Balancing of Equation.pptx

Balancing of
Equations
Presented By: Mr. Faisal Imran
Department of Chemistry
APSACS Ord. Rd. RWP

Chemistry
REDOX REACTIONS: Balancing by the ion-electron method (acid medium)
What are they?
REDOX balancing
The ion-electron
method
Calculating oxidation no.
Identifying of semi-reactions
Ionic half-reactions
Balancing Half-reactions
Multiplying half-reactions
Global ionic reaction
Molecular reaction

What are they?
 A redox process is a reaction in which we find an electron transfer.
In this reactions the element or compound (A) increases its oxidation no. (from n to n+1)
so it is said that it has been oxidized.
And then another element or compound (B) diminishes its oxidation no. (from m to m-1)
so it is said that it has been reduced.
THIS PROCES THEN COULD BE UNDERSTOOD AS THE ADDITION OF TWO HALF-REACTION S:
OXIDATION:
REDUCTION:
Where A transfers an electron to B causing its
reduction and we would say that A is the
REDUCING AGENT
However, if it is B which gets the electron from A, it
becomes oxidized, then we would say that B is the
OXIDIZING AGENT
INDEX

Balancing redox reactions
 The redox reactions adjustment lie in the balance of the chemical equation:
2KMnO4 16HCl  2MnCl2  5Cl2  2KCl 8H2O
The easiest way is the:
ION–ELECTRON (HALF-REACTION) METHOD
INDEX GUIDE ASIC MEDIUM SPAÑOL

1) Calculate the oxidation numbers of each atom in each chemical.
KMnO4  HCl  MnCl2  Cl2  KCl  H2O
RULES:
The oxidation state of a simple one-atom ion is the same as its charge
The ion-electron method (acid medium)
INDEX GUIDE BASIC MEDIUM

1) Calculate the oxidation numbers of each atom in each chemical
+1 +7 -2 +1 -1 +2 -1 0 +1 -1 +1 -2
KMnO4  HCl  MnCl2  Cl2  KCl  H2O
RULES:
The oxidation state of a simple one-atom ion is the same as its charge
The oxidation state of a free element is zero
The oxidation state of H in its combinations is +1, with the exception of the metal
hydrides, where it is -1
The oxidation state of O in its combinations is -2, with the exception of the peroxides,
where it is -1 and in its combination with fluoride, where it will be +2
The oxidation state of the alkaline metals in its combinations is +1 (group 1 of the P.S.)
and +2 for the alkaline earth metals (group 2 of the P.S.)
In the case of the binary salts, the halogen elements behaves with -1, the elements in
the group of O with -2, the elements in the group of N with -3, C and Si -4 and B -3
The algebraic sum of all the oxidation numbers of each atom in a neutral compound is
zero. If the compound is an ion, this addition will be the same as the charge of the ion
INDEX GUIDE BASIC MEDIUM ESPAÑOL

2) Identify which element is reduced and which is oxidized.
+1 -1 +1 -2
KCl  H2O
+1 +7 -2 +1 -1
KMnO4  HCl 
REDUCTION
+2 -1 0
MnCl2  Cl2 
OXIDATION
INDEX GUIDE BASIC MEDIUM ESPAÑO

3) Write down the half-reactions in the ionic mode
+1 -1 +1 -2
KCl  H2O
+1 +7 -2 +1 -1
KMnO4  HCl 
REDUCTION
+2 -1 0
MnCl2  Cl2 
OXIDATION
*¡¡CAREFUL!!: Only the acids, hydroxides and salts are written the ionic mode
The electric charge (q) comes from the addition of oxidation numbers, then as an
example in the premanganate ion:
+7 -2
MnO4
q=7+4·(-2)=-1
4
MnO -
Reduction:
Oxidation:
MnO4
2+
-
 Mn
 Cl2
Cl -

4) Balance each half-reaction
+1 -1 +1 -2
KCl  H2O
+1 +7 -2 +1 -1
KMnO4  HCl 
REDUCTION
Reduction:
Oxidation:
MnO4
2+
-
 Mn
 Cl2
2Cl -
A) First, elements which are not hydrogen or oxygen are balanced
+2 -1 0
MnCl2  Cl2 
OXIDATION

+1 -1 +1 -2
KCl  H2O
+1 +7 -2 +1 -1
KMnO4  HCl 
REDUCTION
MnO4
2+
-
2Cl -
2
 Mn  4H2O
 Cl
+2 -1 0
MnCl2  Cl2 
OXIDATION
B) Oxygen is balanced by adding H2O where necessary
Reduction:
Oxidation:

+1 -1 +1 -2
KCl  H2O
+1 +7 -2 +1 -1
KMnO4  HCl 
REDUCTION
4
-
-
2Cl
MnO  8H
2
2
 Mn
2+
 4H O
 Cl
+2 -1 0
MnCl2  Cl2 
OXIDATION
+
C) Hydrogen is then balanced by adding H+ where necessary
Reduction:
Oxidation:

+1 -1 +1 -2
KCl  H2O
+1 +7 -2 +1 -1
KMnO4  HCl 
REDUCTION
4
-
-
2
D) Then balance the charges by adding electrons (e-) so that at the end the truth charge is the
2Cl Cl
2
2+
Mn  4H O
+
MnO  8H 5e
-
 2e-
+2 -1 0
MnCl2  Cl2 
OXIDATION
same in both parts of the reaction
THIS STEP IS NORMALLY CONFLICTIVE BUT IT IS NOT
DIFFICULT AT ALL, WE JUST NEED TO BARE IN MIND THAT
THE ELECTRONS ARE PLACED ON THE LEFT IN THE
REDUCTION AND ON THE RIGHT IN THE OXIDATION AND
THAT EACH ONE APORTS A NEGATIVE CHARGE. SO THAT
A SIMPOLE EQUATION CAN BE PLANTED:
4
- 2+
2
Mn  4H O
+
M nO  8H  xe
-
  x 
 
 
x=5
C) Hydrogen is then balanced by adding H+ where necessary
Reduction:
Oxidation:

5) Multiply each half-reaction with a coefficient for it to have the same number of electrons
+1 -1 +1 -2
KCl  H2O
4
-
-
(MnO  8H
(2Cl 2
2
+
5e
-
 Cl 2e-
2+
Mn  4H O )·2
)·5
+1 +7 -2 +1 -1
KMnO4  HCl 
REDUCTION
+2 -1 0
MnCl2  Cl2 
OXIDATION
Reduction:
Oxidation:

6) Add up boths semi reactions
+1 -1 +1 -2
KCl  H2O
4
-
-
2
2
+
5e
-
 Cl 2e-
(MnO  8H
(2Cl )·5
2+
Mn  4H O )·2 
4
2+
-

+ -
16H  10Cl 2 2
2MnO  2Mn  8H O  5Cl
+1 +7 -2 +1 -1
KMnO4  HCl 
REDUCTION
+2 -1 0
MnCl2  Cl2 
OXIDATION
Reduction:
Oxidation:

7) Turn the ionic reaction into molecular
+1 -1 +1 -2
KCl  H2O
THIS STEP IS NORMALLY CONFLICTIVE, HOWEVER IT IS
QUITE EASY IF WE BARE IN MIND THAT THE IONIC
REACTION IS ALREADY ADJUSTED, SO THEN WE ADD THE
IONS WE NEED IN THE SAME CUANTITY AS BOTH PARTS
OF THE REACTION
4
-
-
2
2
+
5e
-
 Cl 2e-
(MnO  8H
(2Cl )·5
2+
Mn  4H O )·2 
4
2+
-

+ -
16H  10Cl 2 2
2MnO  2Mn  8H O  5Cl
+1 +7 -2 +1 -1
KMnO4  HCl 
REDUCTION
+2 -1 0
MnCl2  Cl2 
OXIDATION
Reduction:
Oxidation:

+1 -1 +1 -2
KCl  H2O
EACH PERMANGANATE ION (MnO4
-) CARRIES ONE POTASSIUM ION (K+) IN THE ORIGINAL
REACTION
4
-
-
2
2
+
5e
-
 Cl 2e-
(MnO  8H
(2Cl )·5
2+
Mn  4H O )·2 
4
2+
-

+ -
16H  10Cl 2 2
2MnO  2Mn  8H O  5Cl
+1 +7 -2 +1 -1
KMnO4  HCl 
REDUCTION
+2 -1 0
MnCl2  Cl2 
OXIDATION
Reduction:
Oxidation:

+1 -1 +1 -2
KCl  H2O
4
-
-
2
2
+
5e
-
 Cl 2e-
(MnO  8H
(2Cl )·5
2+
Mn  4H O )·2 
4
2+
-

+ -
16H  10Cl 2 2
2MnO  2Mn  8H O  5Cl
2K+
2K+
EACH PERMANGANATE ION (MnO4
-) CARRIES ONE POTASSIUM ION (K+) IN THE ORIGINAL
REACTION
As in each ionic reaction there are two permanganate ions, we add two ions potassium (K+) to
each part of the reaction
+1 +7 -2 +1 -1
KMnO4  HCl 
REDUCTION
+2 -1 0
MnCl2  Cl2 
OXIDATION
Reduction:
Oxidation:

+1 -1 +1 -2
KCl  H2O
4
-
-
2
2
+
5e
-
 Cl 2e-
(MnO  8H
(2Cl )·5
2+
Mn  4H O )·2 
4
2+
-

+ -
16H  10Cl 2 2
2MnO  2Mn  8H O  5Cl
2K+
2K+
THE PROTONS (H+) AND THE CHLORIDE IONS (Cl-) MUST MAKE UP THE HYDROCHLORIC ACID,
AND IT IS USEFUL TO LEAVE THE ACIDS TO THE END
+1 +7 -2 +1 -1
KMnO4  HCl 
REDUCTION
+2 -1 0
MnCl2  Cl2 
OXIDATION
Reduction:
Oxidation:

+1 -1 +1 -2
KCl  H2O
4
-
-
2
2
+
5e
-
 Cl 2e-
(MnO  8H
(2Cl )·5
2+
Mn  4H O )·2 
4
2+
-

+ -
16H  10Cl 2 2
2MnO  2Mn  8H O  5Cl
2K+
2K+
EACH Mn2+ ION CARRIES TWO CHLORIDE IONS (Cl-) IN THE ORIGINAL REACTION
+1 +7 -2 +1 -1
KMnO4  HCl 
REDUCTION
+2 -1 0
MnCl2  Cl2 
OXIDATION
Reduction:
Oxidation:

+1 -1 +1 -2
KCl  H2O
4
-
-
2
2
+
5e
-
 Cl 2e-
(MnO  8H
(2Cl )·5
2+
Mn  4H O )·2 
4
2+
-

+ -
16H  10Cl 2 2
2MnO  2Mn  8H O  5Cl
2K+
4Cl -
4Cl -
2K+
EACH Mn2+ ION CARRIES TWO CHLORIDE IONS (Cl-) IN THE ORIGINAL REACTION
As in the ionic reaction there are two Mn2+ ions, we add four chloride ions (Cl-) to each part of the
reaction
+1 +7 -2 +1 -1
KMnO4  HCl 
REDUCTION
+2 -1 0
MnCl2  Cl2 
OXIDATION
Reduction:
Oxidation:

+1 -1 +1 -2
KCl  H2O
4
-
-
2
2
+
5e
-
 Cl 2e-
(MnO  8H
(2Cl )·5
2+
Mn  4H O )·2 
4
2+
-

+ -
16H  10Cl 2 2
2MnO  2Mn  8H O  5Cl
2K+
4Cl -
4Cl -
2K+
EACH POTASSIUM ION (K+) CARRIES ONE CHLORIDE ION (Cl-) IN THE ORIGINAL REACTION
+1 +7 -2 +1 -1
KMnO4  HCl 
REDUCTION
+2 -1 0
MnCl2  Cl2 
OXIDATION
Reduction:
Oxidation:

+1 -1 +1 -2
KCl  H2O
4
-
-
2
2
+
5e
-
 Cl 2e-
(MnO  8H
(2Cl )·5
2+
Mn  4H O )·2 
4
2+
- + -
16H  10Cl  2 2
2MnO  2Mn  8H O  5Cl
2K+
4Cl -
2Cl -
4Cl -
2K+
2Cl -
EACH POTASSIUM ION (K+) CARRIES ONE CHLORIDE ION (Cl-) IN THE ORIGINAL REACTION
As in the ionic reaction there are two potassium ions (K+), we add two chloride ions (Cl-) to each
part of the reaction
+1 +7 -2 +1 -1
KMnO4  HCl 
REDUCTION
+2 -1 0
MnCl2  Cl2 
OXIDATION
Reduction:
Oxidation:

+1 -1 +1 -2
KCl  H2O
4
-
-
2
2
+
5e
-
 Cl 2e-
(MnO  8H
(2Cl )·5
2+
Mn  4H O )·2 
4
2+
- + -
16H  10Cl  2 2
2MnO  2Mn  8H O  5Cl
2K+
4Cl -
2Cl -
4Cl -
2K+
2Cl -
THE REST OF SPECIES: WATER (H2O) AND CHLORINE (Cl2) BOTH IN THE ORIGINAL AND IN
THE IONIC REACTION ARE THE SAME SO THAT THERE IS NO NEED TO ADD ANYTHING
+1 +7 -2 +1 -1
KMnO4  HCl 
REDUCTION
+2 -1 0
MnCl2  Cl2 
OXIDATION
Reduction:
Oxidation:

+1 -1 +1 -2
KCl  H2O
4
-
-
2
2
+
5e
-
 Cl 2e-
(MnO  8H
(2Cl )·5
2+
Mn  4H O )·2 
4
2+
- + -
16H  10Cl  2 2
2MnO  2Mn  8H O  5Cl
2K+
4Cl -
2Cl -
4Cl -
2K+
2Cl -
NOW WE GROUP UP THE IONS IN ORDER TO FORM THE SPECIES OF THE ORIGINAL
REACTIONS
2KMnO4 16HCl  2MnCl2  Cl2  2KCl  H2O
+1 +7 -2 +1 -1
KMnO4  HCl 
REDUCTION
+2 -1 0
MnCl2  Cl2 
OXIDATION
Reduction:
Oxidation:

+1 -1 +1 -2
KCl  H2O
+1 +7 -2 +1 -1
KMnO4  HCl 
REDUCTION
Reduction:
Oxidation:
4
-
-
2
2
+
5e
-
 Cl 2e-
+2 -1 0
MnCl2  Cl2 
OXIDATION
(MnO  8H
(2Cl )·5
2+
Mn  4H O )·2 
4
2+
- + -
16H  10Cl  2 2
2MnO  2Mn  8H O  5Cl
2K+
4Cl -
2Cl -
4Cl -
2K+
2Cl -
IN THE CASE OF THE MOLECULAR SUBSTANCES WE PRESERVE THE COEFFICIENTS
2KMnO4 16HCl  2MnCl2 5Cl2  2KCl 8H2O

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