Oxidation and reduction are always found to go hand in hand during a redox reaction. Whenever an element or compound is oxidized, another element or another compound must be reduced.

An oxidant is reduced and simultaneously the reductant is oxidized.It follows from our above discussion that we should be able to balance oxidation reduction reactions on the basis of
 Ion  Electron Method of Balancing Oxidation  Reduction Reactions.
 Oxidation Number Method of Balancing Oxidation  Reduction Reactions.
Ion  Electron Method of Balancing Oxidation  Reduction Reactions:
 Ascertain the Reactants and Products, and their chemical formula's.
 Set up the partial equations representing the reduction of the oxidant, and the oxidation of the reductant.
 If the reaction occurs in acid medium use requisite number of H⁺ for balancing the number of atoms involved in the partial equation. for alkaline medium use OH⁻ ions.
 Balancing the charge in the partial equations by adding suitable number of electrons. These electrons indicate the electrons involves in the oxidation or the reduction half reactions.
 Multiply each partial equation by a suitable factor so each of the two partial equations involves the same number of electrons.
 Add the partial equations and cancel out species which appear on both sides of the equations.
In aqueous acid medium potassium permanganate oxidizes a ferrous ion to the ferric state:

In this reaction permanganate ion is the oxidant and ferrous ion the reductant. The left hand side of the ultimate equation will carry, MnO₄⁻ (KMnO₄), H⁺ (H₂SO₄) and Fe²⁺ (FeSO₄).

The Right hand side will have as products, Mn²⁺ (MnSO₄), H₂O and Fe³⁺ [Fe₂(SO₄)₃]. The partial equation representing the reduction of the oxidant MnO₄⁻ will involve,

MnO₄⁻ → Mn²⁺

Since the reaction occurs in acid medium we utilize eight H⁺ ions to balance the four oxygen atoms in MnO₄⁻.

Thus, MnO₄⁻+8H⁺ → Mn²⁺+4H₂O

Since the above partial equation is still unbalanced from the viewpoint of charge, the equation is balanced by bringing in five electrons:

MnO₄⁻ + 8H⁺ + 5e ⇆ Mn²⁺ + 4H₂O

If we now apply the above steps to obtain the partial equation representing the oxidation of reductant we have the following:

Fe⁺² ⇆ Fe⁺³ + e

This equation on multiplication by 5 and then adding to the partial equation of the oxidant gives the final balanced equation:

MnO₄⁻ + 8H⁺ + 5e ⇆ Mn²⁺ + 4H₂O
5×( Fe⁺² ⇆ Fe⁺³ + e)

MnO₄⁻ + 8H⁺ + 5Fe⁺² → Mn²⁺ + 5Fe⁺³ + 4H₂O
The Oxidation of potassium iodide by potassium dichromate in dilute acid medium.

In this reaction dichromate is reduced to (+6) state to (+3) state and iodide ion is oxidised to elementary iodine.
Taking dichromate side, balancing the atoms and charges provides the partial equation:

Cr₂O₇⁻² + 14H⁺ + 6e ⇆ 2Cr⁺³ + 7H₂O
(Reduction of Oxidant)

Considering the case of iodide ion we get:

2I⁻ ⇆ I₂ + 2e
(Oxidation of Reductant)

Multiplying the above equation by 3 and adding these two partial equation we have the following final equation:
Cr₂O₇⁻² + 14H⁺ + 6e ⇆ 2Cr⁺³ + 7H₂O 
3 (2I⁻ ⇆ I₂ + 2e) 
Cr₂O₇⁻²+14H⁺+6I⁻→2Cr⁺³+3I₂+7H₂O 
The reaction of permanganate ion with Sodium Stannite (Na₂SnO₂) in alkaline solution in which manganese dioxide (MnO₂) and Sodium Stannate(Na₂SnO₃) are formed.

MnO₄⁻ → MnO₂

The Partial equation representing the reduction of oxidant is, Since the medium is alkaline we put the requisite number of OH⁻ ions to effect atom balance as:

MnO₄⁻ + 2H₂O → MnO₂ + 4OH⁻

To balance the charge on both sides three electrons are added on the left hand side:

MnO₄⁻ + 2H₂O + 3e ⇆ MnO₂ + 4OH⁻

For the reductant the balance partial equation is:

SnO₂⁻² + OH⁻ ⇆ SnO₃⁻² + H₂O + 2e

To balance the number of electrons involved in the two partial equations the first equation is multiplied by 2 and the second by 3 and on the adding we have the final form of the ionic equation:

2MnO₄⁻ + 4H₂O + 6e ⇆ 2MnO₂ + 8OH⁻
3SnO₂⁻² + 3OH⁻ ⇆ 3SnO₃⁻² + 3H₂O + 6e

2MnO₄⁻+3SnO₂⁻²+H₂O→2MnO₂+3SnO₃⁻²+2OH⁻
Oxidation of Mn⁺² to MnO₄⁻ by PbO₂ or NaBiO₃ in Acid Medium:
Ion electron method for balancing Oxidation Reduction reaction 
Oxidation Number Method of Balancing Oxidation  Reduction Reactions:
 Assuming that the oxidation number of oxygen remains unchanged manganese in MnO₄⁻ has an oxidation number +7. Calculation of the decrease and increase in oxidation number gives:
 Putting right factors the decreases and increases in oxidation numbers are balanced, giving,

MnO₄⁻ + 5Fe⁺² → Mn⁺² + 5Fe⁺³
 Since the reaction medium is acidic and the changes are not equal on the two sides of the above expression, H⁺ ions are added and requisite number of H₂O written:
 MnO₄⁻+8H⁺+5Fe⁺²→Mn⁺²+ 5Fe⁺³+4H₂O
Representation of Decreases and Increases of Oxidation Number 
The Oxidation of iodide ion by dichromate ion in acid medium:

The reaction represented as,
Cr₂O₇⁻² + I⁻ → 2Cr⁺³ + I₂

Oxidation number of Cr decreases by 2 and oxidation number of Sn increases by 1. Equalizing the increase and decrease in oxidation number,

Cr₂O₇⁻² + 6I⁻ → 2Cr⁺³ + 3I₂

The medium is acidic and because the ionic charges are not equal on the two sides, fourteen H⁺ ions are added to the left hand side and requisite number of H₂O added to the left hand side.

Cr₂O₇⁻² + 6I⁻ + 8H⁺ → 2Cr⁺³ + 3I₂ +7H₂O
Oxidation of sodium stannite to stannate in alkaline medium:

This reaction represented as,
MnO₄⁻ + SnO₂⁻² → MnO₂ + SnO₃⁻²

Oxidation number of Mn decreases by 3 and oxidation number of Sn increases by 2. Equalizing the increase and decrease in oxidation number,

2MnO₄⁻ + 3SnO₂⁻² → 2MnO₂ + 3SnO₃⁻²

Because the medium is alkaline and because the ionic charges are not equal on the two sides, two OH⁻ ions are added to the right hand side.

2MnO₄⁻ + 3SnO₂⁻² → 2MnO₂ + 3SnO₃⁻² + 2OH⁻

Since there are two hydrogen atoms on the right and none on the left, H₂O is added to make atom balance. The final equation is:

2MnO₄⁻ + 3SnO₂⁻² + H₂O → 2MnO₂ + 3SnO₃⁻² + 2OH⁻
 Problem 1:

Express by ion electron method the reduction of permagnate to manganous stat by hydrogen peroxide in acid medium.
 Answer:

This Reaction occurs in acid medium and the partial equation is:
MnO₄⁻ + 8H⁺ + 5e ⇆ Mn⁺² + 4H₂O
(Reduction of Oxidant)

In acid medium H₂O₂ will give O₂ and the partial equation being:
H₂O₂ ⇆ 2H⁺ + O₂ +2e
(Oxidation of Reductant)

First equation is multiplying by 2 and the second is 5 to have electron balanced. We have,
2MnO₄⁻ + 16H⁺ + 10e ⇆ 2Mn⁺² + 8H₂O 5H₂O₂ ⇆ 10H⁺ + 5O₂ +10e 
2MnO₄⁻+5H₂O₂+6H⁺⇆2Mn⁺²+8H₂O+5O₂ 
 Problem 2:

Express by ion  electron method the reduction of nitrate ion to ammonia by aluminium in aqueous NaOH.
 Answer:

In alkaline medium we use H₂O and OH⁻ ion according to convenience. The partial equation with charge balance is:

NO₃⁻ + 6H₂O + 8e ⇆ NH₃ + 9OH⁻

Metallic aluminium will go over to aluminate ion, the partial equation being:

Al + 4OH⁻ ⇆ AlO₂⁻ + 2H₂O +3e

Multiplying by right factors for electron balance we have the balanced equation:

3NO₃⁻ + 18H₂O + 24e ⇆ 3NH₃ + 27OH⁻
8Al + 32OH⁻ ⇆ 8AlO₂⁻ + 16H₂O +24e
3NO₃⁻+8Al+2H₂O+5OH⁻→3NH₃+8AlO₂⁻
 Problem 3:

Use Oxidation Number method to balance the reaction of iodide ion and iodate ion in acid medium to liberate iodine.
 Answer:

The reaction represented as,
I⁻ + IO₃⁻ → I₂

Representation of the above equation with oxidation number of iodine,
I⁻(1) + IO₃⁻(+5) → I₂(0)

Since I⁻ (1) oxidation number of iodide increases by 1 and IO₃⁻(+5) oxidation number decreases by 5. Putting the right factors the decrease and increase in oxidation number balanced:

5I⁻ + IO₃⁻ → 3I₂

Since the medium is acidic and the charges are unequal on two sides of the above expression we put Six H⁺ and write three H₂O to balanced the equation:

I⁻ + IO₃⁻ + 6H⁺ → I₂ + 3H₂O
 Problem 4:

Use oxidation number method to balance the reaction between Sulphurus acid and dichromate in acidic medium.
 Answer:

The reaction represented as,
SO₃⁻² + Cr₂O₇⁻² → SO₄⁻² + 2Cr⁺³

Oxidation of two Cr decreases by 2 × (+3) = 6 and oxidation number of S increases by 2. Equalizing the above equation increase and decrees in oxidation number,

3SO₃⁻² + Cr₂O₇⁻² → 3SO₄⁻² + 2Cr⁺³

Because of the medium is acidic and the ionic charges are not equal on the two sides, eight H⁺ ions are added to the Left hand side.

3SO₃⁻² + Cr₂O₇⁻² + 8H⁺ → 3SO₄⁻² + 2Cr⁺³

Since there are eight hydrogen ion on the left and none on the right, four H₂O added to make atom balance. Thus the final equation is:

3SO₃⁻²+Cr₂O₇⁻²+8H⁺→3SO₄⁻²+2Cr⁺³+4H₂O