1. A guide for A level students KNOCKHARDY PUBLISHING
REDOX
A guide for A level students
2008 SPECIFICATIONS
KNOCKHARDY PUBLISHING

2. 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...
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3. 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

4. Before you start it would be helpful to…
REDOX
Before you start it would be helpful to…
Recall the layout of the periodic table
Be able to balance simple equations

5. OXIDATION OXIDATION & REDUCTION - Definitions 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

6. REDUCTION OXIDATION & REDUCTION - Definitions GAIN OF HYDROGEN
C2H 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

7. 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

8. OIL - Oxidation Is the Loss of electrons
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

9. OXIDATION STATES ATOMS AND SIMPLE IONS
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 = neutral already ... no need to add any electrons
cations Na in Na+ = need to add 1 electron to make Na+ neutral
anions Cl in Cl¯ = need to take 1 electron away to make Cl¯ neutral

10. OXIDATION STATES ATOMS AND SIMPLE IONS
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 = neutral already ... no need to add any electrons
cations Na in Na+ = need to add 1 electron to make Na+ neutral
anions Cl in Cl¯ = 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) Fe c) Fe2+
d) O2- e) He f) Al3+

11. OXIDATION STATES ATOMS AND SIMPLE IONS
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 = neutral already ... no need to add any electrons
cations Na in Na+ = need to add 1 electron to make Na+ neutral
anions Cl in Cl¯ = 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) Al (+3)

12. The SUM of the oxidation states adds up to ZERO
MOLECULES
The SUM of the oxidation states adds up to ZERO
ELEMENTS H in H2 = both are the same and must add up to Zero
COMPOUNDS C in CO2 = +4
O in CO2 = x +4 and 2 x -2 = Zero

13. The SUM of the oxidation states adds up to ZERO
MOLECULES
The SUM of the oxidation states adds up to ZERO
ELEMENTS H in H2 = both are the same and must add up to Zero
COMPOUNDS C in CO2 = +4
O in CO2 = 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

14. OXIDATION STATES MOLECULES
The SUM of the oxidation states adds up to ZERO
ELEMENTS H in H2 = both are the same and must add up to Zero
COMPOUNDS C in CO2 = +4
O in CO2 = 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

15. OXIDATION STATES MOLECULES
The SUM of the oxidation states adds up to ZERO
ELEMENTS H in H2 = both are the same and must add up to Zero
COMPOUNDS C in CO2 = +4
O in CO2 = 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

16. The SUM of the oxidation states adds up to THE CHARGE
COMPLEX IONS
The SUM of the oxidation states adds up to THE CHARGE
e.g. NO3- sum of the oxidation states = - 1
SO42- sum of the oxidation states = - 2
NH4+ sum of the oxidation states = +1
Examples
in SO the oxidation state of S = +6 there is ONE S
O = -2 there are FOUR O’s
+6 + 4(-2) = -2 so the ion has a 2- charge

17. OXIDATION STATES COMPLEX IONS Examples
The SUM of the oxidation states adds up to THE CHARGE
e.g. NO3- sum of the oxidation states = - 1
SO42- sum of the oxidation states = - 2
NH4+ sum of the oxidation states = +1
Examples
What is the oxidation state (OS) of Mn in MnO4¯ ?
the oxidation state of oxygen in most compounds is - 2
there are 4 O’s so the sum of its oxidation states - 8
overall charge on the ion is
therefore the sum of all the oxidation states must add up to - 1
the oxidation states of Mn four O’s must therefore equal - 1
therefore the oxidation state of Mn in MnO4¯is +7
+7 + 4(-2) = - 1

18. 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 atom (H) and molecule (H2)
-1 hydride ion, H¯ in sodium hydride NaH
OXYGEN except 0 atom (O) and molecule (O2)
-1 in hydrogen peroxide, H2O2
+2 in F2O
FLUORINE except 0 atom (F) and molecule (F2)

19. Q. Give the oxidation state of the element other than O, H or F in...
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 atom (H) and molecule (H2)
-1 hydride ion, H¯ in sodium hydride NaH
OXYGEN except 0 atom (O) and molecule (O2)
-1 in hydrogen peroxide, H2O2
+2 in F2O
FLUORINE 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- ?

20. 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 = 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 = x -2 = -14 overall neutral Cl = (14/2)
NO3¯ O = x -2 = - 6 overall -1 N = +5
NO2¯ O = x -2 = - 4 overall -1 N = +3
SO32- O = x -2 = - 6 overall -2 S = +4
S2O32- O = x -2 = - 6 overall -2 S = (4/2)
S4O62- O = x -2 = -12 overall -2 S = +2½ ! (10/4)
MnO42- O = 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)

21. 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 Al is +3
• where there are several possibilities the
values go no higher than the Group No. Sn can be +2 or +4
Mn can be +2,+4,+6,+7
NON-METALS • mostly negative based on their usual ion Cl usually -1
• can have values up to their Group No. Cl or +7

22. 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 Al is +3
• where there are several possibilities the
values go no higher than the Group No. Sn can be +2 or +4
Mn can be +2,+4,+6,+7
NON-METALS • mostly negative based on their usual ion Cl usually -1
• can have values up to their Group No. Cl or +7
Q. 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

23. 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 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
USUAL or +5
MAXIMUM

24. CALCULATING OXIDATION STATE - 2
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

25. CALCULATING OXIDATION STATE - 2
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 Cl = -1 O = -2

26. THE ROLE OF OXIDATION STATE IN NAMING SPECIES
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

27. THE ROLE OF OXIDATION STATE IN NAMING SPECIES
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

28. 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

29. 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)

30. OXIDATION AND REDUCTION IN TERMS OF ELECTRONS
REDOX REACTIONS
OXIDATION AND REDUCTION IN TERMS OF ELECTRONS
REDUCTION in O.S. INCREASE in O.S.
Species has been REDUCED Species has been OXIDISED
Q State if the changes involve oxidation (O) or reduction (R) or neither (N)
Fe2+ —> Fe3+
I2 —> I¯
F2 —> F2O
C2O42- —> CO2
H2O —> O2
H2O2 —> H2O
Cr2O72- —> Cr3+
Cr2O72- —> CrO42-
SO42- —> SO2

31. OXIDATION AND REDUCTION IN TERMS OF ELECTRONS
REDOX REACTIONS
OXIDATION AND REDUCTION IN TERMS OF ELECTRONS
REDUCTION in O.S. INCREASE in O.S.
Species has been REDUCED Species has been OXIDISED
Q State if the changes involve oxidation (O) or reduction (R) or neither (N)
Fe2+ —> Fe3+ O +2 to +3
I2 —> I¯ R to -1
F2 —> F2O R to -1
C2O42- —> CO2 O +3 to +4
H2O —> O2 O to 0
H2O2 —> H2O R to -2
Cr2O72- —> Cr3+ R to +3
Cr2O72- —> CrO42- N to +6
SO42- —> SO2 R to +4

32. BALANCING REDOX HALF EQUATIONS
1 Work out formulae of the species before and after the change; balance if required
2 Work out oxidation state of the element before and after the change
3 Add electrons to one side of the equation so that the oxidation states balance
4 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

33. BALANCING REDOX HALF EQUATIONS
1 Work out formulae of the species before and after the change; balance if required
2 Work out oxidation state of the element before and after the change
3 Add electrons to one side of the equation so that the oxidation states balance
4 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 Iron(II) being oxidised to iron(III)
Step 1 Fe2+ ——> Fe3+
Step
Step 3 Fe2+ ——> Fe 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

34. BALANCING REDOX HALF EQUATIONS
1 Work out formulae of the species before and after the change; balance if required
2 Work out oxidation state of the element before and after the change
3 Add electrons to one side of the equation so that the oxidation states balance
4 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 MnO4¯ being reduced to Mn2+ in acidic solution

35. BALANCING REDOX HALF EQUATIONS
1 Work out formulae of the species before and after the change; balance if required
2 Work out oxidation state of the element before and after the change
3 Add electrons to one side of the equation so that the oxidation states balance
4 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 MnO4¯ being reduced to Mn2+ in acidic solution
Step MnO4¯ ———> Mn2+
No need to balance Mn; equal numbers

36. BALANCING REDOX HALF EQUATIONS
1 Work out formulae of the species before and after the change; balance if required
2 Work out oxidation state of the element before and after the change
3 Add electrons to one side of the equation so that the oxidation states balance
4 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 MnO4¯ being reduced to Mn2+ in acidic solution
Step MnO4¯ ———> Mn2+
Step
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 + (4x -2) = -1]

37. BALANCING REDOX HALF EQUATIONS
1 Work out formulae of the species before and after the change; balance if required
2 Work out oxidation state of the element before and after the change
3 Add electrons to one side of the equation so that the oxidation states balance
4 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 MnO4¯ being reduced to Mn2+ in acidic solution
Step MnO4¯ ———> Mn2+
Step
Step MnO4¯ e¯ ———> Mn2+
The oxidation states on either side are different; +7 —> (REDUCTION)
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

38. BALANCING REDOX HALF EQUATIONS
1 Work out formulae of the species before and after the change; balance if required
2 Work out oxidation state of the element before and after the change
3 Add electrons to one side of the equation so that the oxidation states balance
4 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 MnO4¯ being reduced to Mn2+ in acidic solution
Step MnO4¯ ———> Mn2+
Step
Step MnO4¯ e¯ ———> Mn2+
Step 4 MnO4¯ e¯ H+ ———> Mn2+
Total charges on either side are not equal; LHS = 1- and = 6-
RHS = 2+
Balance them by adding 8 positive charges to the LHS [ (8 x 1+) = 2+ ]
You must ADD 8 PROTONS (H+ ions) to the LHS of the equation

39. BALANCING REDOX HALF EQUATIONS
1 Work out formulae of the species before and after the change; balance if required
2 Work out oxidation state of the element before and after the change
3 Add electrons to one side of the equation so that the oxidation states balance
4 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 MnO4¯ being reduced to Mn2+ in acidic solution
Step MnO4¯ ———> Mn2+
Step
Step MnO4¯ e¯ ———> Mn2+
Step 4 MnO4¯ e¯ H+ ———> Mn2+
Step 5 MnO4¯ e¯ H+ ———> Mn H2O now balanced
Everything balances apart from oxygen and hydrogen O LHS = 4 RHS = 0
H LHS = 8 RHS = 0
You must ADD 4 WATER MOLECULES to the RHS; the equation is now balanced

40. 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 Cr2O72- being reduced to Cr3+ in acidic solution
Step 1 Cr2O72- ———> Cr there are two Cr’s on LHS
Cr2O72- ———> 2Cr both sides now have 2
Step both Cr’s are reduced
Step Cr2O e¯ ——> 2Cr3+ each Cr needs 3 electrons
Step Cr2O e¯ H ——> 2Cr3+
Step Cr2O e¯ H ——> 2Cr H2O now balanced

41. BALANCING REDOX HALF EQUATIONS
Q. Balance the following half equations...
Na —> Na+
Fe —> Fe3+
I —> I¯
C2O —> CO2
H2O —> O2
H2O —> H2O
NO —> NO
NO —> NO2
SO —> SO2
REMINDER
1 Work out the formula of the species before and after the change; balance if required
2 Work out the oxidation state of the element before and after the change
3 Add electrons to one side of the equation so that the oxidation states balance
4 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
5 If the equation still doesn’t balance, add sufficient water molecules to one side

42. BALANCING REDOX HALF EQUATIONS
Q. Balance the following half equations...
Na —> Na e-
Fe —> Fe e-
I e —> 2I¯
C2O —> 2CO e-
H2O —> O H e-
H2O H e- —> 2H2O
NO H e- —> NO H2O
NO H+ + e- —> NO H2O
SO H e- —> SO H2O

43. 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 Write out the two half equations
Step 2 Multiply the equations so that the number of electrons in each is the same
Step 3 Add the two equations and cancel out the electrons on either side
Step 4 If necessary, cancel any other species which appear on both sides

44. 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 Write out the two half equations
Step 2 Multiply the equations so that the number of electrons in each is the same
Step 3 Add the two equations and cancel out the electrons on either side
Step 4 If necessary, cancel any other species which appear on both sides
The reaction between manganate(VII) and iron(II)

45. 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 Write out the two half equations
Step 2 Multiply the equations so that the number of electrons in each is the same
Step 3 Add the two equations and cancel out the electrons on either side
Step 4 If necessary, cancel any other species which appear on both sides
The reaction between manganate(VII) and iron(II)
Step 1 Fe2+ ——> Fe e¯ Oxidation
MnO4¯ e¯ H+ ——> Mn H2O Reduction

46. 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 Write out the two half equations
Step 2 Multiply the equations so that the number of electrons in each is the same
Step 3 Add the two equations and cancel out the electrons on either side
Step 4 If necessary, cancel any other species which appear on both sides
The reaction between manganate(VII) and iron(II)
Step 1 Fe2+ ——> Fe e¯ Oxidation
MnO4¯ e¯ H+ ——> Mn H2O Reduction
Step Fe ——> 5Fe e¯ multiplied by 5
MnO4¯ e¯ H+ ——> Mn H2O multiplied by 1

47. 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 Write out the two half equations
Step 2 Multiply the equations so that the number of electrons in each is the same
Step 3 Add the two equations and cancel out the electrons on either side
Step 4 If necessary, cancel any other species which appear on both sides
The reaction between manganate(VII) and iron(II)
Step 1 Fe2+ ——> Fe e¯ Oxidation
MnO4¯ e¯ H+ ——> Mn H2O Reduction
Step Fe ——> 5Fe e¯ multiplied by 5
MnO4¯ e¯ H+ ——> Mn H2O multiplied by 1
Step 3 MnO4¯ + 5e¯ + 8H Fe2+ ——> Mn H2O + 5Fe e¯

48. 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 Write out the two half equations
Step 2 Multiply the equations so that the number of electrons in each is the same
Step 3 Add the two equations and cancel out the electrons on either side
Step 4 If necessary, cancel any other species which appear on both sides
The reaction between manganate(VII) and iron(II)
Step 1 Fe2+ ——> Fe e¯ Oxidation
MnO4¯ e¯ H+ ——> Mn H2O Reduction
Step Fe ——> 5Fe e¯ multiplied by 5
MnO4¯ e¯ H+ ——> Mn H2O multiplied by 1
Step 3 MnO4¯ + 5e¯ + 8H Fe2+ ——> Mn H2O + 5Fe e¯
Step MnO4¯ H Fe2+ ——> Mn H2O Fe3+

49. 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 Write out the two half equations
Step 2 Multiply the equations so that the number of electrons in each is the same
Step 3 Add the two equations and cancel out the electrons on either side
Step 4 If necessary, cancel any other species which appear on both sides
The reaction between manganate(VII) and iron(II)
Step 1 Fe2+ ——> Fe e¯ Oxidation
MnO4¯ e¯ H+ ——> Mn H2O Reduction
Step Fe ——> 5Fe e¯ multiplied by 5
MnO4¯ e¯ H+ ——> Mn H2O multiplied by 1
Step 3 MnO4¯ + 5e¯ + 8H Fe2+ ——> Mn H2O + 5Fe e¯
Step MnO4¯ H Fe2+ ——> Mn H2O Fe3+
SUMMARY

50. 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 Write out the two half equations
Step 2 Multiply the equations so that the number of electrons in each is the same
Step 3 Add the two equations and cancel out the electrons on either side
Step 4 If necessary, cancel any other species which appear on both sides
Q. 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¯

51. ANSWERS BALANCING REDOX EQUATIONS Mg ——> Mg2+ + 2e¯ (x1)
H+ + e¯ ——> ½ H2 (x2)
Mg + 2H+ ——> Mg H2
Cr2O H+ + 6e¯ ——> 2Cr H2O (x1)
Fe2+ ——> Fe3+ + e¯ (x6)
Cr2O H+ + 6Fe2+ ——> 2Cr Fe H2O
MnO4¯ e¯ H+ ——> Mn H2O (x2)
H2O2 ——> O H e¯ (x5)
2MnO4¯ H2O H+ ——> 2Mn O H2O
MnO4¯ e¯ H+ ——> Mn H2O (x2)
C2O42- ——> 2CO e¯ (x5)
2MnO4¯ + 5C2O H+ ——> 2Mn CO2 + 8H2O
2S2O ——> S4O e¯ (x1)
½ I e¯ ——> I¯ (x2)
2S2O I2 ——> S4O I¯
Cr2O H+ + 6e¯ ——> 2Cr H2O (x1)
½ I e¯ ——> I¯ (x6)
Cr2O H I2 ——> 2Cr I ¯ H2O
ANSWERS

52. What should you be able to do?
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? YES NO

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REDOX
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© 2008 JONATHAN HOPTON & KNOCKHARDY PUBLISHING