Articles by "OXIDATION REDUCTION REACTION"

Showing posts with label OXIDATION REDUCTION REACTION. Show all posts

Oxidation Number and Concept of Oxidation and Reduction:

The definition of oxidation and reduction based on loss or gain of electrons is limited to the scope. The definition holds goods for ionic compounds. 
For Examples, formation of water from hydrogen and oxygen, can not be covered by electronic concept since water is not an ionic compound.
2H2 + O2 2H2O
It may recall classically we could still say that hydrogen is oxidised to H2O. In the same seance burning of magnesium in oxygen is considered oxidation. Similarly hydrogen and chlorine react react to form a covalent molecule hydrogen chloride.
2H2 + Cl2  2HCl
The Above reaction hydrogen is oxidised or chlorine is reduced but the resulting compound is covalent one, thus the reaction can not be covered by the electronic concept. In order to cover such reactions also under oxidation-reduction the concept of oxidation number is developed and it is defined as,

Oxidation Number:

The oxidation number of an element in a compound is the formal charge (positive or negative) which would be assigned to the element, if all the bonds in the compounds were ionic bonds.
All the compounds are treated as though they were ionic merely because of the case of counting oxidation numbers. The oxidation number therefore is arbitrary. 
Electronegativity concept is utilized in adducing a formal charge to an atom. The less electronegative of the partners of a binary compound is arbitrary assigned a positive oxidation number and more electronegative one a negative oxidation number. Florine, being the most electronegative, has always a negative oxidation number. The alkali metals are highly electropositive, have low electronegativity, hence have been assigned positive oxidation numbers.
Method of finding out the Oxidation Number:
The following general rules are to be observed for the assignment of oxidation numbers.
1. Atoms of diatomic molecules like H2, Cl2, O2 etc or of metallic elements are assigned zero oxidation numbers since same elements of similar electronegativity are involved in the bonding.
H ➖H
The oxidation number of the above molecules are zero because two hydrogen atom of same electronegativity are involves for bonding.
2. Except of metal hydrides oxidation number of hydrogen in hydrogen compound is +1.
In alkali metal hydrides, LiH, NaH, CsH etc, the oxidation number of hydrogen, for reasons of varying electronegativity, is -1 
Examples:
NaH Na+ + H- 
(Here oxidation number of H is  -1)
HCl H+ + Cl- 
(Here oxidation number of H is +1)
3. The oxidation number of metal is positive.
For Examples, CuO Cu+2 + O-2 
(Here Oxidation number of Cu is +2)
4. Oxygen has normally an oxidation number -2. In hydrogen peroxide the oxidation number of oxygen is -1 since hydrogen has to be assigned +1.
In peroxides and super oxides the oxidation number of oxygen is -1 and -1/2 respectively. 
In fluorine monoxide(F2O) oxygen has an oxidation number +1 because fluorine is more electronegative than oxygen. 
For Examples, CuO  Cu+2 + O-2 
(Here Oxidation number of O is -2).
In H2O, Oxidation Number of Oxygen is (-2), 
but in H2O2, Oxidation Number of H is (+1) and Oxidation Number of Oxygen is (-1). 
In BaO2, Oxidation Number of Ba is (+2) and thus the Oxidation Number of Oxygen is (-1). 
In Na2O, Oxidation Number of Na is (+1) and Oxidation Number of oxygen is (-2). 
Oxidation number of P in Ba(H2PO2)2 is - (a)+3, (b)+2, (c) +1, (d) -1.
(c) +1
Oxidation number of Ba is +2, oxidation number of hydrogen is +1 and oxidation number of oxygen is -2.
Let the oxidation number of P is x
∴ (+2)+2{2(+1)+x+2(-2)} = 0 
or, 2x-2=0; 
or, x=+1
5. The oxidation number of an ion is equal to its charge.
For examples, NaCl Na+ + Cl-
The charge and oxidation number of Na+ is +1 and the charge and oxidation number of Cl- is -1.
Similar way, MgBr2 Mg+2 + 2Br-
Here the charge of Mg+2 is +2 and the oxidation number also +2 and the charge of Br- is -1, thus the oxidation number also -1.
6. The algebraic sum of the oxidation numbers of all the atoms in a compound must be zero.
For Examples:
In HCl oxidation number of hydrogen is +1 and oxidation number of chlorine is -1.
And the sum of these = (+1) + (-1) = 0.
Oxidation Number of an Element in a compound:
1. Oxidation Number of Mn in KMnO4:
Let the oxidation number of Mn in KMnO4 is x.
Thus according to the above rule, 
(+1) + x + 4(-2) = 0
or, x = +7
Thus, the oxidation number of Mn in KMnO4 is +7.
2. Oxidation Number of Mn in MnO4-2:
Let the oxidation number of Mn in MnO4-2 is x and the oxidation number of oxygen is -2(according to the above rule).
Thus the sum of the oxidation number of MnO4-2 = Charge of the MnO4-2.
x +4 (-2) = -2 
or, x = +6
Thus, the oxidation number of Mn in MnO4-2 is +6
3. Oxidation Number of Cr in Cr2O7-2 :
Let the oxidation number of Cr in Cr2O7-2 is x.
∴ 2x + 7(-2) = -2
or, x = +6
Thus, the oxidation number of Cr in Cr2O7-2 is +6.
4. Oxidation Number of S in H2SO4 :
Let the oxidation number of S in H2SO4 is x. According to the rule oxidation number of hydrogen is +1 and oxygen is -2
∴ 2(+1) + x + 4(-2) = 0
or, x = +6
Thus, the oxidation number of S in H2SO4 is +6.
5. Oxidation Number of C in CH3COCH3:
Let the oxidation number of C in CH3COCH3 is x. 
And the oxidation number of hydrogen and oxygen is +1 and -2 respectively.
∴ 3x + 6(+1) + (-2) = 0 
or, x = -(4/3)
Thus, the oxidation number of C in CH3COCH3 is 4/3.
6. Oxidation Number of S in Na2S2O3:
Let the oxidation number of S in Na2S2O3 is x
∴ 2(+1) + 2x + 3(-2) = 0
or, x = +2
Thus, the oxidation number of S in Na2S2O3 is +2.
7. Oxidation Number of S in Na2S4O6:
Let the oxidation number of S in Na2S4O6 is x
∴ 2(+1) + 4x + 6(-2) = 0
or, x = +2.5
Thus, the oxidation number of S in Na₂S₄O₆ is +2.5.
8. Oxidation Number of Cr in [Cr(NH3)6]Cl3: 
Let the oxidation number of Cr in [Cr(NH3)6]Cl3 is x. NH3 is neutral thus the oxidation number is zero and the oxidation number of Chlorine is -1
∴ x + 0 +3(-1) = 0
or, x = +3
Thus, the oxidation number of Cr in [Cr(NH3)6]Cl3 is +3.
Calculate the oxidation number of Iron in [Fe(H2O)5(NO)+]SO4.
H2O is neutral thus the oxidation number is zero, oxidation number of (NO)+ is +1 and the oxidation number of SO4 is -2.
Let the oxidation number of Fe in [Fe(H2O)5(NO)+]SO4 is x.
∴ x + 0 + (+1) + (-2) = 0 
or, x-1 = 0 
or, x = +1
Thus, the oxidation number of Fe in [Fe(H2O)5(NO)+]SO4 is +1.
Oxidation Number of an element in a compound is zero:
Some organic compound where the oxidation number of carbon on this compound is zero.
Compound Formula Oxidation Number
Sugar C12H22O11 0
Glucose C6H12O6 0
Formaldehyde HCHO 0
Let, the oxidation number of carbon in Glucose (C12H22O11) is x
∴ 6x + 12(+1) + 6(-2) = 0
or, x = 0
Calculate the oxidation number of the element marked with blue in the following compounds, (i) K2CrO4, (ii) HOCl, (iii) BaO2, (iv) ClNO, (v) NaNH2, (vi) NaN3, (vii) CH2Cl2, (viii) Ca(OCl)Cl, (ix) Ba(MnO4)2 (x) CaH2.
Compound Element Oxidation Number
K2CrO4 Cr +6
HOCl Cl +1
BaO2 Ba +2
ClNO N +3
NaNH2 N -3
NaN3 N -1/3
CH2Cl2 C 0
Ca(OCl)Cl Cl +1
Ba(MnO4)2 Mn +7
CaH2 Ca -1
What is the Oxidation state of chromium in Cr2O5?
Due to peroxy linkage oxidation state of Cr in Cr2O5 is +6.

Oxidation Number and concept of Oxidation and Reduction:

The oxidation number of CH4, CH3Cl, CH2Cl2, CHCl3, CCl4 and CO2 are -4, -2, 0, +2, +4 and +4 respectively. In H2, the oxidation number of hydrogen is zero but in H2O it is +1. Similarly magnesium in elementary state has a zero oxidation number but in MgCl2, the oxidation number is +2.
From the above we can define Oxidation and Reduction according to the Oxidation number increase or decrees,
Oxidation:
Oxidation may be defined as a process in which an increase in oxidation number by atoms or ions occurs. A reagent which can increase the oxidation number of an element or ion is called an oxidising agent. 
Reduction:
Reduction is a process in which a decrease in oxidation number by atoms or ions occurs. A reagent that lowers the oxidation number of an element or ion is called a reducing agent.

Schematic Representation of Oxidation and Reduction According to Electronic Concept and Oxidation Number Concept:

Oxidation and reduction are always found to go hand to hand during a radox reaction. Whenever an element or a compound is oxidised, another element or another compound must be simultaneously reduced. 
An oxidant is reduced and simultaneously the reductant is oxidised.
Oxidation Number and Concept of Oxidation and Reduction.
Schematic Representation of Oxidation and Reduction.
| Oxidation
H2S + Cl2 2HCl + S
Reductant Oxidant Oxidant Reductant
| Reduction
1.Magnesium metal burns in oxygen to produce magnesium oxide:
Before the reaction oxidation number of magnesium and oxygen both zero. But after the reaction the oxidation number of magnesium and oxygen is +2 and -2 respectively. Thus the oxidation number of magnesium is increases and oxidation number of oxygen is decreases. So magnesium is oxidised and oxygen is reduced.
0 0 +2  -2
Mg + O2 2MgO
2. Reaction between Sodium and Chlorine:
From same way we can explain this reaction. Here oxidation number of Sodium and Chlorine is zero before the reaction and +1 and -1 after the reaction. Thus sodium oxidised and chlorine reduced.
0 0 +1  -1
Na + Cl2 2NaCl
Why sulphur dioxide has properties of Oxidation and reduction?
This can be explain by the oxidation state of sulphur. The oxidation numbers of sulphur in the compounds H2S, SO2, and SO3 are -2, +4 and +6 respectively. Thus the highest oxidation state of sulphur is +6 and lowest is -2. The oxidation state of free sulphur element is 0. In SO2, the oxidation state of sulphur is +4, this oxidation state is middle of 0 and +6.
Thus, the oxidation state can increases from +4 to +6 and decreases from +4 to 0.
| Oxidation Number Decreases
Reduction
SO2 + H2S 2H2O + 3S
| Oxidation Number Increases
Oxidation



Method of Balancing Oxidation - Reduction Reactions:

Oxidation and reduction are always found to go hand in hand during a redox reaction. Whenever an element or compound is oxidized, another element or another compound must be reduced. An oxidant is reduced and simultaneously the reductant is oxidized.It follows from our above discussion that we should be able to balance oxidation reduction reactions on the basis of 
(1) Ion - Electron Method of Balancing Oxidation - Reduction Reactions.
(2) Oxidation Number Method of Balancing Oxidation - Reduction Reactions.

1. Ion - Electron Method of Balancing Oxidation - Reduction Reactions:

(i) Ascertain the Reactants and Products, and their chemical formulas. 
(ii) Set up the partial equations representing the reduction of the oxidant, and the oxidation of the reductant. 
(iii) If the reaction occurs in acid medium use requisite number of H⁺ for balancing the number of atoms involved in the partial equation. for alkaline medium use OH⁻ ions. 
(iv) Balancing the charge in the partial equations by adding suitable number of electrons. These electrons indicate the electrons involves in the oxidation or the reduction half reactions. 
(v) Multiply each partial equation by a suitable factor so each of the two partial equations involves the same number of electrons. 
(vi) Add the partial equations and cancel out species which appear on both sides of the equations.
Some examples will now needed to elucidate the rules.
In aqueous acid medium potassium permanganate oxidizes a ferrous ion to the ferric state.
In this reaction permanganate ion is the oxidant and ferrous ion the reductant.
The left hand side of the ultimate equation will carry, MnO₄⁻ (KMnO₄), H⁺ (H₂SO₄) and Fe²⁺ (FeSO₄).
The Right hand side will have as products, Mn²⁺ (MnSO₄), H₂O and Fe³⁺ [Fe₂(SO₄)₃].
The partial equation representing the reduction of the oxidant MnO₄⁻ will involve, 
 MnO₄⁻ Mn²⁺
Since the reaction occurs in acid medium we utilize eight H⁺ ions to balance the four oxygen atoms in MnO₄⁻.
Thus,  MnO₄⁻+8H⁺ → Mn²⁺+4H₂O
Since the above partial equation is still unbalanced from the viewpoint of charge, the equa-tion is balanced by bringing in five electrons:
MnO₄⁻ + 8H⁺ + 5e Mn²⁺ + 4H₂O
Note that MnO₄⁻ Mn²⁺ is indeed a five electron reduction, Mn⁷⁺ + 5e Mn²⁺.
If we now apply the above steps to obtain the partial equation representing the oxidation of reductant we have the following: 
Fe⁺² Fe⁺³ + e
This equation on multiplication by 5 and then adding to the partial equation of the oxidant gives the final balanced equation:
MnO₄⁻
+
8H⁺
+
5e
Mn²⁺
+
4H₂O


(
Fe⁺²
Fe⁺³
+
e)

MnO₄⁻+8H⁺+5Fe⁺²Mn²⁺+5Fe⁺³+4H₂O
The Oxidation of potassium iodide by potassium dichromate in dilute acid medium.
In this reaction dichromate is reduced to (+6) state to (+3) state and iodide ion is oxidised to elementary iodine.
Taking dichromate side, balancing the atoms and charges provides the partial equation:
Cr₂O₇⁻² + 14H⁺ + 6e 2Cr⁺³ + 7H₂O  
(Reduction of Oxidant)
Considering the case of iodide ion we get:
2I⁻ I₂ + 2e 
(Oxidation of Reductant)
Multiplying the above equation by 3 and adding these two partial equation we have the following final equation:
 Cr₂O₇⁻² + 14H⁺ + 6e  2Cr⁺³ + 7H₂O
3 (2I⁻  I₂ + 2e)

Cr₂O₇⁻²+14H⁺+6I⁻2Cr⁺³+3I₂+7H₂O

Express by ion electron method the reduction of permagnate to manganous stat by hydrogen peroxide in acid medium.

This Reaction occurs in acid medium and the partial equation is:
MnO₄⁻ + 8H⁺ + 5e  Mn⁺² + 4H₂O (Reduction of Oxidant)
In acid medium H₂O₂ will give O₂ and the partial equation being:
H₂O₂  2H⁺ + O₂ +2e 
(Oxidation of Reductant)
First equation is multiplying by 2 and the second is 5 to have electron balanced. 
We have,
2MnO₄⁻ + 16H⁺ + 10e  2Mn⁺² + 8H₂O
5H₂O₂ 10H⁺ + 5O₂ +10e

2MnO₄⁻+5H₂O₂+6H⁺2Mn⁺²+8H₂O+5O₂
The reaction of permanganate ion with Sodium Stannite (Na₂SnO₂) in alkaline solution in which manganese dioxide (MnO₂) and Sodium Stannate(Na₂SnO₃) are formed.
The Partial equation representing the reduction of oxidant is,
MnO₄⁻  MnO₂.
Since the medium is alkaline we put the requisite number of OH⁻ ions to effect atom balance as:
MnO₄⁻ + 2H₂O MnO₂ + 4OH⁻
To balance the charge on both sides three electrons are added on the left hand side:
MnO₄⁻ + 2H₂O + 3e  MnO₂ + 4OH⁻
For the reductant the balance partial equation is:
SnO₂⁻² + OH⁻ SnO₃⁻² + H₂O + 2e
To balance the number of electrons involved in the two partial equations the first equation is multiplied by 2 and the second by 3 and on the adding we have the final form of the ionic equation:
2MnO₄⁻ + 4H₂O + 6e  2MnO₂ + 8OH⁻
3SnO₂⁻² + 3OH⁻  3SnO₃⁻² + 3H₂O + 6e

2MnO₄⁻+3SnO₂⁻²+H₂O2MnO₂+3SnO₃⁻²+2OH⁻
Oxidation of Mn⁺² to MnO₄⁻ by PbO₂ or NaBiO₃ in Acid Medium:
Balancing Oxidation Reduction Reactions
Representation Oxidation Reduction Reaction

Express by ion - electron method the reduction of nitrate ion to 
ammonia by aluminium in aqueous NaOH.

In alkaline medium we use H₂O and OH⁻ ion according to convenience. The partial equation with charge balance is:
NO₃⁻ + 6H₂O + 8e NH₃ + 9OH⁻
Metallic aluminium will go over to aluminate ion, the partial equation being:
Al + 4OH⁻ AlO₂⁻ + 2H₂O +3e
Multiplying by right factors for electron balance we have the balanced equation:
3NO₃⁻ + 18H₂O + 24e 3NH₃ + 27OH⁻
8Al + 32OH⁻ 8AlO₂⁻ + 16H₂O +24e

3NO₃⁻+8Al+2H₂O+5OH⁻3NH₃+8AlO₂⁻

Oxidation Number Method of Balancing Oxidation - Reduction Reactions:

(i) Assuming that the oxidation number of oxygen remains unchanged manganese in MnO₄⁻ has an oxidation number +7. Calculation of the decrease and increase in oxidation number gives:
Balancing Oxidation Reduction Reactions
Representation of Decreases and Increases of
Oxidation Number
Putting right factors the decreases and increases in oxidation numbers are balanced, giving,
MnO₄⁻ + 5Fe⁺² Mn⁺² + 5Fe⁺³
Since the reaction medium is acidic and the changes are not equal on the two sides of the above expression, H⁺ ions are added and requisite number of H₂O written:
MnO₄⁻+8H⁺+5Fe⁺²Mn⁺²+ 5Fe⁺³+4H₂O

Use Oxidation Number method to balance the reaction of iodide ion 
and iodate ion in acid medium to liberate iodine.

The reaction represented as, I⁻ + IO₃⁻ I₂
Representation of the above equation with oxidation number of iodine,
I⁻(-1) + IO₃⁻(+5) I₂(0)
Since I⁻ (-1) oxidation number of iodide increases by 1 and IO₃⁻(+5) oxidation number decreases by 5.
Putting the right factors the decrease and increase in oxidation number balanced:
5I⁻ + IO₃⁻ 3I₂
Since the medium is acidic and the charges are unequal on two sides of the above expression we put Six H⁺ and write three H₂O to balanced the equation:
I⁻ + IO₃⁻ + 6H⁺ I₂ + 3H₂O
The Oxidation of iodide ion by dichromate ion in acid medium:
The reaction represented as, 
Cr₂O₇⁻ + I⁻ Cr⁺³ + I₂
Balancing Oxidation Reduction Reactions
The Oxidation of Iodide ion by dichromate
ion in Acid medium
Oxidation of sodium stannite to stannate in alkaline medium:
This reaction represented as,
MnO₄⁻ + SnO₂⁻² MnO₂ + SnO₃⁻²
Oxidation number of Mn decreases by 3 and oxidation number of Sn increases by 2.
Equalizing the increase and decrease in oxidation number,
2MnO₄⁻ + 3SnO₂⁻² 2MnO₂ + 3SnO₃⁻²
Because the medium is alkaline and because the ionic charges are not equal on the two sides, two OH⁻ ions are added to the right hand side.
2MnO₄⁻+3SnO₂⁻²2MnO₂+3SnO₃⁻²+2OH⁻
Since there are two hydrogen atoms on the right and none on the left, H₂O is added to make atom balance. The final equation is:
2MnO₄⁻+3SnO₂⁻²+H₂O2MnO₂+3SnO₃⁻²+2OH⁻

Use oxidation number method to balance the reaction between 
Sulphurus acid and dichromate in acidic medium.

The reaction represented as,
SO₃⁻² + Cr₂O₇⁻² SO₄⁻² + 2Cr⁺³
Oxidation of two Cr decreases by 2 × (+3) = 6 and oxidation number of S increases by 2.
Equalizing the above equation increase and decrees in oxidation number,
We have, 3SO₃⁻² + Cr₂O₇⁻²  3SO₄⁻² + 2Cr⁺³
Because of the medium is acidic and the ionic charges are not equal on the two sides, eight H⁺ ions are added to the Left hand side. 
3SO₃⁻² + Cr₂O₇⁻² + 8H⁺  3SO₄⁻² + 2Cr⁺³
Since there are eight hydrogen ion on the left and none on the right, four H₂O added to make atom balance.
Thus the final equation is:
3SO₃⁻²+Cr₂O₇⁻²+8H⁺3SO₄⁻²+2Cr⁺³+4H₂O



Author Name

Contact Form

Name

Email *

Message *

Powered by Blogger.