1. IB Chemistry Power Points
Topic 09
Oxidation and Reduction
www.pedagogics.ca

2. Much taken from
REDOX
A guide for A level students
KNOCKHARDY PUBLISHING
2008
SPECIFICATIONS

3. OXIDATION & REDUCTION – Simplified Definitions
OXIDATION
is a GAIN OF OXYGEN
2Mg + O2 ——> 2MgO
magnesium has been oxidised as it has gained oxygen
is the REMOVAL (LOSS) OF HYDROGEN
C2H5OH ——> CH3CHO + H2
ethanol has been oxidised as it has ‘lost’ hydrogen

4. OXIDATION & REDUCTION – Simplified Definitions
REDUCTION
is a GAIN OF HYDROGEN
C 2 H4 + H 2 ——> C2H6
ethene has been reduced as it has gained hydrogen
is the 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

5. OXIDATION & REDUCTION – Better 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

6. OIL - Oxidation Is the Loss of electrons
RIG - Reduction Is the Gain of electrons

7. OXIDATION NUMBERS (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
FOR ATOMS AND SIMPLE IONS
The number of electrons which must be added or removed to become neutral
Atoms Na = 0 neutral already ... no need to add any electrons
Cations Na+ = +1 need to add 1 electron to make Na+ neutral
Anions 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)

8. 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 = +4
O in CO2 = -2 1 x +4 and 2 x -2 = Zero
Explanation
• because CO2 is a neutral molecule, the sum of the oxidation states must be ?
• for this, one element must have a positive OS and the other must be ?
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 (oxygen is almost always -2 etc)
HOW DO YOU DETERMINE WHICH IS THE POSITIVE ONE?
• the more electronegative species will have the negative value

9. OXIDATION STATES
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
Example SO42-
in SO42- 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

10. OXIDATION STATES
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
Example
What is the oxidation number 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 -1
• 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

11. 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 atom (H) and molecule (H2)
-1 hydride ion, H¯ in sodium hydride NaH
OXYGEN -2 except 0 atom (O) and molecule (O2)
-1 in hydrogen peroxide, H2O2
+2 in F2O
HALOGENS -1 except 0 atom (X) and molecule (X2)
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- ?

12. 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
MnO42- O = -2 4 x -2 = - 8 overall -2 Mn = +6

13. 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 +1 +3 +5 or +7

14. 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 +1 +3 +5 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

15. 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
+1 +5 +2 +4 +6 +7 +6
What will be the usual and the maximum oxidation state in compounds of?
Li Br Sr O B N
USUAL +1 -1 +2 -2 +3 -3 or +5
MAXIMUM +1 +7 +2 +6 +3 +5

16. 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

17. 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

18. 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
+7
OXIDATION Removal (loss) of electrons ‘OIL’
species will get less negative or more positive O +6 R
X +5 E
REDUCTION Gain of electrons ‘RIG’ I +4 D
species will become more negative or less positive D +3 U
A +2 C
T +1 T
I 0 I
O -1 O
N -2 N
-3
-4

19. 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
+7
OXIDATION Removal (loss) of electrons ‘OIL’
species will get less negative or more positive O +6 R
X +5 E
REDUCTION Gain of electrons ‘RIG’ I +4 D
species will become more negative or less positive D +3 U
A +2 C
T +1 T
I 0 I
REDUCTION in O.N. Species has been REDUCED -1
O O
e.g. Cl is reduced to Cl¯ (0 to -1)
N -2 N
-3
INCREASE in O.N. Species has been OXIDISED -4
e.g. Na is oxidised to Na+ (0 to +1)

20. 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
H2O2 —> O2
H2O2 —> H2O
Cr2O72- —> Cr3+
Cr2O72- —> CrO42-
SO42- —> SO2

21. 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 0 to -1
F2 —> F2O R 0 to -1
C2O42- —> CO2 O +3 to +4
H2O2 —> O2 O -1 to 0
H2O2 —> H2O R -1 to -2
Cr2O72- —> Cr3+ R +6 to +3
Cr2O72- —> CrO42- N +6 to +6
SO42- —> SO2 R +6 to +4

22. OXIDATION STATES - Review
CALCULATING OXIDATION STATE – MOST IMPORTANT
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

23. 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 = +5 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

25. END OF PART ONE

26. WRITING & BALANCING REDOX HALF EQUATIONS

27. WRITING & BALANCING REDOX HALF EQUATIONS
Example 1 Iron(II) being oxidised to iron(III)
Step 1 Fe2+ ——> Fe3+
Step 2 +2 +3
Step 3 Fe2+ ——> 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)

28. In the case of reactions occurring in acidic solutions, more effort is required to
write the half reactions.
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
If equation still doesn’t balance, add sufficient water molecules to one side
Example 2 MnO4¯ being reduced to Mn2+ in acidic solution
No need to balance
Step 1 MnO4¯ ———> Mn2+
Mn; equal numbers

29. In the case of reactions occurring in acidic solutions, more effort is required to
write the half reactions.
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
If equation still doesn’t balance, add sufficient water molecules to one side
Example 2 MnO4¯ being reduced to Mn2+ in acidic solution
Step 1 MnO4¯ ———> Mn2+
Step 2 +7 +2
Overall charge on MnO4¯ is -1; sum of the ON’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]

30. In the case of reactions occurring in acidic solutions, more effort is required to
write the half reactions.
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
If equation still doesn’t balance, add sufficient water molecules to one side
Example 2 MnO4¯ being reduced to Mn2+ in acidic solution
Step 1 MnO4¯ ———> Mn2+
Step 2 +7 +2
Step 3 MnO4¯ + 5e¯ ———> Mn2+
The oxidation states on either side are different; +7 —> +2 (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

31. In the case of reactions occurring in acidic solutions, more effort is required to
write the half reactions.
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
If equation still doesn’t balance, add sufficient water molecules to one side
Example 2 MnO4¯ being reduced to Mn2+ in acidic solution
Step 1 MnO4¯ ———> Mn2+
Step 2 +7 +2
Step 3 MnO4¯ + 5e¯ ———> Mn2+
Step 4 MnO4¯ + 5e¯ + 8H+ ———> Mn2+
Total charges on either side are not equal; LHS = 1- and 5- = 6-
RHS = 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

32. In the case of reactions occurring in acidic solutions, more effort is required to
write the half reactions.
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
If equation still doesn’t balance, add sufficient water molecules to one side
Example 2 MnO4¯ being reduced to Mn2+ in acidic solution
Step 1 MnO4¯ ———> Mn2+
Step 2 +7 +2
Step 3 MnO4¯ + 5e¯ ———> Mn2+
Step 4 MnO4¯ + 5e¯ + 8H+ ———> Mn2+
Step 5 MnO4¯ + 5e¯ + 8H+ ———> Mn2+ + 4H2O 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

33. 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 Cr2O72- being reduced to Cr3+ in acidic solution
Step 1 Cr2O72- ———> Cr3+ there are two Cr’s on LHS
Cr2O72- ———> 2Cr3+ both sides now have 2
Step 2 2 Cr @ +6 2 Cr @ +3 both Cr’s are reduced
Step 3 Cr2O72- + 6e¯ ——> 2Cr3+ each Cr needs 3 electrons
Step 4 Cr2O72- + 6e¯ + 14H+ ——> 2Cr3+
Step 5 Cr2O72- + 6e¯ + 14H+ ——> 2Cr3+ + 7H2O now balanced

34. BALANCING REDOX HALF EQUATIONS
Q. Balance the following half equations...
Na —> Na+
Fe2+ —> Fe3+
I2 —> I¯
C2O42- —> CO2
H2O2 —> O2
H2O2 —> H2O
NO3- —> NO
NO3- —> NO2
SO42- —> 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

35. BALANCING REDOX HALF EQUATIONS
Q. Balance the following half equations...
Na —> Na+ + e-
Fe2+ —> Fe3+ + e-
I2 + 2e- —> 2I¯
C2O42- —> 2CO2 + 2e-
H2O2 —> O2 + 2H+ + 2e-
H2O2 + 2H+ + 2e- —> 2H2O
NO3- + 4H+ + 3e- —> NO + 2H2O
NO3- + 2H+ + e- —> NO2 + H2O
SO42- + 4H+ + 2e- —> SO2 + 2H2O

36. 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)

37. 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+ ——> Fe3+ + e¯ Oxidation
MnO4¯ + 5e¯ + 8H+ ——> Mn2+ + 4H2O Reduction

38. 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+ ——> Fe3+ + e¯ Oxidation
MnO4¯ + 5e¯ + 8H+ ——> Mn2+ + 4H2O Reduction
Step 2 5Fe2+ ——> 5Fe3+ + 5e¯ multiplied by 5
MnO4¯ + 5e¯ + 8H+ ——> Mn2+ + 4H2O multiplied by 1

39. 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+ ——> Fe3+ + e¯ Oxidation
MnO4¯ + 5e¯ + 8H+ ——> Mn2+ + 4H2O Reduction
Step 2 5Fe2+ ——> 5Fe3+ + 5e¯ multiplied by 5
MnO4¯ + 5e¯ + 8H+ ——> Mn2+ + 4H2O multiplied by 1
Step 3 MnO4¯ + 5e¯ + 8H+ + 5Fe2+ ——> Mn2+ + 4H2O + 5Fe3+ + 5e¯

40. OXIDIZING AGENT AND REDUCING AGENT
In the reaction between manganate(VII) and iron(II)
MnO4¯ + 5e¯ + 8H+ + 5Fe2+ ——> Mn2+ + 4H2O + 5Fe3+ + 5e¯
The manganate (VII) ion is reduced (gains electrons) and is called
the oxidizing agent.
The iron(II) ion is oxidized (loses electrons) and is called the
reducing agent.

41. 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+ ——> Fe3+ + e¯ Oxidation
MnO4¯ + 5e¯ + 8H+ ——> Mn2+ + 4H2O Reduction
Step 2 5Fe2+ ——> 5Fe3+ + 5e¯ multiplied by 5
MnO4¯ + 5e¯ + 8H+ ——> Mn2+ + 4H2O multiplied by 1
Step 3 MnO4¯ + 5e¯ + 8H+ + 5Fe2+ ——> Mn2+ + 4H2O + 5Fe3+ + 5e¯
Step 4 MnO4¯ + 8H+ + 5Fe2+ ——> Mn2+ + 4H2O + 5Fe3+

42. 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 (OX) and H+ (RED)
Fe2+ (OX) and Cr2O72- (RED)
MnO4¯ (RED) and H2O2 (OX)
MnO4¯ (RED) and C2O42- (OX)
S2O32- (OX) and I2 (RED)
Cr2O72- (RED) and I- (OX)

43. BALANCING Mg ——> Mg2+ + 2e¯ (x1)
H+ + e¯ ——> ½ H2 (x2)
REDOX
Mg + 2H+ ——> Mg2+ + H2
EQUATIONS
Cr2O72- + 14H+ + 6e¯ ——> 2Cr3+ + 7H2O (x1)
Fe2+ ——> Fe3+ + e¯ (x6)
Cr2O72- + 14H+ + 6Fe2+ ——> 2Cr3+ + 6Fe2+ + 7H2O
ANSWERS
MnO4¯ + 5e¯ + 8H+ ——> Mn2+ + 4H2O (x2)
H2O2 ——> O2 + 2H+ + 2e¯ (x5)
2MnO4¯ + 5H2O2 + 6H+ ——> 2Mn2+ + 5O2 + 8H2O
MnO4¯ + 5e¯ + 8H+ ——> Mn2+ + 4H2O (x2)
C2O42- ——> 2CO2 + 2e¯ (x5)
2MnO4¯ + 5C2O42- + 16H+ ——> 2Mn2+ + 10CO2 + 8H2O
2S2O32- ——> S4O62- + 2e¯ (x1)
½ I2 + e¯ ——> I¯ (x2)
2S2O32- + I2 ——> S4O62- + 2I¯
Cr2O72- + 14H+ + 6e¯ ——> 2Cr3+ + 7H2O (x1)
I¯ ——> ½ I2 + e¯ (x6)
Cr2O72- + 14H+ + 6I ¯ ——> 2Cr3+ + 3I2 + 7H2O

44. END OF PART TWO

45. Reactivity Series

46. Displacement reactions
A displacement reaction is one where a MORE REACTIVE metal will DISPLACE
a LESS REACTIVE metal from a compound.
Magnesium Copper sulphate
Mg Cu SO4
The magnesium
DISPLACES the copper from
copper sulphate
Mg SO4 Cu
Magnesium sulphate Copper

47. We have looked at several reactions: K
Fe + Cu(NO3)2 Cu + Fe(NO3)2 Na
Li + H2O LiOH + H2 Li
No, Ni is Ca
Yes, Li is
below Na
Such experiments reveal trends. The activity series ranks the
Mg
relative reactivity of metals.
above Zn
Al
It allows us to predict if certain chemicals will undergo Zn
Yes, Al is
single displacement reactions when mixed: metals near the
Fe
top are most reactive and will displace metals near the
above Cu
bottom. Ni
Sn
Q: Which of these reactions occur? Yes, Fe is Pb
Fe + CuSO4 Cu + Fe2(SO4)3 Cu
above H
Ni + NaCl NR (no reaction) Cu
Li + ZnCO3 Zn + Li2CO3 Hg
Al + CuCl2 Cu + AlCl3 Ag
Au

49. Trends in Oxidation and Reduction
Active metals:
Lose electrons easily
Are easily oxidized
Are strong reducing agents
Active nonmetals:
Gain electrons easily
Are easily reduced
Are strong oxidizing agents