Ion – Electron Method, Half Reaction Method, Concept of Limiting Reagent

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Ion - electron method or Half reaction method.

The various steps involved are

(i) Write the skeletal equation and indicate the oxidation number (O. N.) of each element.
(ii) Identify the elements which has undergone change in oxidation state.
(iii) Divide the skeletal equation into two half reactions i.e. oxidation half reaction and reduction half reaction. In each half reaction balance the atom which undergo change in oxidation number.
(iv) Add electron to whichever side is necessary to make up the difference in oxidantion number in each half reaction.
(v) Balance oxygen atoms by addition of proper number of H2O molecules to side deficient in 'O' atoms.
(vi) For ionic equations - It involves the balancing of H atoms in each half reaction. (after O is balanced in each half reaction).


(a) For acidic medium: Add proper number of H+ ions to the side falling short of H atoms.
(b) For basic medium: Add proper number of Hp molecules to the side falling short of H atoms and equal number of OH- ions to the other side.
(vii) Equalise the number of electrons lost and gained by multiplying the half reactions with suitable integer and add them to get the final equation.

Note: It may be remembered that in balanced equation the number of atoms and also the electrical charges must be equal on both sides.

Concept of Limiting Reagent

When ever, the ratios of reactant molecules actually used in an experiment are different from those given by co-efficients of the balance equation, a surplus of one reagent is left over after the reaction is completed.

Thus the extent to which a reaction takes place, depends on the reactant that is present in limiting amount - the limiting reactant. The other reactant is said to be the excess reactant.

Consider the following chemical reaction N2 + 3H2 -> 2NH3

From the reaction, it is clear that 1 mol N2 reacts with 3 mol H2 to from 2 mol NH3.

If we take 1 mol N2 and 4 mol H2, even then 2 mol of NH3 are formed since 1 mol H2 is left unreacted. So in this case N2 is limiting reagent.

Now, if we take 2 mol N2 and 3 mol H2 even then 2 mol of NH3 are formed and 1 mol N2 is left unreacted. 50 in this case H2 is limiting reagnet.



Illustration 5: A sample of Ca3(PO4)2 contains 3.1 g phosphorus, the weight of Ca in the sample is

(A) 6 g
(B) 4 g
(C) 2 g
(D) 5 g 1
Solution: (A). Mole of phosphorus = 3.1/31=1/10
Mole of Ca 3/20
Weight of Ca 3/20×40=6 gm

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