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PostSubject: redox reaction   Tue Jul 07, 2009 8:04 am

[b]redox reaction: click here
mediafire.com ?ygg0mbr10ec[u]
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PostSubject: Re: redox reaction   Tue Jul 07, 2009 1:07 pm

sir...the link you provided doesnot contain the redox reaction notes.the link is incorrect sir...please check it
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Aaqib Ahmed
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PostSubject: Re: redox reaction   Tue Jul 07, 2009 6:13 pm

mediafire.com ?ygg0mbr10ec
Above is the correct link for redox reaction
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PostSubject: Re: redox reaction   Tue Jul 07, 2009 8:41 pm

REDOX REACTIONS

REDOX REACTIONS IN TERMS OF ELECTRON TRANSFER REACTIONS

1. Reduction is a chemical process in which there is a gain of electrons
2. Oxidation is a chemical process in which there is loss of electrons.
3. Oxidising agent : Acceptor of electron(s) (OR) The substance which accepts the electrons during the chemical process is called oxidising agent.
4. Reducing agent : Donor of electron(s) (OR) The substance which lose the electrons during the chemical process is called Reducing Agent.


• Na loosing electron undergo oxidation and by which it donating electrons act as reducing agents
• Chlorine gains electrons undergo reduction by which accepting electron act as oxidizing agent
Oxidation Number


1. The term oxidation number explains the phenomenon of oxidation-reduction in covalent and ionic substances.
2. Oxidation number is defined as the charge, which an atom appears to have when all other atoms are removed from it as ions.
3. The oxidation number is a fictitious charge in case of covalent species. It can have positive, zero or negative values depending upon their state of combination.

Determination of Oxidation Number of an Atom
State the rules for the calculation of oxidation number.
The rules used for the calculation of oxidation number of an atom in a molecule are:
1. The oxidation number of an element in the free or elementary state is always zero. For example, oxidation numbers of helium in He, hydrogen in H2, oxygen in O2, iron in Fe, bromine in Br2, phosphorus in P4, Sulphur in S8, Na, Mg, Al has the oxidation number zero.

2. The oxidation number of an element in a single (monoatomic) ion is the same as the charge on the ion. For example, oxidation number of K+ is +1, of Ca2+ is + 2, of Al3+ is +3, Similarly, the oxidation numbers of Cl-, SO4-2 and PO4--3 are -1, -2 and -3 respectively,
3. In all compounds of hydrogen, the oxidation number of hydrogen is +1except in hydrides of active metals such as LiH, NaH, KH, MgH2 CaH2, etc., where hydrogen has the oxidation number of -1.
4. In compounds containing oxygen, the oxidation number of oxygen is -2
EExcept
• In peroxides such as Na2O2 (-1)
• In superoxides (e.g., KO2, RbO2) each oxygen atom is assigned an oxidation number of –(½).
• When oxygen is bonded to fluorine. In such compounds ( oxygen difluoride (OF2) and dioxygen difluoride (O2F2) the oxygen is assigned an oxidation number of +2 and +1, respectively.

5. For neutral molecule, the sum of the oxidation numbers of all the atoms is equal to zero. For example, in NH3 the sum of the oxidation numbers of nitrogen atom and hydrogen atoms is equal to zero.
6. For a complex ion, the sum of the oxidation number of all the atoms is equal to charge on the ion. For example, in SO ion, the sum of the oxidation numbers of sulphur atom and 4 oxygen atoms must be equal to -2.
Oxidation Number and Nomenclature


When an element forms two monoatomic cations (representing different oxidation states), the two ions are distinguished by using the ending 'ous' and 'ic' at the end. The suffix 'ous' is used for the cation with lower oxidation state and the suffix 'ic' is used for the cation with higher oxidation state.
For example:
Cu+ (oxidation number +1) cuprous : Cu2+ (oxidation number +2) cupric
However, Albert Stock proposed a new system known as Stock system. In this system, the oxidation states are indicated by Roman numeral written in parentheses immediately after the name of the element. For example,
Cu2O Copper (I) oxide FeCl3 Iron (III) chloride
CuO Copper (II) oxide Mn2O7 Manganese (VII) oxide
SnO Tin (II) oxide K2Cr2O7 Potassium dichromate (VI)
SnO2 Tin (IV) oxide Na2CrO4 Sodium chromate (VI)
FeCl2 Iron (II) chloride V2O5 Vanadium (V) oxide

Stock system is not used for non-metals



1. Write formulas for the following compounds:


• Oxidation: An increase in the oxidation number of the element in the given substance.
• Reduction: A decrease in the oxidation number of the element in the given substance.
• Oxidising agent: A reagent which can increase the oxidation number of an element in a given substance. These reagents are called as oxidants.
• Reducing agent: A reagent which lowers the oxidation number of an element in a given substance. These reagents are also called as reductants.
• Redox reactions: Reactions which involve change in oxidation number of the interacting species.

2. Justify that the following reactions are redox reactions:



Types of Redox Reactions

1. Combination reactions
A combination reaction may be denoted in the manner:

example : All combustion reactions,



2. Decomposition reactions

Decomposition reactions are the opposite of combination reactions. Precisely, a decomposition reaction leads to the breakdown of a compound into two or more components at least one of which must be in the elemental state. Examples of this class of reactions are:



Note : all decomposition reactions are not redox reactions. For example, decomposition of calcium carbonate is not a redox reaction as no change in oxidation number


Displacement reactions

In a displacement reaction, an ion (or an atom) in a compound is replaced by an ion (or an
atom) of another element. It may be denoted as:

Displacement reactions fit into two categories: metal displacement and non-metal displacement.
(a) Metal displacement:

b) Non-metal displacement:
1. metal above in the reactivity series displace the hydrogen from its compound depending on its reactivity .

• Like metals, activity series also exists for the halogens.
• The power of these elements as oxidising agents decreases as we move down from fluorine to iodine in group 17 of the periodic table.
• In group-17 above in the group can displace below below halogen from its compound
• This implies that fluorine is so reactive that it can replace chloride, bromide and iodide ions in solution. In fact, fluorine is so reactive that it attacks water and displaces the oxygen of water :




Disproportionation reactions

In a disproportionation reaction an element in one oxidation state is simultaneously oxidised and reduced.
Here the oxygen of peroxide, which is present
in –1 state, is converted to zero oxidation state
in O2 and decreases to –2 oxidation state in H2O. that disproportionates.

Phosphorous, sulphur and chlorine undergo disproportionation in the alkaline medium
because in this oxoanion chlorine is present in its highest oxidation state that

Balancing Oxidation-Reduction Reactions


Steps for balancing redox equations by oxidation number method
1 Write the skeleton redox reaction.
2 Indicate the oxidation number of atoms in each compound above the symbol of the element.
3 Identify the element or elements, which undergo a change in oxidation number, one whose oxidation number increases (reducing agent) and the other whose oxidation number decreases (oxidizing agent).
4 Calculate the increase or decrease in oxidation numbers per atom. Multiply this number of increase/decrease of oxidation number, with the number of atoms, which are undergoing change.
5 Equate the increase in oxidation number with decrease in oxidation number on the reactant side by multiplying the formulae of the oxidizing and reducing agents.
6 Balance the equation with respect to all other atoms except hydrogen and oxygen.
7 Finally, balance hydrogen and oxygen.
8 For reactions taking place in acidic solutions, add H+ ions to the side deficient in hydrogen atoms.
9 For reactions taking place in basic solutions, add H2O molecules to the side deficient in hydrogen atoms and simultaneously add equal number to OH- ions on the other side of the equation.
Let us discuss the above method stepwise with the help of reaction between zinc and hydrochloric acid.
Step 1
The skeleton equation is:


Step 2
Oxidation number of various atoms involved in the reaction


Step 3
The oxidation number of zinc has increased from 0 to +2 while that of hydrogen has decreased from +1 to 0. However, the oxidation number of chlorine remains same on both sides of the equation. Therefore, zinc is reducing agent while HCl is oxidizing agent in reaction and the changes are shown as:


Step 4
The increase and decrease in oxidation number per atom can be indicated as: O.N. increases by 2 per atom


Step 5
The increase in oxidation number of 2 per atom can be balanced with decrease in oxidation number of 1 per atom if Zn atoms are multiplied by 1 and HCl by 2. The equation will be:


Problem
12. Copper reacts with nitric acid. A brown gas is formed and the solution turns blue. The equation may be written as:


Balance the equation by oxidation number method.
Solution
Step 1
Skeleton equation


Step 2
Writing oxidation numbers of each atom


Step 3:


The oxidation number of copper has increased from 0 to +2 while that of nitrogen has decreased from +5 to +4.
Step 4
Show, the increase/decrease of oxidation number


Step 5
Balance the increase/decrease in oxidation number by multiplying NO3- by 2 and Cu by 0.


Step 6
Balance other atoms except H and O as


Step 7
Reaction takes place in acidic medium, so add H+ ions to the side deficient in H+ and balance H and O atoms:


In this case MnO-4 is reduced (oxidation number of Mn decreases from +7 to +2) and Cl is oxidized (oxidation number increases from -1 to 0). Thus, MnO-4 acts as an oxidizing agent

Try this
Balance the following redox reactions by ion – electron method

Balance the following equations in basic medium by ion-electron method and oxidation number methods and identify the oxidising agent and the reducing agent.
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PostSubject: Re: redox reaction   

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