the Group No.
NON-METALS
Al is +3
Sn can be +2 or +4
Mn can be +2,+4,+6,+7
• mostly negative based on their usual ion
Cl
usually -1
• can have values up to their Group No.
Cl
+1 +3 +5 or +7
OXIDATION STATES
CALCULATING OXIDATION STATE - 2
The position of an element in the periodic table can act as a guide
• have positive values in compounds
• value is usually that of the Group Number
METALS
• where there are several possibilities the
values go no higher than the Group No.
NON-METALS
Q.
Al is +3
Sn can be +2 or +4
Mn can be +2,+4,+6,+7
• mostly negative based on their usual ion
Cl
usually -1
• can have values up to their Group No.
Cl
+1 +3 +5 or +7
What is the theoretical maximum oxidation state of the following elements?
Na
P
Ba
Pb
S
Mn
Cr
What will be the usual and the maximum oxidation state in compounds of?
Li
Br
Sr
O
B
N
+1
OXIDATION STATES
CALCULATING OXIDATION STATE - 2
The position of an element in the periodic table can act as a guide
A.
What is the theoretical maximum oxidation state of the following elements?
Na
+1
P
+5
Ba
+2
Pb
+4
S
+6
Mn
+7
Cr
+6
What will be the usual and the maximum oxidation state in compounds of?
USUAL
MAXIMUM
Li
Br
Sr
O
B
N
+1
+1
-1
+7
+2
+2
-2
+6
+3
+3
-3 or +5
+5
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
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)
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
C2O42H2O2
H2O2
Cr2O72Cr2O72SO42-
—>
—>
—>
—>
—>
—>
—>
—>
I¯
F2O
CO2
O2
H2O
Cr3+
CrO42SO2
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
C2O42H2O2
H2O2
Cr2O72Cr2O72SO42-
—>
—>
—>
—>
—>
—>
—>
—>
I¯
F2O
CO2
O2
H2O
Cr3+
CrO42SO2
R
R
O
O
R
R
N
R
0 to -1
0 to -1
+3 to +4
-1 to 0
-1 to -2
+6 to +3
+6 to +6
+6 to +4
YOUR TASK:
Set 5 equations are redox reactions.
Which of the following is oxidised or reduced?
Cu or C ?
Al or Fe ?
Look at the oxidation states:
Cu2+O2- + C0→ 2Cu0 + C4+O22
Al0 + Fe3+2O32- → Al3+2O32- + 2Fe0
During oxidation and reduction, the oxidation numbers of
atoms change.
If an atom is oxidized, its oxidation number increases
(ie it becomes more +ve or less –ve)
If an atom is reduced, its oxidation number decreases
(ie it becomes less +ve or more –ve)
These ideas can be summarized in the following table:
DISPROPORTIONATION REACTIONS
In a disproportionation reaction, a species is simultaneously
oxidised and reduced.
OXIDISING and REDUCING AGENTS
OXIDISING and REDUCING AGENTS
An oxidizing agent, or oxidant, gains electrons and is
reduced in a chemical reaction.
A reducing agent, or reductant, loses electrons and is
oxidized in a chemical reaction.
YOUR TASK:
Complete the table about the following reactions.
BALANCING REDOX HALF EQUATIONS
Watch out for cases when the species is present in different amounts on
either side of the equation ... IT MUST BE BALANCED FIRST
Example 3
Step 1
Cr2O72Cr2O72-
Step 2
2 @ +6
Step 3
Cr2O72- being reduced to Cr3+ in acidic solution
———>
———>
there are two Cr’s on LHS
both sides now have 2
Cr3+
2Cr3+
both Cr’s are reduced
2 @ +3
Cr2O72- + 6e¯
——>
2Cr3+
Step 4
Cr2O72- + 6e¯ + 14H+
——>
2Cr3+
Step 5
Cr2O72- + 6e¯ + 14H+
——>
2Cr3+ +
each Cr needs 3 electrons
7H2O now balanced
BALANCING REDOX HALF EQUATIONS CLASSWORK
Q.
1
2
3
4
5
Balance the following half equations...
Na
—> Na+
Fe2+
—>
Fe3+
I2
—>
I¯
C2O42-
—>
CO2
H2O2
—>
O2
H2O2
—>
H2O
NO3-
—>
NO
NO3-
—>
NO2
SO42-
—>
SO2
REMINDER
Work out the formula of the species before and after the change; balance if required
Work out the oxidation state of the element before and after the change
Add electrons to one side of the equation so that the oxidation states balance
If the charges on all the species (ions and electrons) on either side of the equation do
not balance then add sufficient H+ ions to one of the sides to balance the charges
If the equation still doesn’t balance, add sufficient water molecules to one side
BALANCING REDOX HALF EQUATIONS
Q.
Balance the following half equations...
Na
—> Na+
+
e—>
Fe2+
I2
+
2e-
Fe3+
+
e-
2e-
—> 2I¯
C2O42-
—> 2CO2
+
H2O2
—>
+ 2H+ +
H2O2 + 2H+ + 2e-
—> 2H2O
O2
NO3- + 4H+ + 3e- —>
NO
+
2H2O
NO3- + 2H+ + e-
NO2
+
H2O
—> SO2
+
2H2O
SO42- + 4H+ + 2e-
—>
2e-
COMBINING HALF EQUATIONS
A combination of two ionic half equations, one involving oxidation and the other
reduction, produces a REDOX equation. The equations are balanced as follows...
Step 1
Step 2
Step 3
Step 4
Write out the two half equations
Multiply the equations so that the number of electrons in each is the same
Add the two equations and cancel out the electrons on either side
If necessary, cancel any other species which appear on both sides
The reaction between manganate(VII) and iron(II)
Step 1
Fe2+ ——>
MnO4¯ + 5e¯ + 8H+ ——>
Fe3+ + e¯
Mn2+ + 4H2O
Oxidation
Reduction
Step 2
5Fe2+
MnO4¯ + 5e¯ + 8H+
——>
——>
5Fe3+ + 5e¯
Mn2+ + 4H2O
multiplied by 5
multiplied by 1
Step 3
MnO4¯ + 5e¯ + 8H+ + 5Fe2+ ——>
Step 4
MnO4¯ + 8H+ + 5Fe2+ ——>
Mn2+ + 4H2O + 5Fe3+ + 5e¯
Mn2+ + 4H2O + 5Fe3+
SUMMARY
COMBINING HALF EQUATIONS CLASSWORK
A combination of two ionic half equations, one involving oxidation and the other
reduction, produces a REDOX equation. The equations are balanced as follows...
Step 1
Step 2
Step 3
Step 4
Q.
Write out the two half equations
Multiply the equations so that the number of electrons in each is the same
Add the two equations and cancel out the electrons on either side
If necessary, cancel any other species which appear on both sides
Construct balanced redox equations for the reactions between...
Mg/ Mg2+ and H+ / H2
Cr2O72- / Cr3+
and
Fe2+ /Fe3+
and
MnO4¯ /Mn2+
C2O42- / CO2 and
MnO4¯ /Mn2+
H2O2 / O2
S2O32- /S4O62
Cr2O72- / Cr3+
and
I 2 / I¯
and I2 /I¯
BALANCING
REDOX
EQUATIONS
ANSWERS
Mg ——> Mg2+ + 2e¯
H+ + e¯ ——> ½ H2
Mg + 2H+ ——> Mg2+ + H2
(x1)
(x2)
Cr2O72- + 14H+ + 6e¯ ——> 2Cr3+ + 7H2O
Fe2+ ——> Fe3+ + e¯
Cr2O72- + 14H+ + 6Fe2+ ——> 2Cr3+ + 6Fe2+ + 7H2O
(x1)
(x6)
MnO4¯ + 5e¯ + 8H+ ——>
Mn2+ + 4H2O
H2O2 ——> O2 + 2H+ + 2e¯
2MnO4¯ + 5H2O2 + 6H+ ——> 2Mn2+ + 5O2 + 8H2O
(x2)
(x5)
MnO4¯ + 5e¯ + 8H+ ——>
Mn2+ + 4H2O
C2O42- ——> 2CO2 + 2e¯
2MnO4¯ + 5C2O42- + 16H+ ——> 2Mn2+ + 10CO2 + 8H2O
(x2)
(x5)
2S2O32- ——> S4O62- + 2e¯
½ I2 + e¯ ——> I¯
2S2O32- + I2 ——> S4O62- + 2I¯
(x1)
(x2)
Cr2O72- + 14H+ + 6e¯ ——> 2Cr3+ + 7H2O
½ I2 + e¯ ——> I¯
Cr2O72- + 14H+ + 3I2 ——> 2Cr3+ + 6I ¯ + 7H2O
(x1)
(x6)
• TASK,,