REDOX
A guide for A level students
KNOCKHARDY PUBLISHING
2008
SPECIFICATIONS
KNOCKHARDY PUBLISHING
REDOX
INTRODUCTION
This Powerpoint show is one of several produced to help students understand
selected topics at AS and A2 level Chemistry. It is based on the requirements of
the AQA and OCR specifications but is suitable for other examination boards.
Individual students may use the material at home for revision purposes or it
may be used for classroom teaching if an interactive white board is available.
Accompanying notes on this, and the full range of AS and A2 topics, are
available from the KNOCKHARDY SCIENCE WEBSITE at...
www.knockhardy.org.uk/sci.htm
Navigation is achieved by...
either
or
clicking on the grey arrows at the foot of each page
using the left and right arrow keys on the keyboard
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
REDOX
Before you start it would be helpful to…
• Recall the layout of the periodic table
• Be able to balance simple 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
...
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
OIL - Oxidation Is the Loss of electrons
RIG - Reduction Is the Gain of electrons
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
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+
c) Fe2+
d) O2-
e) He
f) Al3+
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 (0)
b) Fe3+ (+3)
c) Fe2+ (+2)
d) O2- (-2)
e) He (0)
f) Al3+ (+3)
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
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 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- ?
OXIDATION STATES
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
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 C = +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
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
BALANCING REDOX HALF EQUATIONS
1
2
3
4
Work out formulae of the species before and after the change; balance if required
Work out 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 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
5 If equation still doesn’t balance, add sufficient water molecules to one side
BALANCING REDOX HALF EQUATIONS
1
2
3
4
Work out formulae of the species before and after the change; balance if required
Work out 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 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
5 If equation still doesn’t balance, add sufficient water molecules to one side
Example 1
Step 1
Step 2
Step 3
Fe2+
+2
Fe2+
Iron(II) being oxidised to iron(III)
——>
——>
Fe3+
+3
Fe3+
+
e¯
now balanced
An electron (charge -1) is added to the RHS of the equation...
this balances the oxidation state change i.e. (+2) ——> (+3)
+ (-1)
As everything balances, there is no need to proceed to Steps 4 and 5
BALANCING REDOX HALF EQUATIONS
1
2
3
4
Work out formulae of the species before and after the change; balance if required
Work out 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 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
5 If equation still doesn’t balance, add sufficient water molecules to one side
Example 2
MnO4¯ being reduced to Mn2+ in acidic solution
BALANCING REDOX HALF EQUATIONS
1
2
3
4
Work out formulae of the species before and after the change; balance if required
Work out 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 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
5 If equation still doesn’t balance, add sufficient water molecules to one side
Example 2
Step 1
MnO4¯ being reduced to Mn2+ in acidic solution
MnO4¯
———>
Mn2+
No need to balance
Mn; equal numbers
BALANCING REDOX HALF EQUATIONS
1
2
3
4
Work out formulae of the species before and after the change; balance if required
Work out 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 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
5 If equation still doesn’t balance, add sufficient water molecules to one side
Example 2
Step 1
Step 2
MnO4¯ being reduced to Mn2+ in acidic solution
MnO4¯
+7
———>
Mn2+
+2
Overall charge on MnO4¯ is -1; sum of the OS’s of all atoms must add up to -1
Oxygen is in its usual oxidation state of -2; four oxygen atoms add up to -8
To make the overall charge -1, Mn must be in oxidation state +7 ... [+7 + (4x -2) = -1]
BALANCING REDOX HALF EQUATIONS
1
2
3
4
Work out formulae of the species before and after the change; balance if required
Work out 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 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
5 If equation still doesn’t balance, add sufficient water molecules to one side
Example 2
Step 1
Step 2
Step 3
MnO4¯ being reduced to Mn2+ in acidic solution
MnO4¯
+7
MnO4¯ + 5e¯
———>
———>
Mn2+
+2
Mn2+
The oxidation states on either side are different;
+7 —> +2
To balance; add 5 negative charges to the LHS
[+7 + (5 x -1) = +2]
You must ADD 5 ELECTRONS to the LHS of the equation
(REDUCTION)
BALANCING REDOX HALF EQUATIONS
1
2
3
4
Work out formulae of the species before and after the change; balance if required
Work out 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 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
5 If equation still doesn’t balance, add sufficient water molecules to one side
Example 2
Step 1
Step 2
Step 3
Step 4
MnO4¯ being reduced to Mn2+ in acidic solution
MnO4¯ ———>
+7
MnO4¯ + 5e¯ ———>
MnO4¯ + 5e¯ + 8H+ ———>
Mn2+
+2
Mn2+
Mn2+
Total charges on either side are not equal;
LHS = 1- and 5RHS = 2+
Balance them by adding 8 positive charges to the LHS
[ 6- + (8 x 1+) = 2+ ]
You must ADD 8 PROTONS (H+ ions) to the LHS of the equation
=
6-
BALANCING REDOX HALF EQUATIONS
1
2
3
4
Work out formulae of the species before and after the change; balance if required
Work out 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 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
5 If equation still doesn’t balance, add sufficient water molecules to one side
Example 2
Step 1
Step 2
Step 3
Step 4
Step 5
MnO4¯ being reduced to Mn2+ in acidic solution
MnO4¯
+7
MnO4¯ + 5e¯
MnO4¯ + 5e¯ + 8H+
MnO4¯ + 5e¯ + 8H+
———>
———>
———>
———>
Mn2+
+2
Mn2+
Mn2+
Mn2+ +
Everything balances apart from oxygen and hydrogen
4H2O
now balanced
O
LHS = 4
RHS = 0
H
LHS = 8
RHS = 0
You must ADD 4 WATER MOLECULES to the RHS; the equation is now balanced
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
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+
+
e2e-
—> 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
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)
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
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
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+ ——>
Mn2+ + 4H2O + 5Fe3+ + 5e¯
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+
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
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 and H+
Cr2O72- and Fe2+
H2O2 and MnO4¯
C2O42- and MnO4¯
S2O32- and I2
Cr2O72- and 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)
REVISION CHECK
What should you be able to do?
Recall the definitions for oxidation and reduction in terms of oxygen, hydrogen and electrons
Write balanced equations representing oxidation and reduction
Know the trend in electronegativity across periods
Predict the oxidation state of elements in atoms, simple ions, compounds and complex ions
Recognize, in terms of oxidation state, if oxidation or reduction has taken place
Balance ionic half equations
Combine two ionic half equations to make a balanced redox equation
CAN YOU DO ALL OF THESE?
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NO
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