Question

What is the equivalent mass $IO^{-}_{4}$ when it is converted into $I_{2}$ in acid medium?
A.M/6
B.M/7
C.M/5
D.M/4

Answer 1

Equivalent weight of a given mole of substance is generally defined as molar mass divided by n-factor. It there are x equivalents, its weight would be
$\dfrac{x\times M}{n-factor}$
N-factor is the capability of accepting or releasing electrons, which is reflected as change in oxidation number.

Complete answer:
- Equivalent weight has got units of mass and dimensions where in case of atomic weight it is dimensionless.
- Now, let us consider the conversion of $IO^{-}_{4}$ into $I_{2}$ in acidic medium
$2IO^{-}_{4}+16H^{+}+14e^{-}\rightarrow I_{2}+8H_{2}O$
Oxidation number of I = X x 4(-2) = -1
I = +7
Here in LHS the oxidation state of I is +7 and in RHS the oxidation state of $I_{2}$ is 0. This clearly indicates that iodine is reduced to 0 oxidation state in the reaction.
Equivalent weight is given as- 14 electrons are involved in the reaction and 2 moles of $IO^{-}_{4}$ is been used thus
$EqWt.=\dfrac{2\times M}{14}$
$EqWt.=\dfrac{M}{7}$

Therefore, the correct answer is option (B).

Note:
In case of acid base titration the equivalent weight of acid or base is the mass that reacts or supplies one mole of hydrogen ions and in case of redox reactions the equivalent weight of each reactant reacts or supplies one mole of electrons. In precipitation reactions, the number of ions which will precipitate in the given reaction is shown by the equivalence factor.