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BALNCING REDOX REACTIONS

Quite often one has to write a balanced chemical equation while dealing with the problem on stoichiometry. After writing reactant and products, the balancing can be carried out by hit and trial method. However, systematic methods are available for balancing redox reactions. Before describing these methods, the rules governing the computation of oxidation state of an element are described in the following.

Rules to Compute Oxidation Number

The oxidation number of an element is the number assigned to it by following the arbitrary rules given below:

(i) A free element (regardless of whether it exists in monatomic or polyatomic form, e.g. Hg, H₂, P₄ and S₈) is assigned an oxidation number of zero.

(ii) A free monatomic ion is assigned an oxidation number equal to the charge it carries. For example, the oxidation number of Al³⁺ + S^2- and Cl^- are +3, -2 and -1 respectively.

(iii) In their compounds, the alkali and alkaline earth metals are assigned oxidation numbers of +1 and +2 respectively.

(iv) The oxidation number of hydrogen in its compounds is generally +1 except the ionic hydrides such as LiH. LiAIH₄ where its 'oxidation number is -1.

(v) The oxidation number of fluorine in all its compounds is -1. The oxidation number of all other halogens is -1 in all compound except those with oxygen (e.g. CIO₄^-) and halogens having a lower atomic number (e.g. ICI₃^-). The oxidation number of the latter is determined via oxygen and halogen of lower atomic number.

(vi) The oxidation numbers of both oxygen and sulphur in their normal oxides (e.g. Na₂O) and sulphides (e.g. CS₂) is -2. The exception are the peroxides (e.g. H₂O₂ and Na₂O₂), superoxides (e.g. KO2) and the compound OF₂. Their oxidation numbers are determined by the rules 3, 4 and 5.

(vii) The algebraic sum of oxidation numbers of atoms in a chemical species (compound or ion) is equal to the net charge on the species.

A few examples of the computation of oxidation number of atoms N in various compounds are given as follows.

If x is the oxidation number of N, we 'have

NH₃ x+3(+1) = 0 whichgiv.esx = -3
HN₃ +1 + 3(x) = 0 which gives x = -1/3
N₂H₄ 2x + 4(+1) = 0 which gives x =-2
NO₂ x + 2(-2) = 0 which gives x = 4
N₂O₄ 2x + 4(-2) = 0 which gives x = 4
NO₂2_- x + 2(-2) = -1 which gives x = 3
NH₂OH x + 2(+1) + 1(-2) + 1 = 0 which gives x =-1
NO x + 1(-2) = 0 which gives x = 2
HNO₃ +1 + x + 3(-2) = 0 which gives x = 5
H₂O 2x + 1(-2) = 0 which gives x = 1
HCN + 1 + 4 + x = 0 which gives x = -5


Balancing Redox Reactions Via Oxidation Numbers

The steps involved in this method are as follows:

1. Identify the elements in the unbalanced equation whose oxidation number are changed.
2. Balance the number of atoms of each element whose oxidation number is changed.
3. Find out the total change in oxidation number for each of oxidant and reductant and make them equal by multiplying by small coefficients.
4. Balance the remainder of atoms by inspection and add, if necessary H⁺(acidic medium) or OH^-(alkaline medium) or H2O (to balance oxygen) as the case may be.

Balancing Redox Reactions Via Oxidation Numbers

The steps involved in this method are as follows:

1. Identify the elements in the unbalanced equation whose oxidation number are changed.

2. Balance the number of atoms of each element whose oxidation number is changed.

3. Find out the total change in oxidation number for each of oxidant and reductant and make them equal by multiplying by small coefficients.

4. Balance the remainder of atoms by inspection and add, if necessary H+(acidic medium) or OH^-(alkaline medium) or H₂O (to balance oxygen) as the case may be.

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