Transcript Redox reactions - SALEM-Immanuel Lutheran College

Early Definitions of Redox Reactions
Redox Reactions in terms of Loss and Gain of Oxygen
Oxidation :
The addition of oxygen to a substance
Reduction : The removal of oxygen from a substance
1
oxidation
CuO(s)
+
H2(g)
 Cu(s)
reduction
CuO is the oxidizing agent
H2 is the reducing agent
2
+
H2O(g)
Q.1
oxidation
3Fe(s) + 4H2O(g)

Fe3O4(s) + 4H2(g)
reduction
Oxidizing agent : H2O(g)
Reducing agent : Fe(s)
3
Redox Reactions in terms of Loss and Gain of Hydrogen
Oxidation :
The removal of hydrogen from a substance
Reduction : The addition of hydrogen to a substance
4
reduction
H2S(g)
+
Cl2(g)  S(s)
oxidation
Oxidizing agent : Cl2(g)
Reducing agent : H2S(g)
5
+
2HCl(g)
Q.2
oxidation
4NH3(g) +
3O2(g)  2N2(g) + 6H2O(g)
reduction
Oxidizing agent : O2(g)
Reducing agent : NH3(g)
6
The application is limited to reactions
involving O and/or H
 Broader definitions were developed
7
Modern Definitions of Redox Reactions
1. Redox Reactions in terms of Electron
Transfer
2. Redox Reactions in terms of Changes in
Oxidation States or Oxidation Numbers
8
Redox Reactions in terms of Electron Transfer
Oxidation : The loss of electrons
Reduction : The gain of electrons
9
oxidation
2Na(s)
+
Cl2(g)

2NaCl(s)
reduction
2Na(s)

2Na+(s) + 2e
Cl2(g) + 2e  2Cl(s)
10
Q.3
oxidation
Mg(s) + Cu2+(aq)  Mg2+(aq) + Cu(s)
reduction
Mg(s)

Cu2+(aq) +
11
Mg2+(aq)
2e

+
Cu(s)
2e
Q.4
oxidation
Mg(s) + H2O(g)  MgO(s) + H2(g)
reduction
Mg(s)
H2O(g)
12

Mg2+(s)
+ 2e

?
+
2e
H2(g)
13
•
oxidation and reduction always
accompany each other
•
Acid-base reactions are
competitions for protons
•
Redox reactions are competitions
for electrons
Competitions for protons
Gain of H+
stronger acid
weaker acid
Cl(aq) +
HCl(aq) + H2O(l)
stronger base
weaker base
Gain of H+
14
H3O+(aq)
Competitions for electrons
Gain of e
stronger R.A.
weaker R.A.
Mg(s) + Cu2+(aq)
Mg2+(aq) +
stronger
stronger O.A
O.A.
weaker
weaker O.A.
O.A.
Gain of e
15
Cu(s)
Redox Reactions in terms of Changes in O.N.
Oxidation : Increase in O.N.
Reduction : Decrease in O.N.
16
Definition of Oxidation Number : the presumed electrical charge of
the atom or ion in the substance
17
 The bond pair in any non-polar bond
or slightly polar bond between like
atoms is divided equally between the
two atoms.

the oxidation number of an atom in
an element is zero.
 E.g. H  H , bond pair is equally
shared  presumed electrical charge
on H is 0.  O.N. of H is 0
18
 The bond pair in any polar covalent
bond (including ionic bond) between
two unlike atoms is presumed to
belong to the more electronegative
atom.

the oxidation number of an ion in an
ionic compound is equal to its actual
electrical charge.

E.g. Na+Cl , O.N. of Na+ is +1 ;
19
O.N. of Cl is –1
 the oxidation number of an atom
in a compound with polar covalent
bonds is equal to the charge it
would have if it existed as an ion
in that compound.
 E.g. H  Cl , since Cl is more
electronegative, the presumed
electrical charges and thus O.N. of
Cl and H are 1 and 1 respectively.
20
-
+
-
O
C
O
Since O is more electronegative,
the presumed charge on O is –2
O.N. of O = –2
O.N. of C = +4
21
22
•
To work out O.N., information
about electronegativity and bond
arrangement of the species
are required.
•
To make it simple, a set of rules
has been devised for assignment
of O.N.
Rules for Assigning Oxidation States :  Oxidation state of an atom in an element
is 0
 Sum of oxidation states of a neutral
compound is 0
 Sum of oxidation states of a simple or
polyatomic ion = charge on the ion
23
Elements with Fixed Oxidation States in Their Compounds : 
Group I metals 
+1

Group II metals 
+2

Al  +3

H
+1
in covalent compounds with non-metals;
1
when combined with Group I or II metals.
E.g. metal hydrides such NaH,
MgH2.
24

F  1

O  2,
except in peroxides(1) , superoxides(0.5)
and compounds with F where it is +ve.

Cl  1
except in compounds with F and O.
Q.5 (1)
HClO4
(+1) + Cl + 4(-2) = 0
Cl = +7
Chloric(VII) acid
HClO3
(+1) + Cl + 3(-2) = 0
Cl = +5
25
Chloric(V) acid
Q.5 (1)
HClO2
(+1) + Cl + 2(-2) = 0
Cl = +3
Chloric(III) acid
HClO
(+1) + Cl + (-2) = 0
Cl = +1
26
Chloric(I) acid
Q.5 (2)
Na2SO4 2(+1) + S + 4(-2) = 0
S = +6
Sodium sulphate(VI)
Na2SO3 2(+1) + S + 3(-2) = 0
S = +4
Sodium sulphate(IV)
27
Q.5 (3)
KNO3
(+1) + N + 3(-2) = 0
N = +5
Potassium nitrate(V)
KNO2
(+1) + N + 2(-2) = 0
N = +3
Potassium nitrate(III)
28
Q.5 (4)
KMnO4 (+1) + Mn + 4(-2) = 0
Mn = +7
Potassium manganate(VII)
K2MnO4 2(+1) + Mn + 4(-2) = 0
Mn = +6
Potassium manganate(VI)
29
Q.5 (5)
FeCl3
Fe + 3(-1) = 0
Fe = +3
Iron(III) chloride
FeCl2
Fe + 2(-1) = 0
Fe = +2
Iron(II) chloride
30
Q.5 (6)
Na2CO3 2(+1) + C + 3(-2) = 0
C = +4
Sodium carbonate*
There is only one type of carbonate, O.N.
of carbon needs not be indicated.
31
Q.5 (6)
CO
C + (-2) = 0
C = +2
Carbon monoxide
CO2
C + 2(-2) = 0
C = +4
Carbon dioxide
32
Q.5 (7)
CH4
C + 4(+1) = 0
 C = -4
Methane
H
H
C
H
33
H
C is more electronegative
than H,
O.N. of C is – 4
Q.5 (7)
C2H6
2C + 6(+1) = 0
C = -3
Ethane
34
A bond pair is equally shared between two C atoms
H
-3
H
35
H
-3
C
C
H
H
H
Q.5 (7)
C3H8
3C + 8(+1) = 0
8
C
3
propane
36
Electrons between C atoms are equally shared
H
37
H
H
-3
C
-2
C
H
H
H
-3
C
H
H
Q.5 (8)
S2O32
2S + 3(-2) = -2
S = +2
Thiosulphate
38
Electrons between S atoms are equally shared
0
S
0
+4 +4
S
-2 O
-2
O -2-2
O -2
-2
39
Q.5 (8)
S4O62
4S + 6(-2) = -2
S = +2.5
Tetrathionate
40
Electrons between S atoms are equally shared
O
+5 S
O
0
S
0
S
O
+5
S
O
O
41
O
Q.5 (8)
S2O82
2S + 8(-2) = -2
S = +7
The highest O.N. of S is +6
Peroxodisulphate(VI)
42
The highest O.N. = Group number
The lowest O.N. = Group number – 8
and  -4
E.g. N,
Highest O.N. = +5;
Lowest O.N. = -3
Cl, Highest O.N. = +7;
Lowest O.N. = -1
43
-2
O
+6
-2 O
S
-1
O
-1
O
O-2
-2
O
+6
S
O -2
44
-2
O
Q.5 (9)
00
00
S
C
00
S
Electrons are equally shared between C and S
45
Q.5 (9)
C2H4O3 2C + 4(+1) + 3(-2) = 0
C = +1
Ozonide
46
+1
H
-2
H
O
+1
H
0
C
-1 O
47
+1
+1
C
0
O-1
H
Q.6 (1)
+3
+2
Fe3O4(s) + 8H+(aq)  Fe2+(aq) + 2Fe3+(aq) + 4H2O(l)
FeO
Fe2O3
mixed oxide  no change in O.N.
 non-redox reaction
48
Q.6 (2)
+4
+2
Pb3O4(s) + 4H+(aq)  PbO2(s) + 2Pb2+(aq) + 2H2O(l)
2PbO
PbO2
mixed oxide  no change in O.N.
 non-redox reaction
49
0
Q.6 (3)
S2O32-(aq) +
2H+(aq)
0
+4
-2 O
50
0
+4 +4
 S(s) + SO2(g)
+
H2O(l)
S
No change in O.N.
S
O -2  non-redox reaction
O -2
Internal Redox Reactions
When the oxidizing agent and the reducing
agent of a redox reaction are the same
substance, the redox reaction is known as
an internal redox reaction
51
oxidation
-2
2KClO3(s)

+5
reduction
52
0
2KCl(s)
-1
+
3O2(g)
oxidation
-2
2Pb(NO3)2(s)
+5

0
2PbO(s) + 4NO2(g) + O2(g)
+4
reduction
53
Q.7(a)
R.A. : NH4+(s)
oxidation
-3
NH4NO3(s)

+1
N2O(g)
+
2H2O(g)
+5
reduction
54
O.A. : NO3(s)
Q.7(b)
R.A. : Fe2+(s)
oxidation
+2
2FeSO4(s) 
+3
Fe2O3(s) + SO2(g) + SO3(g)
+6
+4
reduction
55
O.A. : SO42(s)
Q.7(c)
R.A. : NH4+(s)
oxidation
-3
0
(NH4)2Cr2O7(s)  Cr2O3(s) + 4H2O(g) + N2(g)
+6
+3
reduction
56
O.A. : Cr2O72(s)
Disproportionation
A redox reaction in which an element
(combined or uncombined) in an
intermediate oxidation state is
simultaneously reduced and oxidized
57
oxidation
-1
2H2O2(aq)

2H2O(l)
-2
reduction
58
+
0
O2(g)
oxidation
0
Cl2(aq) + H2O(l)  HCl(aq) +
-1
reduction
59
+1
HOCl(aq)
Q.8 (a)
oxidation
+4
+5
2NO2(g) + H2O(l)  HNO3(aq) + HNO2(aq)
+3
reduction
60
Q.8 (b)
oxidation
+3
+4
(COOH)2(s)  CO(g) + CO2(g) + H2O(l)
+2
reduction
O
+3
OH
C
C
HO
61
+3
O
Q.8 (c)
oxidation
+1
2CuI(s)

Cu(s)
0
reduction
62
+
+2
CuI2(aq)
Balancing Redox Equations
Two systematic methods : -
63

The Oxidation Number Method

The Half Reaction Method
Unlike the method of inspection, which is
a general method of balancing chemical
equations,
• the oxidation number method only
applies to redox reactions
• the half reaction method only
applies to electron-transfer reactions.
64
The Oxidation Number Method
Principle : Total increase in O.N. = total decrease in O.N.
65
Example 1
All fixed  RHS needs 6 H  RHS needs 3H2O
Step 3 : Balance the Rest of the Equation by Inspection
-2
0
C2H5OH(l) + 3 O2(g) 
+4 -2
2 CO2(g)
-2
+ 3 H2O(g)
Step
1 :O.N.
Work: C,
out -2
theChanges
inIn
O.N.
 In
+4

O.N. : O, 0  -2
Step
: Balance
Changes
in O.N.
Total2increase
in the
O.N.
= 2  (+6)
= +12
Total decrease in O.N. = 12 = 6  (2)
 six oxygen atoms are needed.
66
Example 2
14
7 hydrogen
oxygen
atoms
atoms
on
on
thethe
left
right
 
7 water
14 hydrogen
molecules
ions
are are
needed
needed
The
charges
are
automatically
balanced
fixed
Step fixed
3 : Balance the Rest of the Equation
by Inspection
+3
+2
+3
Cr2O72(aq) + 6 Fe2+(aq) +14H+(aq)  2 Cr3+(aq) + 6 Fe3+(aq) + 7H2O(l)
+6
Step 1 : Work out the Changes in O.N.
 In O.N. : Fe, +2  +3
 In O.N. : Cr, +6  +3
Step
: Balance
Changes
in O.N.
Total2decrease
inthe
O.N.
= 2  (-3)
= -6
Total increase in O.N. = +6 = 6  (+1)
 six Fe2+ ions are needed
67
Q.9 (a)
All fixed  RHS needs 2H2O
0
+5
+6
+4
HNO3(aq)  H2SO4(aq) + 6NO
NO2(g) + 2 H2O(l)
S(s) + 6HNO
 In O.N. : 0  +6
 In O.N. : +5  +4
Total  in O.N. = +6
Total  in O.N. = -6 = 6(-1)
 6 HNO3 are needed
68
Q.9 (b)
All fixed  RHS needs 1.5 H2O
4NH3(g) + 5O2(g)  4NO(g) + 6H2O(l)
-3
NH3(g) +
5 0
O (g)
4 2
+2 -2
 NO(g) +
 In O.N. : -3  +2
 In O.N. : 0  -2
Total  in O.N. = +5
Total  in O.N. = -5 = 2.5(-2)
 1.25 O2 are needed
69
-2
3
H O(l)
2 2
Q.9 (c)
All fixed  RHS needs 2H2O
0
C(s)
+6
+ 2 H2SO4(aq)
+4
 CO2(g)
 In O.N. : 0  +4
+4
+ 2 SO2(g)
 In O.N. : +6  +4
Total  in O.N. = +4
Total  in O.N. = -4 = 2(-2)
 2 H2SO4 are needed
70
+ 2 H2O(l)
Q.9 (d)
Six HNO3 should be added to LHS to balance

the six
Cu(NO
All fixed
 NO
RHS
needs
4H
3 in
3)22O
0
3Cu(s)
2 +5
2+2
+2
+ 28 HNO3(aq)  3Cu(NO3)2(aq) + 2NO(g) + 4 H2O(l)
3
3
 In O.N. : 0  +2
 In O.N. : +5  +2
Total  in O.N. = +2
Total  in O.N. = -2 = (2/3)(-3)
 (2/3) HNO3 is needed
71
Q.9 (e)
4FeS2(s) + 11O2(g)
+2 -1
2 FeS2(s)
11 0
+ O2(g)
2
 2Fe2O3(s) + 8SO2(g)
+3 -2
 Fe2O3(s) + 4 SO2(g)
 In O.N. : Fe, +2  +3
 In O.N. : O, 0  -2
S, -1  +4
Total  in O.N. = 2(+1) + 4(+5) = +22
Total  in O.N. = -22 = 11(-2)
72
+4 -2
 11 oxygen atoms are required
Q.9 (e)
Inspection method
11
2
2FeS2(s) + O2(g)
 Fe2O3(s) + 4SO2(g)
Balancing Fe
Balancing S
Balancing O
4FeS2(s) + 11O2(g)
73
 2Fe2O3(s) + 8SO2(g)
Q.9 (f)
2Na(s)
+
0
2NH3(g)
+1
Na(s)
+
NH3(g)
 2NaNH2(s) + H2(g)
+1
 NaNH2(s)
 In O.N. : Na, 0  +1
Total  in O.N. = 1(+1) = +1
Total  in O.N. = -1
 1 hydrogen atom are needed
 0.5 H2 is needed
74
10
+
H2(g)
2
 In O.N. : H, +1  0
Q.9 (g)
All fixed  0.5 H2O on the right
4MnO4(aq) + 4H+(aq)  4MnO2(s) + 2H2O(l) + 3O2(g)
+7 -2
MnO4(aq) +
H+(aq)
+4
 MnO2(s)
 In O.N. : O, -2  0
 In O.N. : Mn, +7  +4
Total  in O.N. = 1(-3) = -3
Total  in O.N. = +3 = 1.5(+2)
 1.5 oxygen atoms are needed
 0.75 O2 is needed
75
1
30
+ H2O(l) + O2(g)
2
4
Q.9(h)
Highly exothermic, glycerol burns with a purple flame
+7
-2/3
+4
+3
+4
KMnO4(s) + C3H5(OH)3(l)K2CO3(s)+Mn2O3(s)+CO2(g)+ H2O(l)
 In O.N. : C, -2/3  +4  In O.N. : Mn, +7  +3
Total  in O.N. = 3(4+2/3) = +14
Total  in O.N. = 7/2(-4) = -14
7/2KMnO4(s) + C3H5(OH)3(l)K2CO3(s)+7/4Mn2O3(s)+CO2(g)+ H2O(l)
76
Q.9(h)
7/2KMnO4(s) + C3H5(OH)3(l)K2CO3(s)+7/4Mn2O3(s)+CO2(g)+ H2O(l)
14KMnO4(s) + 4C3H5(OH)3(l)K2CO3(s)+7Mn2O3(s)+CO2(g)+ H2O(l)
14KMnO4(s) + 4C3H5(OH)3(l)7K2CO3(s)+7Mn2O3(s)+CO2(g)+ H2O(l)
14KMnO4(s) + 4C3H5(OH)3(l)7K2CO3(s)+7Mn2O3(s)+5CO2(g)+ H2O(l)
14KMnO4(s) + 4C3H5(OH)3(l)7K2CO3(s)+7Mn2O3(s)+5CO2(g)+ 16H2O(l)
77
The Half Reaction Method
In Acidic Solution : H+ and H2O are used to balance the half equations
In Basic Solution : OH and H2O are used to balance the half equations
78
In Acidic Solution : - Example
MnO4(aq) + Fe2+(aq) 
Mn2+(aq) +
Fe3+(aq)
+2
+7
(aq)  Mn2+(aq)
Reduction
half
equation
:
MnO
Step 1 : Identify and Write Equations
4 for the Half Reactions
+2
+3
Oxidation half equation : Fe2+(aq)  Fe3+(aq)
Step 2 : Balance Each Half Equation
79
For the reduction half equation : +7
(1) MnO4(aq) + 8H+(aq) + 5e 
+2
Mn2+(aq) + 4H2O(l)
O.N. of Mn decreases by 5 units  5 electrons are gained per
(3)
Balance
the Hydrogen
Hydrogen
Ion
(1) Add
electrons
according
toAdding
the change
in O.N.
(2)
Oxygen
byby
Adding
Water
Mn atom
(Acidic
Solution)
For the
oxidation
half equation : (2) Fe2+(aq)
 Fe3+(aq) + e
2+
Step
Balance
the
Electron
in the
Half
Equations
and
(1)3 :O.N.
of Fe
increases
byTransfer
1 unit
1
electron
is
lost per Fe
(2)/(3)
No
need
to
balance
oxygen
and
hydrogen
(1) + 5Add
 (2)
the Half Equations
5Fe2+(aq)+MnO4(aq)+8H+(aq)  5Fe3+(aq)+Mn2+(aq)+4H2O(l)
80
In Basic Solution : - Example 1
MnO4(aq)
+
Fe(OH)2(s)  Fe(OH)3(s)
+7
+ Mn(OH)2(s)
+2
Reduction half equation : MnO4(aq)  Mn(OH)2(s)
Step 1 : Identify and Write Equations for the Half Reactions
+2
+3
Oxidation half equation : Fe(OH)2(s)  Fe(OH)3(s)
Step 2 : Balance Each Half Equation
81
For the reduction half equation : +7
(1) MnO4(aq) + 4H2O(l)
+2
+ 5e  Mn(OH)2(s) + 6OH(aq)
O.N.
of
Mn oxidation
decreases
by
5ofunits
to5 the
electrons
arein
gained
per

(3)
(1)
Balance
Add
electrons
the
Rest
according
the
Equation
change
by
Adding
O.N.
Water
For
the
half
equation
:
(2) Balance the charge by adding
OH
Mn atom
(2) Fe(OH)2(s) + OH(aq)  Fe(OH)3(s)
+ e
2+
(1)(2)
O.N.
of
Fe
increases
by
1
unit
1
electron
is
lost
per
Fe

Step
3
:
Balance
the
Electron
Transfer
in
the
Half
Equations
and
(1) +Balance
5  (2)the charge by adding OH
Add
the Half Equations


MnO4 (aq)+4H2O(l)+5Fe(OH)2(s)  Mn(OH)2(s)+OH (aq)+5Fe(OH)3(s)
82
In Basic Solution : - Example 2
Ag(s) + CN(aq) + O2(g)
0
 Ag(CN)2(aq)
Reduction half equation : O2(g) 
-2
?
Step 1 : Identify and Write Equations for the Half Reactions
0
+1
Oxidation half equation : Ag(s)  Ag(CN)2(aq)
Step 2 : Balance Each Half Equation
83
For the reduction half equation : (1)
0
O2(g) + 2H2O(l) + 4e 
-2
4OH(aq)
O.N.
decreases
by
2 Rest
units
perthe
oxygen

2 change
electrons
are
(3)
Balance
the
of
Equation
by
Adding
Water
(1)
Add
electrons
according
to
the
in
O.N.

(2) Balance the charge by adding gained
OH per O atom
For the oxidation half equation : (2)
Ag(s) + 2CN(aq)  Ag(CN)2(aq) + e
(1)
O.N.
of Ag increases
by 1 unit
1
electron
per Ag atom
is lost
Step
3 : Balance
the
Electron
Transfer
in the
Half
Equations
and
(2)
Balance
the
charge
by
adding
CN
(1) + 4 Add
 (2)the Half Equations
O2(g) + 2H2O(l) + 4Ag(s) + 8CN(aq) 4OH(aq) + 4Ag(CN)2(aq)
84
Q.10 (a)
Cr2O72(aq) + C2H5OH(aq)  Cr3+(aq) + CO2(g)
+6
+3
Reduction half equation : Cr2O72(aq)  Cr3+(aq)
Step 1 : Identify and Write Equations for the Half Reactions
+4
-2
Oxidation half equation : C2H5OH(aq)  CO2(g)
Step 2 : Balance Each Half Equation
85
For the reduction half equation : (1) Cr2O72(aq) + 6e + 14H+(aq)
 2 Cr3+(aq) + 7H2O(l)
O.N.
of
Cr
byby
3Adding
units
3 electrons
are
gained
per
Balance
thedecreases
Hydrogen
Ion (Acidic
Solution)
For
the
oxidation
halfHydrogen
equation
:
Balance
the
Oxygen
by
Adding
Water
Cr atom
(2) C2H5OH(aq) + 3H2O(l)  2CO2(g) + 12e
+ 12H+(aq)
Balance
the
Electron
Transfer
in
the
Half Equations
and
Add
the
Balance
the
Oxygen
by
Adding
Water
O.N.
of
C
increases
by
6
unit

6
electrons
are
lost
per
C
atom
Balance
the
Hydrogen
by
Adding
Hydrogen
Ion
(Acidic
Solution)
2(1)
(2)
Half+Equations
2 Cr2O72(aq)+16H+(aq)+C2H5OH(aq) 4Cr3+(aq)+11H2O(l) + 2CO2(g)
86
Q.10 (b)
Zn(s) + VO2+(aq)

Zn2+(aq) + V2+(aq)
+5
Reduction half equation : VO2+(aq)

+2
V2 +(aq)
Step 1 : Identify and Write Equations for the Half Reactions
+2
0
Oxidation half equation : Zn(s)  Zn2+(aq)
Step 2 : Balance Each Half Equation
87
For the reduction half equation : (1) VO2+(aq)
+ 3e + 4H+(aq)
 V2 +(aq) + 2H2O(l)
O.N.Balance
ofthe
V decreases
by
3by
units
Water
3 electrons
are gained
per
Balance
Hydrogen
the
Oxygen
by
Adding
Adding
Hydrogen
Ion
(Acidic
Solution)
For the oxidation half equation
:V atom
(2) Zn(s)
 Zn2+(aq) + 2e
Balance
theincreases
Electronby
Transfer
Half Equations
the
O.N.
Zn
2 unit in
 the
2 electrons
are lostand
per Add
Zn atom
2(1) +of3(2)
Half Equations
2VO2+(aq)+ 8H+(aq)+3Zn(s) 2V2+(aq)+4H2O(l)+3Zn2+(aq)
88
Q.11 (a)
Cr2O72(aq) + C2H5OH(aq)  Cr(OH)3(s) + CO32(aq)
+6
Reduction half equation : Cr2O72(aq)
+3
 Cr(OH)3(s)
Step 1 : Identify and Write Equations for the Half Reactions
+4
-2
Oxidation half equation : C2H5OH(aq)  CO32(aq)
Step 2 : Balance Each Half Equation
89
For the reduction half equation : (1) Cr2O72(aq) + 6e + 7H2O(l)
 2Cr(OH)3(s) + 8OH(aq)
O.N.
of Crthe
decreases
by 3Equation
units 
3 electrons
are gained per
Balance
Rest
of
the
by
Adding
Water
(aq)
Balance
the
Charge
by
Adding
OH
Cr
atom : For the oxidation half equation
(2) C2H5OH(aq)+ 16OH(aq) 2CO32(aq) + 12e + 11H2O(l)
(aq)Equations
Balance
the
Electron
Transfer
in the
Half
andCAdd
the
Balance
Balance
the
the
Rest
Charge
ofby
the
Equation
Adding
OH
Adding
Water
O.N.
of C
increases
6by
unit

6by
electrons
are lost per
atom
Half+Equations
2(1)
(2)
2Cr2O72(aq)+ 3H2O(l)+ C2H5OH(aq)  4Cr(OH)3(s)+ 2CO32(aq)
90
Q.11 (b)
MnO4(aq) + C2O42(aq)  MnO2(s) + CO32(g)
+7
Reduction half equation : MnO4(aq)
+4
 MnO2(s)
Step 1 : Identify and Write Equations for the Half Reactions
+4
+3
Oxidation half equation : C2O42(aq)  CO32(g)
Step 2 : Balance Each Half Equation
91
For the reduction half equation : (1) MnO4(aq) + 3e + 2H2O(l)
 MnO2(s) + 4OH(aq)
O.N.
of Mn
decreases
byEquation
3 units by
 Adding
3 electrons
are gained per
Balance
the
Rest
of
the
Water
(aq)
Balance
the
Charge
by
Adding
OH
Mn
atom : For the oxidation half equation
(2) C2O42(aq) + 4OH(aq)  2CO32(g) + 2e
+ 2H2O(l)
(aq)Equations
Balance
theincreases
Electron
inbythe
Half
Add the
Balance
Rest
the Charge
of the
Equation
Adding
Adding
Water
O.N.Balance
of the
C
byTransfer
1by
unit

1OH
electron
is lost perand
C atom
Half
Equations
2(1)
+ 3(2)
2MnO4(aq)+3C2O42(aq)+ 4OH(aq)  2MnO2(s)+6CO32(g)+ 2H2O(l)
92
Q.12 (a)
CxHyOz + Cr2O72
 CO2 + Cr3+
+6
+3
Reduction half equation : Cr2O72(aq)  Cr3+(aq)
Step 1 : Identify and Write Equations for the Half Reactions
+4
?
Oxidation half equation : CxHyOz(aq)  CO2
x(?) y(1)  z(2)  0
2z - y
?
x
93
For the reduction half equation : (1) Cr2O72(aq) + 6e + 14H+(aq)
 2 Cr3+(aq) + 7H2O(l)
O.N.
ofthe
Crthe
decreases
byby
3 units
Water
3 electrons
gained
per
Balance
Balance
the
Hydrogen
Oxygen
by
Adding
Adding
Hydrogen
Ion are
(Acidic
Solution)
For
oxidation
half
equation
:
Cr atom
(2) CxHyOz(aq) + [2(z-2x)-y]H+(aq)  xCO2  x 4 

2z  y  
e
x 
+ (z-2x)H2O(l)
Balance
Transfer
in
the
Half Equations
and
Add the
Balancethe
theElectron
Oxygen by
Hydrogen
by
Adding
Adding
Water
Hydrogen
Ion (Acidic
Solution)
Half Equations
O.N. of C increases by 4 

94
2z  y 

2z  y 
 electrons are lost
 units   4 
x
x 


per C atom
(1) Cr2O72(aq) + 6e + 14H+(aq)
 2 Cr3+(aq) + 7H2O(l)
(2) CxHyOz(aq) + [2(z-2x)-y]H+(aq)  xCO2  x 4  2z  y e  + (z-2x)H2O(l)

x

2z  y 

x 4 
(1)  6(2)
x 

2z  y 
2z  y  + + 6 C H O (aq)


+ 6[2(z-2x)-y]H+(aq)
x 4 
 Cr2O72(aq)  14x 4 
H
x y z
x 
x 


2z  y  3+
2z  y 


2x 4 
 Cr (aq)  7x 4 
 H2O(l) + 6xCO2
x
x 



95
+ 6(z-2x)H2O(l)
Q.12 (b)
2z  y 
2z  y  + + 6 C H O (aq)


+ 6[2(z-2x)-y]H+(aq)
x 4 
 Cr2O72(aq)  14x 4 
H
x y z
x 
x 


2z  y  3+
2z  y 


2x 4 
 Cr (aq)  7x 4 
 H2O(l) + 6xCO2
x 
x 


nCr O 2
2
7
nCxHyOz
96
+ 6(z-2x)H2O(l)
2z  y 

 4x  2z  y 
x 4 
 x

4x  y  2z
x 
x






6
6
6
19.1 Redox Reactions (SB p.185)
(a) By using oxidation numbers, determine whether the
underlined species is an oxidizing agent or a reducing
agent.
2Na(s) + 2HCl(aq)  2NaCl(aq) + H2(g)
(a) Na(s) is the reducing agent as the
oxidation number of Na is increased from
0 for Na(s) to +1 for NaCl(aq).
97
Answer
19.1 Redox Reactions (SB p.185)
(b) By using oxidation numbers, determine whether the
underlined species is an oxidizing agent or a reducing
agent.
Cu2+(aq) + Ni(s)  Cu(s) + Ni2+(aq)
(b) Cu2+(aq) is the oxidizing agent as the
oxidation number of Cu is decreased from
+2 for Cu2+(aq) to 0 for Cu(s).
98
Answer
19.1 Redox Reactions (SB p.185)
(c) By using oxidation numbers, determine whether the
underlined species is an oxidizing agent or a reducing
agent.
2MnO4–(aq) + 5Cu(s) + 16H+(aq)
 2Mn2+(aq) + 5Cu2+(aq) + 8H2O(l)
(c) MnO4–(aq) is the oxidizing agent as the
oxidation number of Mn is decreased from
+7 for MnO4–(aq) to +2 for Mn2+(aq).
99
Answer
19.1 Redox Reactions (SB p.185)
(d) By using oxidation numbers, determine whether the
underlined species is an oxidizing agent or a reducing
agent.
2SO2(g) + O2(g)  2SO3(g)
(d) SO2(g) is the reducing agent as the
oxidation number of S is increased from
+4 for SO2(g) to +6 for SO3(g).
Back
100
Answer
19.2 Balancing Redox Equations (SB p.187)
Chlorine reacts with iron(II) sulphate solution to
form chloride ions and iron(III) sulphate. Write the
two half equations, and hence the overall equation
for the reaction. State the reducing agent and
oxidizing agent, and explain your answers.
Answer
101
19.2 Balancing Redox Equations (SB p.187)
Back
Oxidation: Fe2+(aq)  Fe3+(aq) + e– .......................…................ (1)
Reduction: Cl2(g) + 2e–  2Cl–(aq) ............................................ (2)
Multiplying equation (1) by 2, we obtain:
2Fe2+(aq)  2Fe3+(aq) + 2e– ...............…………...................... (3)
Combining equations (2) and (3) together and eliminating the
electrons on both sides, we have the overall equation:
2Fe2+(aq) + Cl2(g)  2Fe3+(aq) + 2Cl–(aq)
Assigning oxidation numbers to Fe and Cl, we can identify the
oxidizing agent and reducing agent of the reaction.
2Fe2+(aq) + Cl2(g)  2Fe3+(aq) + 2Cl–(aq)
+2
0
+3
–1
Fe2+(aq) is the reducing agent as the oxidation number of Fe is
increased from +2 for Fe2+(aq) to +3 for Fe3+(aq) in the reaction. Cl2(g)
is the oxidizing agent as the oxidation number of Cl is decreased
102
from 0 for Cl2(aq) to –1 for Cl–(aq) in the reaction.
19.2 Balancing Redox Equations (SB p.190)
Peroxodisulphate(VI) ions (S2O82–) react with
manganese(II) ions in an alkaline medium to form
sulphate ions (SO42–) and manganese(IV) oxide (MnO2).
Write the overall ionic equation for the reaction by using
the half equation method. State the oxidizing agent and
reducing agent of the reaction.
Answer
103
19.2 Balancing Redox Equations (SB p.190)
Back
The half equation for oxidation is:
Mn2+(aq) + 4OH–(aq)  MnO2(s) + 2H2O(l) + 2e–
The half equation for reduction is:
S2O82–(aq) + 2e–  2SO42–(aq)
Combining the two half equations and eliminating the electrons on both
sides, we obtain:
Mn2+(aq) + 4OH–(aq) + S2O82–(aq)  MnO2(s) + 2H2O(l) + 2SO42–(aq)
+2
+7
+4
+6
S2O82–(aq) is the oxidizing agent as the oxidation number of S is
decreased from +7 for S2O82–(aq) to +6 for SO42–(aq) in the reaction.
Mn2+(aq) is the reducing agent as the oxidation number of Mn is
increased from +2 for Mn2+(aq) to +4 for MnO2(s).
104
19.2 Balancing Redox Equations (SB p.190)
Sodium sulphate(IV) solution turns acidified
potassium dichromate(VI) solution from orange to
green. Write the overall ionic equation for the
reaction by using the half equation method. State
the oxidizing agent and reducing agent of the
reaction.
Answer
105
19.2 Balancing Redox Equations (SB p.190)
Back
The half equation for oxidation is:
SO32–(aq) + H2O(l)  SO42–(aq) + 2H+(aq) + 2e–.........……...…..... (1)
The half equation for reduction is:
Cr2O72–(aq) + 14H+(aq) + 6e–  2Cr3+(aq) + 7H2O(l) ............……... (2)
Multiplying equation (1) by 3, we obtain:
3SO32–(aq) + 3H2O(l)  3SO42–(aq) + 6H+(aq) + 6e– .......……......... (3)
Combining equations (2) and (3) and eliminating the electrons on both
sides, we have:
Cr2O72–(aq) + 8H+(aq) + 3SO32–(aq)
+6
+4
 2Cr3+(aq) + 3SO42–(aq) + 4H2O(l)
+3
+6
Sodium sulphate(IV) is the reducing agent as the oxidation number of S is
increased from +4 for SO32–(aq) to +6 for SO42–(aq). Acidified potassium
dichromate(VI)
is the oxidizing agent as the oxidation number of Cr is
106
decreased from +6 for Cr2O72–(aq) to +3 for Cr3+(aq).
19.2 Balancing Redox Equations (SB p.191)
Using the half equation method, write the overall ionic
equations for the following reactions. State the oxidizing
agent and reducing agent of the reaction, and explain your
answers.
(a) Reaction between lead and silver nitrate solution
Answer
107
19.2 Balancing Redox Equations (SB p.191)
(a)
108
The half equation for oxidation is:
Pb(s)  Pb2+(aq) + 2e– ................................... (1)
The half equation for reduction is:
Ag+(aq) + e–  Ag(s) ...................................... (2)
Multiplying equation (2) by 2, we obtain:
2Ag+(aq) + 2e–  2Ag(s) ............................... (3)
Combining equations (1) and (3) and eliminating electrons on both
sides, we obtain:
Pb(s) + 2Ag+(aq)  Pb2+(aq) + 2Ag(s)
0
+1
+2
0
Ag+(aq) is the oxidizing agent as the oxidation number of Ag is
decreased from +1 for Ag+(aq) to 0 for Ag(s). Pb(s) is the reducing
agent as the oxidation number of Pb is increased from 0 for Pb(s) to
+2 for Pb2+(aq).
19.2 Balancing Redox Equations (SB p.191)
Using the half equation method, write the overall ionic
equations for the following reactions. State the oxidizing
agent and reducing agent of the reaction, and explain your
answers.
(b) Reaction between acidified potassium manganate(VII)
solution and iron(II) sulphate solution
Answer
109
19.2 Balancing Redox Equations (SB p.191)
Back
(b)
110
The half equation for oxidation is:
Fe2+(aq)  Fe3+(aq) + e– ........................………………........ (1)
The half equation for reduction is:
MnO4–(aq) + 8H+(aq) + 5e–  Mn2+(aq) + 4H2O(l) ............... (2)
Multiplying equation (1) by 5, we obtain:
5Fe2+(aq)  5Fe3+(aq) + 5e– ....................…………………... (3)
Combining equations (2) and (3) and eliminating electrons on both
sides, we obtain:
MnO4–(aq) + 8H+(aq) + 5Fe2+(aq)  Mn2+(aq) + 4H2O(l) + 5Fe3+(aq)
+7
+2
+2
+3
MnO4–(aq) is the oxidizing agent as the oxidation number of Mn is
decreased from +7 for MnO4–(aq) to +2 for Mn2+(aq). Fe2+(aq) is the
reducing agent as the oxidation number of Fe is increased from +2 for
Fe2+(aq) to +3 for Fe3+(aq).
19.2 Balancing Redox Equations (SB p.191)
Balance the following redox equation by using the
oxidation number method.
Cr2O72–(aq) + SO2(aq)  2Cr3+(aq) + SO42–(aq)
Answer
111
19.2 Balancing Redox Equations (SB p.191)
1. Assigning oxidation numbers for the oxidizing agent, reducing agent
and the products, we have:
Cr2O72–(aq) + SO2(aq)  2Cr3+(aq) + SO42–(aq)
+6
+4
+3
+6
2. The oxidation number of each Cr atom is decreased by 3, so the
oxidation number of two Cr atoms is decreased by 6 totally. Besides,
the oxidation number of S is increased by 2.
O.N. increased by 2
Cr2O72–(aq) + SO2(aq)  2Cr3+(aq) + SO42–(aq)
+6
+4
+3
+6
O.N. decreased by (3  2) = 6
112
19.2 Balancing Redox Equations (SB p.191)
3. Since the total increase in oxidation number of the reducing agent is
equal to the total decrease in oxidation number of the oxidizing agent,
Cr2O72–(aq) and SO2(aq) are in the ratio of 1 : 3.
Cr2O72–(aq) + 3SO2(aq)  2Cr3+(aq) + 3SO42–(aq)
4. Balancing the number of O atoms by adding H2O(l) molecules, we have:
Cr2O72–(aq) + 3SO2(aq)  2Cr3+(aq) + 3SO42–(aq) + H2O(l)
5. Balancing the number of H atoms by adding H+(aq) ions, then the
balanced redox equation is obtained:
Cr2O72–(aq) + 3SO2(aq) + 2H+(aq)  2Cr3+(aq) + 3SO42–(aq) + H2O(l)
Back
113
19.2 Balancing Redox Equations (SB p.191)
Balance the following redox equation by using the oxidation
number method.
(a) MnO4–(aq) + Fe2+(aq)  Mn2+(aq) + Fe3+(aq)
Answer
114
19.2 Balancing Redox Equations (SB p.191)
(a)
1. Assigning the oxidation numbers for the oxidizing agents, reducing
agents and products, we obtain:
MnO4–(aq) + Fe2+(aq)  Mn2+(aq) + Fe3+(aq)
+7
+2
+2
+3
2. The oxidation number of Mn is decreased from +7 for MnO4–(aq) to
+2 for Mn2+(aq) (i.e. decreased by 5). The oxidation number of Fe
is increased from +2 for Fe2+(aq) to +3 for Fe3+(aq) (i.e. increased
by 1).
O.N. increased by 1
MnO4–(aq) + Fe2+(aq)  Mn2+(aq) + Fe3+(aq)
+7
+2
+2
+3
O.N. decreased by 5
115
19.2 Balancing Redox Equations (SB p.191)
3.
Since the total increase in oxidation number of the reducing agent is
equal to the total decrease in oxidation number of the oxidizing agent,
MnO4–(aq) and Fe2+(aq) are in the ratio of 1 : 5.
MnO4–(aq) + 5Fe2+(aq)  Mn2+(aq) + 5Fe3+(aq)
4. Balancing the number of O atoms by adding H2O(l) molecules, we have:
MnO4–(aq) + 5Fe2+(aq)  Mn2+(aq) + 5Fe3+(aq) + 4H2O(l)
5. Balancing the number of H atoms by adding H+(aq) ions, then the
following balanced equation is obtained.
MnO4–(aq) + 5Fe2+(aq) + 8H+(aq)  Mn2+(aq) + 5Fe3+(aq) + 4H2O(l)
116
19.2 Balancing Redox Equations (SB p.191)
Balance the following redox equation by using the oxidation
number method.
(b) IO4–(aq) + I–(aq)  I2(aq)
117
Answer
19.2 Balancing Redox Equations (SB p.191)
(b) 1. Assigning the oxidation numbers for the oxidizing agents, reducing
agents and products, we obtain:
IO4–(aq) + I–(aq)  I2(aq)
+7
–1
0
2. The oxidation number of I is decreased from +7 for IO4–(aq) to 0 for
I2(aq) (i.e. decreased by 7). On the other hand, the oxidation
number of I is increased from –1 for I–(aq) to 0 for I2(aq) (i.e.
increased by 1).
O.N. increased by 1
IO4–(aq) + I–(aq)  I2(aq)
+7
–1
0
118
O.N. decreased by 7
19.2 Balancing Redox Equations (SB p.191)
3.
Since the total increase in oxidation number of the reducing agent is
equal to the total decrease in oxidation number of the oxidizing agent,
IO4–(aq) and I–(aq) are in the ratio of 1 : 7.
IO4–(aq) + 7I–(aq)  4I2(aq)
4. Balancing the number of O atoms by adding H2O(l) molecules, we have:
IO4–(aq) + 7I–(aq)  4I2(aq) + 4H2O(l)
5. Balancing the number of H atoms by adding H+(aq) ions, then the
following balanced equation is obtained.
IO4–(aq) + 7I–(aq) + 8H+(aq)  4I2(aq) + 4H2O(l)
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