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__Balancing Oxidation Reduction Reactions:
__

__Balancing Oxidation Reduction Reactions:__

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

(1)

__Ion - Electron Method of Balancing Oxidation - Reduction Reactions__
(2)

__Oxidation Number Method of Balancing Oxidation - Reduction Reactions.__###
__1. Ion - Electron Method of Balancing Oxidation - Reduction Reactions__:

__1. Ion - Electron Method of Balancing Oxidation - Reduction Reactions__:

(i) Ascertain the Reactants and Products, and their chemical formulas.

(ii) Set up the partial equations representing the reduction of the oxidant, and the oxidation of the reductant.

(iii) If the reaction occurs in acid medium use requisite number of

**Hᐩ**for balancing the number of atoms involved in the partial equation. for alkaline medium use**OH⁻**ions.
(iv) 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.

(v) Multiply each partial equation by a suitable factor so each of the two partial equations involves the same number of electrons.

(vi) Add the partial equations and cancel out species which appear on both sides of the equations.

__Examples:__
Some examples will now needed to elucidate the rules.

(a) In aqueous acid medium potassium permanganate oxidizes a ferrous ion to the ferric state. The following Rule is useful for the balancing the Oxidation - Reduction Reactions by Ion - Electron method.

__this reaction permanganate ion is the oxidant__and

__ferrous ion the reductant__.

The left hand side of the ultimate equation will carry,MnO₄⁻ (KMnO₄), Hᐩ (H₂SO₄)

and Fe²ᐩ (FeSO₄). The Right hand side will have as products, Mn²ᐩ (MnSO₄), H₂O and Fe³ᐩ [Fe₂(SO₄)₃].

The partial equation representing the

__reduction of the oxidant__**MnO₄⁻**will involve,**MnO₄⁻ → Mn²ᐩ**

Since the reaction occurs in

__acid medium__we utilize__eight__**Hᐩ**__ions to balance the four oxygen atoms__in MnO₄⁻.
Thus, MnO₄⁻ + 8Hᐩ

**→**Mn²ᐩ + 4H₂O
Since the above partial equation is still

__unbalanced from the viewpoint of charge__, the equa-tion is balanced by bringing in five electrons:
MnO₄⁻ + 8Hᐩ + 5e

Note that MnO₄⁻ → Mn²ᐩ is indeed a five electron reduction,
Mn⁷ᐩ + 5e → Mn²ᐩ.**⇆**Mn²ᐩ + 4H₂O
If we now apply the above steps to obtain the partial equation representing the oxidation of reductant we have the following:

Feᐩ²

**⇆**Feᐩ³ + e
This equation on multiplication by

**5**and then adding to the partial equation of the oxidant gives the final balanced equation:Balanced Oxidation Reduction Reaction |

(b) 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:

This Reaction occurs in acid medium and the partial equation is:

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)

Final Equation: Cr₂O₇⁻² + 14H⁺ + 6I⁻ → 2Cr⁺³ + 3I₂ + 7H₂O

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

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

Adding These two equation, 2MnO₄⁻ + 5H₂O₂ + 16H⁺ ⇆ 2Mn⁺² + 8H₂O +5O₂

(c) The reaction of permanganate ion with Sodium Stannite (Na₂SnO₂) in alkaline solution in which manganese dioxide (MnO₂) and Sodium Stannate(Na₂SnO₃) are formed.

The Partial equation representing the reduction of oxidant is,

MnO₄⁻ → MnO₂.

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⁻

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⁻

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__Oxidation Number Method of Balancing Oxidation - Reduction Reactions:__

__Oxidation Number Method of Balancing Oxidation - Reduction Reactions:__

(i) 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:

Representation of Decreases and Increases ofOxidation Number |

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

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

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**

(ii)

**The Oxidation of iodide ion by dichromate ion in acid medium:**
The reaction represented as,

**Cr₂O₇⁻ + I⁻ → Cr⁺³ + I₂**The Oxidation of Iodide ion by dichromateion in Acid medium |

__(iii) 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⁻**

Use oxidation number method to balance the reaction between Sulphurus acid and dichro-

mate in acidic medium.

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,

We have, 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**