Mrs.Vaishali Mahajan

MOLE CONCEPT and

OXIDATION REDUCTION

1 dozens mangoes = 12 mangoes

1 mole of any substance =

6.02 2x 1023

particles

The mass of a mole of atoms = the atomic mass in grams in

the Periodic Table

1 mole of C atoms is 12.01 grams

1 mole of Na atoms is 22.99 grams

1 mole of any Gas = Molar mass of it = 22.4

litres

HOW MANY MOLES OF HELIUM ARE PRESENT IN 6.46 GRAMS OF

IT ?

Atomic mass of Helium is 4.003 grams

Therefore 1 mole of Helium = 4.003 grams

4.003 grams = 1 mole of Helium

6.460 grams = ?

= 6.460 X 1 /4.OO3

= 1.61 Mole of Helium

Thus 1.61 moles of Helium atoms in 6.460 grams of it.

HOW MANY GRAMS OF LEAD (PB) ARE THERE IN 12.4 MOLE OF

LEAD ?

Atomic mass/ molar mass of lead = 207.2 grams

Thus 1 mole of lead = 207.2 g lead

12.4 mole of lead = ?

= 12.4 X 207.2 / 1

= 2596.2 g Lead

Thus there are 2596.2 g Lead in 12.4 moles of it.

CONVERT 26.5 G OF H2SO4 IN MOLES.

Molar mass of H2SO4 = 2(H) + S + 4 (O)

= 2(1) + 32 + 4(16)

= 98

98 g of H2SO4 = 1 mole of H2SO4

26.5 g of H2SO4 = ?

= 26.5 X 1 / 98

= 0.27 mole of H2SO4

Thus 0.27 moles of H2SO4 in 26.5 g of it.

WHAT IS THE VOLUME IN LITRES OCUPIED BY 45.6 G OF H2S

AT STP?

Molar mass of H2S = 1+1+32 = 34

34 g of H2S = 1 mole = 22.4 litres of H2S

45.6 G OF H2S = ?

= 45.6 X 22.4 / 34

= 30.05 litres of H2S

Thus volume occupied by 45.6 g of H2S is 30.05 litres

OXIDATION REDUCTION REACTIONS

The term oxidation was originally used to describe reactions in which an element combines with oxygen.

Example: The reaction between magnesium metal and oxygen to form magnesium oxide involves the oxidation of magnesium.

2 Mg(s) + O2(g) 2 MgO(s)

The term reduction comes from the Latin stem

meaning "to lead back." Anything that that leads

back to magnesium metal therefore involves

reduction.

The reaction between magnesium oxide and

carbon at 2000C to form magnesium metal and

carbon monoxide is an example of the reduction of

magnesium oxide to magnesium metal.

MgO(s) + C(s) Mg(s) + CO(g)

THE ROLE OF OXIDATION NUMBERS IN

OXIDATION-REDUCTION REACTIONS

Chemists eventually extended the idea of

oxidation and reduction to reactions that do not

formally involve the transfer of electrons.

CO(g) + H2O(g) CO2(g) + H2(g)

What changes in this reaction is the oxidation

state of these atoms. The oxidation state of

carbon increases from +2 to +4, while the

oxidation state of the hydrogen decreases from

+1 to 0.

• LEO = Loss of Electrons is Oxidation

• GER = Gain of Electrons is Reduction

• Another way is to simply remember that reduction is to reduce the oxidation number.

• Therefore, oxidation must increase the value.

• Oxidizing Agent - that substance which oxidizes somebody else. It is reduced in the process.

• Reducing Agent - that substance which reduces somebody else. It is oxidized in the process.

RULES FOR ASSIGNING OXIDATION NUMBERS

1. All free, uncombined elements have an oxidation number of zero. This includes diatomic elements such as O2 or others like P4

and S8.

2. Hydrogen, in all its compounds except hydrides, has an oxidation number of +1 (positive one)

3. Oxygen, in all its compounds except peroxides, has an oxidation number of -2 (negative two).

WHAT IS A HALF-REACTION?

A half-reaction is simply one which shows either

reduction OR oxidation, but not both. Here is an

example redox reaction:

Ag+ + Cu ---> Ag + Cu2+

It has BOTH a reduction and an oxidation in it.

That is why we call it a redox reaction, from

REDuction and OXidation.

• What you must be able to do is look at a redox reaction and separate out the two half-reactions in it. To do that, identify the atoms which get reduced and get oxidized. Here are the two half-reactions from the example:

Ag+ ---> Ag

Cu ---> Cu2+

• The silver is being reduced, its oxidation number going from +1 to zero. The copper's oxidation number went from zero to +2, so it was oxidized in the reaction. In order to figure out the half-reactions, you MUST be able to calculate the oxidation number of an atom.

• When you look at the two half-reactions, youwill see they are already balanced for atomswith one Ag on each side and one Cu on eachside. So, all we need to do is balance thecharge.

• To do this you add electrons to the morepositive side. You add enough to make the totalcharge on each side become EQUAL.

• To the silver half-reaction, we add one electron:

Ag+ + e¯ ---> Ag

• To the copper half-reaction, we add two electrons:

Cu ---> Cu2+ + 2e¯

HALF-REACTIONS NEVER OCCUR ALONE.

notice that each half-reaction wound up with a

total charge of zero on each side. This is not

always the case. You need to strive to get the

total charge on each side EQUAL, not zero.