Nios + cbse , class 11 and 12 science - biology,chemistry ,physics topper notes. Im a topper who scores above 90, so these are topper notes. nothing is skipped, everything is correct. Just simplified and shortened.

Chem – ELECTROCHEMISTRY
Electrochemistry is the study of the conversion of electrical energy into chemical
energy and chemical energy into electrical energy.examples: found in dry cells,
button cells or lead acid batteries.
When electric current is passed through an aqueous solution or molten salts, it
causes a chemical reaction to occur. This converts electrical energy to chemical
energy.
Oxidation and Reduction as Electron Transfer Process

  Oxidation is the loss of one or more electrons by an atom or ion. Reduction is
the gain of one or more electrons by an atom or ion.





 Reducing agent ( aka reductant): An atom that loses electrons and helps another
atom to get reduced. In this reaction, sodium is the reducing agent. Oxidising agent
(oxidant): A species that gains electrons and helps another species to get
oxidised.here chlorine is the oxidising agent.
 Redox Reactions : 1. Oxidation and reduction always happens in same time,
independently.
A redox reaction is a reaction in which oxidation and reduction occur
simultaneously.

 Oxidation Number: In simple molecules, it is easy to identify oxidation and
reduction. but In polyatomic molecules, it becomes difficult. To solve this, we use
oxidation number. Oxidation number is the apparent charge which an atom appears to
have when each pair of electrons is counted with more elecronegative atom..

 Oxidation number is always assigned to an atom. It is written as a number with a +ve or
−ve sign.


, 

+ve oxidation number - atom that shifts electrons away from itself.

 -ve oxidation number: the more electronegative atom that gains electrons. Oxidation state
(OS) is another term used for oxidation number.

Rules for Assigning Oxidation Number- 8 rules.




 Using these rules, find the oxidation number of S, N and Cl atoms in : (a) H2
SO4 (b)NO– 3 (c) ClO– 4 respectively
a. H2so4 = 1. Let the oxidation number of sulphur be x. Since the
oxidation number of O is –2. Therefore the sum of four O atoms is
equal to –8. (2x4)
2.the oxidation number of each H is +1, two H atoms have total oxidation number
of +2.

, 3. H2 SO4 is a neutral molecule. Therefore the sum of all the oxidation numbers is
equal to zero. Thus +2 + x – 8 = 0 x = + 6 Therefore oxidation number of sulphur in
H2 SO4 is + 6.




 BALANCING REDOX REACTION
The redox reaction can be balanced by any of the following methods : (a)
Oxidation number method. (b) Ion electron method.
a) Balancing by Oxidation Number method – steps to balance redox reactions by
this method.
1.Write the skeletal equation (without coefficients/without balancing).
2. Write the oxidation number of each atom above its symbol.
3. Identify the atoms whose oxidation number changes.
4. Calculate the increase or decrease in oxidation number per atom. If multiple
atoms, multiply by number of atoms for total change.
5. Equalise increase and decrease by multiplying oxidising and reducing agents.
6. Balance all atoms, except H and O.then balance H AND O finally.

7. balance finally by Acidic medium: Add H₂O to side with fewer O atoms. Add H⁺ to
side with fewer H atoms. Basic medium: Add OH⁻ to balance charges. Add H₂O to
balance OH⁻