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# Balancing (the Equation) of Redox Reactions

The two methods for balancing the equation of redox reactions are as follows. i. The oxidation number method ii. Ion-electron method / half reaction method.

Balancing (the Equation) of Redox Reactions

The two methods for balancing the equation of redox reactions are as follows.

i.               The oxidation number method

ii.               Ion-electron method / half reaction method.

Both are based on the same principle: In oxidation - reduction reactions the total number of electrons donated by the reducing agent is equal to the total number of electrons gained by the oxidising agent.

## i. Oxidation number method

In this method, the number of electrons lost or gained in the reaction is calculated from the oxidation numbers of elements before and after the reaction. Let us consider the oxidation of ferrous sulphate by potassium permanganate in acid medium. The unbalanced chemical equation is,

FeSO4 + KMnO4+H2SO4 →Fe2(SO4)3 + MnSO4+ K2SO4+H2O

### Step 1

Using oxidation number concept, identify the reactants (atom) which undergo oxidation and reduction.

a. The oxidation number of Mn in KMnO4 changes from +7 to +2 by gaining five electrons.

b. The oxidation number of Fe in FeSO4 changes from +2 to +3 by loosing one electron.

### Step 2

Since, the total number of electrons lost is equal to the total number of electrons gained, equate, the number of electrons, by cross multiplication of the respective formula with suitable integers on reactant side as below. Here, the product Fe2(SO4)3 contains 2 moles of iron, So, the Coefficients 1e- & 5e- are multiplied by the number '2'

### Step 3 Balance the reactant / Product - Oxidised / reduced

10 FeSO4+2 KMnO4+H2SO4  → Fe2(SO4)3 + MnSO4 + K2SO4 + H2O

Now, based on the reactant side, balance the products (ie oxidised and reduced).The above equation becomes

10FeSO4+2KMnO4+H2SO4  → Fe2(SO4)3 + MnSO4 + K2SO4 + H2O

Step 4 Balance the other elements except H and O atoms. In this case, we have to balance K and S atoms but K is balanced automatically.

Reactant Side : 10 'S' atoms (10 FeSO4)

Product Side : 18 'S' atoms

5Fe(SO4)3 + 2MnSO4 + K2SO4

15S         +     1S    +    2S  = 18S

Therefore the difference 8-S atoms in reactant side, has to be balanced by multiplying H2SO4 by '8' The equation now becomes,

10FeSO4+2KMnO4+8H2SO4  →Fe2(SO4)3 + 2MnSO4 + K2SO4 + H2O

### Step 5

Balancing 'H' and 'O' atoms

Reactant side '16'-H atoms (8H2SO4 i.e. 8 x 2H = 16 'H')

Product side '2' - H atoms (H2O i.e. 1 x 2H = 2 'H')

Therefore, multiply H2O molecules in the product side by '8'

10 FeSO4+2 KMnO4+8 H2SO4  → 5 Fe2(SO4)3 + 2 MnSO4 + K2SO4 + 8H2O

The oxygen atom is automatically balanced. This is the balanced equation.

## ii. Ion - Electron method

This method is used for ionic redox reactions.

### Step 1

Using oxidation number concept, find out the reactants which undergo oxidation and reduction.

### Step 2

Write two separate half equations for oxidation and reduction reaction,

Let us consider the same example which we have already discussed in oxidation number method.

KMnO4 +FeSO4 + H2SO4  → MnSO4+Fe2(SO4)3 + K2SO4+H2O

The ionic form of this reaction is,

The two half reactions are,

Fe2+ → Fe3+ + 1e- ------------------------ (1)

and

MnO4 + 5e- → Mn2+ -------------------- (2)

Balance the atoms and charges on both sides of the half reactions.

Equation (1) No changes i.e.,

Fe2+→ Fe3+ + 1e- ------------------------- (1)

Equation (2) 4'O' on the reactant side, therefore add 4H2O on the product side, to balance 'H' - add, 8H+ in the reactant side

MnO?4 + 5e- + 8H+  → Mn2+ + 4H2O---- (3)

### Step 3

Equate both half reactions such that the number of electrons lost is equal to number of electrons gained.

Addition of two half reactions gives the balanced equation represented by equation (6).

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11th Chemistry : UNIT 1 : Basic Concepts of Chemistry and Chemical Calculations : Balancing (the Equation) of Redox Reactions |