OXIDATION AND REDUCTION:



OXIDATION AND REDUCTION:

A chemical change that takes place as a result of electron transfer is called a redox or oxidation –reduction reaction.

The chemical reactions in which the oxidation number of atoms get changed are termed as redox reactions.

| OXIDATION | REDUCTION |

|It is a process which involves |It is a process which involves |

| | |

|Gain |Gain |

|gain of oxygen |gain of hydrogen |

|gain of electronegative radical |gain of electropositive radical |

|Loss |gain of electron |

|loss of hydrogen |Loss |

|loss of electropositive radical |loss of oxygen |

|loss of electron |loss of electronegative radical |

ILLUSTRATION:

Gain and Loss of Oxygen: i) Burning of hydrogen in oxygen

2H2 + O2 → 2H2O

The hydrogen is oxidized and the oxygen is reduced

ii) The combination of nitrogen and oxygen which occurs at high temperatures follows the same pattern.

N2 + O2 → 2 NO

Here, nitrogen is oxidize and oxygen is reduced

c) Combustion of methane,

CH4 + 2O2 → CO2 + 2H2O

Both carbon and hydrogen are oxidized (gain oxygen) and oxygen is reduced.

OXIDIZING AND REDUCING AGENTS:

OXIDIZING AGENTS:

Substances that oxidize other substances are known as oxidizing agents. , oxidants or oxidizers. Oxidants are usually chemical substances with elements in high oxidation numbers (e.g., H2O2,MnO4-,CrO3, Cr2O72-,etc.) or highly electronegative substances that can gain one or two extra electrons by oxidizing a substance (e.g.O2,F2,Cl2,Br2 etc.)

REDUCING AGENTS:

Substances that reduce other substances are known as reducing agents ,reductants, or reducers. Reductants are electropositive elemental metals like (Li, Na, Mg, Fe, Zn, Al)or hidirdies like NaBH4, LiAlH4 that are widely used in the reduction of carbony compouds to alcohols.Also hydrogen gas, H2 with palladium, platinum,nickel as catqlyst

( The chemical which is oxidised is the reducing agent.

The chemical which is reduced is the oxidising agent.

( Oxidation and reduction always occur side by side that is simultaneously.

OXIDATION NUMBER OR OXIDATION STATE:

“Oxidation number or oxidation state of an element is the charge which it has or appears to have in the combined state.”

The determination of oxidation number or oxidation state of an atom is based on certain rules.

Rules for determining oxidation state:

1. All elements in their elemental state have zero oxidation number Diatomic gases such as O2 and H2 are also in this category.

2. Hydrogen atom in combined state has +1 oxidation number except in case of hydrides where its oxidation number is -1

3. Oxygen atom in combined state has -2 oxidation number except in peroxides where it has -1 oxidation number.

4. Halogens have -1 oxidation number in their combined state. In case of interhalogens the more electronegative atom will possess negative oxidation number whereas the other will have positive oxidation number.

4 Alkali metals have +1 oxidation number in combined state and alkaline earth metals have +2 oxidation number..

6. In radicals or small covalent molecules, the element with the greatest electronegativity has its natural ion charge as its oxidation state.

7. The sum of oxidation numbers of all atoms in a neutral compound is taken as zero.

8. In case of complex ion the sum of oxidation numbers of all atoms present in is equal to the net charge present on it.

OXIDATION AND REDUCTION IN TERMS OF OXIDATION STATE:

A raise (increase) of oxidation state is an oxidation whereas a lower (decrease) of oxidation state is a reduction.

BALANCING OF A CHEMICAL EQUATION:

Balancing of a chemical equation:

Chemical equations can be balanced using oxidation numbers of the species involved. The following steppes are used

1. Write the skeleton equation:

2. Assign oxidation numbers to all atoms

3. Make the number of atoms that change oxidation number the same on both sides by inserting temporary coefficient.

4. Identify oxidized and reduced species

5. Determine the number of electrons lost (oxidation) and gained (reduction)

6. Balance the oxidation number by adding electrons to whatever side required.

7. For reactions in acidic or basic solution, add H+/H2O (acidic) or OH-/H2O (basic) to balance the charges.

8. Multiply these by appropriate factors make electrons lost = electrons gained

9. Complete the equation by final inspection.

Balancing Equations using Oxidation Numbers

e.g.1 Balance the following reaction

Cu (s) + Ag+(aq) → Cu2+(aq) + Ag (s)

Following steps

Initial Ox. No. Final Ox. No.

0 +1 +2 0

Loss of 2 electrons (Oxidation)

Cu(s) + Ag+(aq) → Cu2+(aq) + Ag(s)

Gain of 1 electron (reduction)

It is clear that the number of electrons lost by copper is not equal the number gained by silver. Therefore multiply Ag by 2, giving us a total of two silvers, and multiply copper by one.

1 Cu(s) + 2Ag+(aq) → 1 Cu2+(aq) + 2Ag(s)

It is not necessary to put the "1" in front of copper.

No. of electrons lost by one Cu atom =2

No. of electrons gained by two Ag atoms= 2

Thus the no. of electrons gained = No. of the electrons gained

Therefore the balanced equation is

Cu(s) + 2Ag+(aq) → 1 Cu2+(aq) + 2Ag(s)

( The multipliers to balance electrons become the balancing coefficients in the equation. Thus, we should put a "1" in front of copper and a "2" in front of silver. However, it is not necessary to put the "1" in front of copper.

e.g.2

Balance the following reaction:

KIO3 + KI + H2SO4 -> K2SO4 + H2O + I2

K+ + IO3- + K+ + I- +2H+ + SO42- → 2K+ + SO42- + H2O + I2

However the net ionic reaction may be written as

IO3- + I- + H+ → H2O + I2

Find the oxidtion number of each atom in the equation

Decrease in ox. no. ( reduction)

Ox. No. + 5 -2 -1 +1 +1 -2 0

IO3- + I- + H+ → H2O + I2

Increase in ox. no. ( Oxidation)

Decrease in oxidation no.= 5 by I in IO3-

Increase in oxidation no =1 by I in I-

Now multiply I- by 5 and IO3- by 1, we get

IO3- + 5I- + H+ → H2O + I2

There are six atoms of iodine on the left, so we need three I2 molecules to balance iodine:

1IO3- + 5I- + H+ → H2O + 3I2

Fianally, balancing oxygen, hydrogen and water:

IO3- + 5I- + 6H+ → 3H2O + 3I2

This is the balanced equation.

Balance the following equation:

MnO4-(aq) + Cl-(aq) + H+(aq) → Mn2+(aq) + Cl2(aq) + H2O(l)

Assign oxidation numbers (o.n.) to atoms.

Increase in ox. no. (Oxidation)

Ox.no.

+7 -2 -1 +1 +2 0 +1 -2

MnO4-(aq) + Cl-(aq) + H+(aq) → Mn2+(aq) + Cl2(aq) + H2O(l)

Decrease in ox. no. (Reduction)

No. of electrons gained = 5 by Mn in MnO4-

No. of electrons lost = 1 by Cl in Cl-

Multiply MnO4- by 1 and Cl- by 5 , we have

MnO4-(aq) + 5Cl-(aq) + H+(aq) → Mn2+(aq) + Cl2(aq) + H2O(l)

Balance chlorine atoms on both sides

MnO4-(aq) + 10Cl-(aq) + H+(aq) → Mn2+(aq) + 5Cl2(aq) + H2O(l)

Balance O and H atoms on both sides

MnO4-(aq) + 10Cl-(aq) + 8 H+(aq) → Mn2+(aq) + 5Cl2(aq) + 4H2O(l)

To make eletrone gained = electrons lot

Multiply MnO4- and Mn2+ each by 2 ,thus we have finally the required balanced equation

2 MnO4-(aq) + 10Cl-(aq) + 8 H+(aq) → 2Mn2+(aq) + 5Cl2(aq) + 4H2O(l)

Alternate methood half equation method

Balance the following equation:

MnO4-(aq) + Cl-(aq) + H+(aq) → Mn2+(aq) + Cl2(aq) + H2O(l)

Assign oxidation numbers (o.n.) to atoms.

Increase in ox. no. (Oxidation)

Ox.no.

+7 -2 -1 +1 +2 0 +1 -2

MnO4-(aq) + Cl-(aq) + H+(aq) → Mn2+(aq) + Cl2(aq) + H2O(l)

Decrease in ox. no. (Reduction)

No. of electrons gained = 5 by Mn in MnO4-

No. of electrons lost = 1 by Cl in Cl-

Write two half equations i.e. one for oxidation half equation and other reduction half equation:

| Oxidation half reaction |Reduction half reaction |

| | |

|Cl- → Cl2 |MnO4- → Mn2+ |

|Balancig atoms other than O and H | |

|2Cl- → Cl2 | |

|Balancing oxidation no by adding to whatever side required | |

|2Cl- → Cl2 + 2e- |MnO4- → Mn2+ |

|Balancing of charge by adding H+ ions to whaever side |Balancing oxidation no by adding to whatever side required |

|required ( being acidic medium) |MnO4- + 5e- → Mn2+ |

|2Cl- → Cl2 + 2e- |Balancing of charge by adding H+ ions to whaever side |

|Balancing of O atoms on both sides by adding H2O to whatever |required ( being acidic medium) |

|side required. |MnO4- + 5e- + 8H+ → Mn2+ |

|2Cl- → Cl2 + 2e- ---(1) |Balancing of O atoms on both sides by adding H2O to whatever |

| |side required. |

| |MnO4- + 5e- + 8H+ → Mn2++ 4H2O |

| | |

| |--- (2) |

In order to balance the the number of electrons gained and the number of electrons lost, multiply equation (i) by 5 nd equation by 2 and add the two resultant equations:

2Cl- → Cl2 + 2e- ---(1) x 5

MnO4- + 5e- + 8H+ → Mn2++ 4H2O ---(2) x 2

-----------------------------------------------------------------------------------------

2 MnO4- + 10 Cl- + 16 H+ → 5 Cl2 + 2 Mn2+ + 8 H2O

-------------------------------------------------------------------------------------

EQUIVALENTS IN REDOX PROCESSES:

Equivalent weight : Equivalent of an element is the weight of the element that combines with 1 g hydrogen, 8 g oxygen or 35.5 g chlorine.

In ither words; Equivalent weight of a substance is the number of parts by weight of it (element or compound) that will combine with or replace, directly or indirectly, 1.008 parts by weight of hydrogen, 8.00 parts by weight of oxygen, or 35.5 parts by weight of chlorine(or the equivalent weight of any other element or compound).

( For all elements, the atomic weight is equal to the equivalent weight times a small whole number, called the valence of the element. An element can have more than one valence and therefore more than one equivalent weight.

or

Equivalent weight is the amount of an element that reacts, or is involved in reaction with, 1 mole of electrons.

or

The equivalent weight of an element or radical is equal to its atomic weight or formula weight divided by the valence it assumes in compounds.

The unit of equivalent weight is the atomic mass unit i.e. the amount of a substance in grams numerically equal to the equivalent weight is called a gram equivalent. e.g.

i) Hydrogen has atomic weight 1.008 and always assumes valence 1 in compounds, so its equivalent weight is 1.008

ii). Oxygen has an atomic weight of 15.9994 and always assumes valence 2 in compounds, so its equivalent weight is 7.9997.

iii) The sulphate radical (SO 4 2-) has formula weight 96.0636 and always has valence 2 in compounds, so its equivalent weight is 48.0318.

iv) Some elements exhibit more than one valence in forming compounds and thus have more than one equivalent weight. Iron (atomic weight 55.845) has an equivalent weight of 27.9225 in ferrous compounds (valence 2) and 18.615 in ferric compounds (valence 3).

or

Equivalent weight: (For oxidation-reduction and electrolysis)

The equivalent weight is the weight associated with the loss or gain of 6.02 × 1023 electrons (Avogadro’s number) or 96500 coulombs of electric charge.

This is also the molecular weight divided by the number of electrons lost or gained.

Numerical-1

i) How many equivalents of KMnO4 will be required to react with 60 g of FeSO4 in the reaction?

5 Fe2+ + MnO4- + 8 H+ ⇄ 5 Fe3+ + Mn2+ + 4 H2O

Ans.

Following the relationship:

Equivalents of KmnO4 = Equivalents of FeSO4

Now, Formula weght of FeSO4

Equivalent weight of FeSO4 = ---------------------------------------

Oxidation number change of Fe

151.9

Equivalent weight of FeSO4 = -----------

1

151.9 g FeSO4

Equivalent weight of FeSO4 = ----------------------------

1 equivalent FeSO4

60 g FeSO4

∴ Equivalents FeSO4 = -------------------------------

151.9 g/equialent FeSO4

= 0.40 equivalent FeSO4

Hence, 0.40 equivalent of KMnO4 is required.

( [substances react in the ratio of their equivalent weights]

Numerical- 2.

Find the conentration of 10-3 N KMnO4 in gL-1 in the following process.

2 MnO4- + 10 Cl- + 16 H+ ⇄ 2 Mn2+ + 5 Cl2 + 8 H2O

Ans. A 10-3 N KMnO4 solution contains 10-3 g equivalent of KMnO4 per litre.

Change in oxidation number of Mn in the given reaction = 5 , (+7 to +2)

Molecular weight

∴ Eq.Weight of KMnO4 = --------------------

Ox.No.Change

158

= ------------ = 31.6

5

Now, 1 g equivalent KMnO4 = 31.6 g

∴ 10-3 g equivalent of KMnO4 = 31.6 x 10-3 g

= 0.0316 g L-1

Numerical.3

Calculate the mass of KClO3 required for the preparation of 400 cm3 of 0.20 N KClO3 to be used in the following reaction.

ClO3- + 3 H2SO4 ⇄ Cl- + 3 SO42- + 6 H+

Change in oxidation number of Cl in the given reaction = 6 ( +5 to -1)

Molecular weighr

∴ Eq. weight of KClO3 = -- ----------------------

Ox. No. Change

122.5

= -------- = 20.4 gL-1

6

Now, 1 g eq.wt (1N) of KClO3 = 20.4 gL-1

∴ 0.2N KClO3 = 20.4 x .2 = 4.08 gL-1

i.e. 1000 ml of 0.2N KClO3 = 4.08 g

∴ 400 ml of 0.2N KClO3 = (4.08x400)/1000 = 1.632 g

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