A redox reaction in which a same element present in a particular compound in a definite oxidation state is oxidized as well as reduced simultaneously is a disproportionate reaction.
n factor of overall reaction = n1 × n2 / n1 + n2
MnO4– + 8H+ + 5e → Mn+2 + 4H2O
n factor of KMnO4 = 5
Find the n factor of Na2S2O3 in the following reactions.
(a) Na2S2O3 + I2 → NaI + Na2S4O6
Sodium thio sodium tetra
(b) Na2S2O3 + I2 → NaI + Na2SO4
(a) Change in oxidation number of sulphur=|+2-2.5|=0.5
Change in oxidation number per molecule =0.5×2
∴ n factor of Na2S2O3 = 1
(b) Change in oxidation number of sulphur=|2-6|=4
Change in oxidation number per molecule =4×2
∴ n factor of Na2S2O3 = 8
=> Consider the salt Ax By to undergo a reaction so that the element A undergoes a change in O.S. but is present in more than one product with the same oxidation state i.e. A+cxBy → A+faBy + A+fgH
(the superscripts denote the oxidation state of the respective elements).
=> Salts that react in such a way that more than one type of atom in the salt undergoes oxidation state Change.
A+cxBy → A+faE + JxB-1
In this case both A and B are changing their O.SO. ‘s and both of them are either getting oxidized or reduced. In such a case n factor of the compound is the sum of the individual n factors of A and B. i.e.|xc-xf| + |-xC- (-yi)|. Then n factor of A can be understood which is |xc-xf|. The n factor of B is |-xc-(-yi)|
Law of Chemical equivalence:
It states that in any chemical reaction the equivalents of all the reactants and products must be same.
Eq. of A=Equivalents of ‘B’=Equivalents of ‘C’
Equivalents of ‘A’ = (Weight of ‘A’)/Equivalent weight of ‘A’
Or Equivalents of ‘A’ = no. of moles of ‘A’ x n-factor
A + B + C → product
If A & B do not react with each other but C react with both A & B then
Gm eq. of A + Gm eq. of B= gm eq. of C
Meq. Of A + Meq. Of B = gm Meq. Of C
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