Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# An amount of $2.5\times {{10}^{-3}}$ mole of an ion ${{A}^{n+}}$ exactly requires $1.5\times {{10}^{-3}}$ moles of $MnO_{4}^{-}$ for the oxidation of ${{A}^{n+}}$ to $AO_{3}^{-}$ in acid medium. What is the value of n?

Last updated date: 24th Jul 2024
Total views: 349.8k
Views today: 3.49k
Verified
349.8k+ views
Hint :Redox reactions: These are the type of chemical reactions in which transfer of electrons take place between two chemical species. In these reactions, the oxidation and reduction occur simultaneously during the process and products are formed accordingly.

For finding the value of n, we need to write the balanced chemical equation using a half-cell reaction method. In a redox reaction, when we represent the oxidation and reduction reactions separately then these reactions are referred to as half-cell reactions. The two half reactions combine together to form a reduction-oxidation couple. For the given process, the half-cell reactions are represented as follows:
Oxidation half reaction: - ${{A}^{n+}}+3{{H}_{2}}O\to AO_{3}^{-}+6{{H}^{+}}+(5-n){{e}^{-}}$
Reduction half reaction: - $MnO_{4}^{-}+8{{H}^{+}}+5{{e}^{-}}\to M{{n}^{2+}}+4{{H}_{2}}O$
To balance the given redox reaction, we need to equate the number of electrons involved in  oxidation half reaction to the number of electrons participating in the reduction half reaction.
Therefore,
Total number of electrons participated in the reduction of $MnO_{4}^{-}$ to $M{{n}^{2+}}=$ number of moles of $MnO_{4}^{-}\times$ number of electrons gained during the reaction.
$\Rightarrow {{n}_{red}}=5\times 1.5\times {{10}^{-3}}$
$\Rightarrow {{n}_{red}}=7.5\times {{10}^{-3}}$
Hence, number of electrons lost in the oxidation of ${{A}^{n+}}$ to $AO_{3}^{-}$ will be equal to the ratio of the number of electrons gained in the reduction and the amount of moles of $MnO_{4}^{-}$ present i.e.,
${{n}_{oxi}}=\dfrac{7.5\times {{10}^{-3}}}{1.5\times {{10}^{-3}}}$
$\Rightarrow {{n}_{oxi}}=3$
But according to oxidation half reaction, the number of electrons lost $=5-n$
So, on comparing:
$5-n=3$
$\Rightarrow n=2$
Thus, the value of n for the given conditions is $2$ .

Note :
Ensure while balancing the chemical reaction, the number of atoms of each element in the should be equal to the number of atoms of element in the product. Further, the charge should also be equal on both sides of the reaction.