The document explains the concepts of oxidation and reduction, emphasizing that oxidation involves the loss of electrons while reduction involves the gain of electrons. It illustrates these concepts with chemical reactions, detailing changes in oxidation states of manganese and carbon. Additionally, it provides a method for determining oxidation numbers in compounds.

1. Oxidation is gain of

2. Definition of Reduction

3. ExAmPle

4. Hard tO Remember?
LEO
Lose Electrons
in Oxidation
Gain Electrons
in Reduction

5. VIDEO!!
http://www.youtube.com/watch?v=dxlWtsFinT
M

6. FUN F@CTS!
-My name is BOND…
Convalent Bond!:D
-When an atom or molecule combines
with oxygen, it tends to give up
electrons to the oxygen in forming
a chemical bond.

7. Pls click to the next pg, thank you

8. For example, in the reaction (unbalanced),
MnO4- + H+ -> Mn2+ + H2O
the oxidation of Mn goes from +7 to +2. Thus, Mn
is reduced. In the following reaction (unbalanced),
H2C2O4 + O2 -> CO2 + H2O
C is oxidized, because its oxidation state
the element
changes from +3 to +4 in the reaction.

9. The oxidation state of an uncombined
element is .That's
obviously so, because it hasn't been
either oxidised or reduced yet!
The sum of the oxidation states of
all the atoms or ions in a neutral
compound is zero.

10. • First determine what the oxidation numbers of your known atoms are.
• we know the oxidation number for H is +1.
• Then set this value equal to the overall net charge of the ion.
• In this case, it is +1.
• Our equation now looks like this: 1(4) = 1, You use the multiplier of 4 to indicate that
the ammonium ion has 4 hydrogen.
• Next substitute a variable in the equation for the missing oxidation number:
• +1(4) + N = +1
• Solve for N.
• +1(4) + N = +1
• N + 4 = +1
• N = +1 - 4
• N = -3
• Thus, the oxidation number for Nitrogen is -3.

11. CREDITS: