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Question: \({{I}^{-}}\) reduces \(I{{O}_{3}}^{-}\) to \({{I}_{2}}\) and itself oxidised to \({{I}_{2}}\) in ac...

I−{{I}^{-}} reduces IO3−I{{O}_{3}}^{-} to I2{{I}_{2}} and itself oxidised to I2{{I}_{2}} in acidic medium. Thus, final reaction is:
AI−+IO3−+6H+→I2+3H2O{{I}^{-}}+I{{O}_{3}}^{-}+6{{H}^{+}}\to {{I}_{2}}+3{{H}_{2}}O
B)I−+IO3−→I2+O2{{I}^{-}}+I{{O}_{3}}^{-}\to {{I}_{2}}+{{O}_{2}}
C)5I−+IO3−+6H+→3I2+3H2O5{{I}^{-}}+I{{O}_{3}}^{-}+6{{H}^{+}}\to 3{{I}_{2}}+3{{H}_{2}}O
D) None of the above

Explanation

Solution

The concept of reduction reaction in acidic medium which means with H+{{H}^{+}} ion and the resulting product as iodine and balancing this equation based on valency gives the required answer.

Complete answer:
In the classes of inorganic chemistry, we have come across several reactions that are based on reduction, oxidation, addition and so on. Now, let us see the reduction and self oxidation reaction of iodine in the presence of acidic media. Here, it is given that I−{{I}^{-}} reduces IO3−I{{O}_{3}}^{-} to I2{{I}_{2}} and itself oxidised to I2{{I}_{2}} in acidic medium. Thus, this reaction can be written as,
I−+IO3−→I2+I2{{I}^{-}}+I{{O}_{3}}^{-}\to {{I}_{2}}+{{I}_{2}}
Now, we have to balance the iodine number and this is as follows,
I−+IO3−→12I2+12I2{{I}^{-}}+I{{O}_{3}}^{-}\to \dfrac{1}{2}{{I}_{2}}+\dfrac{1}{2}{{I}_{2}}
Since, balancing oxidation number is to be done, in the above reaction the iodine of IO3−I{{O}_{3}}^{-} has +5 oxidation state and on the product side the first iodine molecule12I2\dfrac{1}{2}{{I}_{2}} is in ‘0’ oxidation state.
Also, I−{{I}^{-}} has -1 oxidation state and the last molecule of iodine 12I2\dfrac{1}{2}{{I}_{2}} has ‘0’ oxidation state.
Therefore, by cross multiplication of values, we get the equation as shown below,
5I−+IO3−→3I25{{I}^{-}}+I{{O}_{3}}^{-}\to 3{{I}_{2}}
Now, since again the oxygen atom on the reactant side is three, this has to be balanced and can be done by adding three moles of water on the product side. Therefore this balanced equation can be depicted as,
5I−+IO3−+6H+→3I2+3H2O5{{I}^{-}}+I{{O}_{3}}^{-}+6{{H}^{+}}\to 3{{I}_{2}}+3{{H}_{2}}O

Thus, the correct answer is option C) 5I−+IO3−+6H+→3I2+3H2O5{{I}^{-}}+I{{O}_{3}}^{-}+6{{H}^{+}}\to 3{{I}_{2}}+3{{H}_{2}}O

Note: Note that when the molecules are present in ionic form, the oxidation numbers on both the side of the reaction that is on the reactant side as well as product side has to be balanced too and not just the number of molecules undergoing the reaction.