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Question: For \({I^ - }\) reduces \(I{O_3}^ - \) to \({I_2}\) and itself oxidised to in acidic medium .Final r...

For I−{I^ - } reduces IO3−I{O_3}^ - to I2{I_2} and itself oxidised to in acidic medium .Final reaction is:
A.I−+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 reactions in which the simultaneous reduction and oxidation reactions takes place are known as redox reactions .Such reactions usually occur in the acidic, basic and neutral medium and the product of the reaction is highly dependent on the nature of the medium.

Complete step by step answer:
The formation of the products depends on the medium in which the reaction is carried out since the two processes the reduction and the oxidation process occurs simultaneously. The reaction conditions have the major influence due to the availability of the electrons and other reaction agents to carry out the process.
Some of the important oxidising agents are: KMnO4KMnO_4, HNO3HNO_3, KClO3KClO_3 etc. also fluorine is the strongest oxidising agent. Some of the important reducing agents are Na, Zn, Fe and Al they are the strongest reducing metals, few non-metals are C, H, S and P. In the solution Lithium is the strongest reducing agent while Caesium is the strongest reducing agent in the absence of water.
Two reactions occur in the electrolytic cell, the reduction half reactions and the oxidation half reactions the two cells are separated by the salt bridge ,which ensures that the solutions do not mix with each other, during the process of electrolysis carried out in the cell.
Iodine in the given process undergoes the reduction and oxidation under the acidic conditions as follows:
2I−(aq)⇌I2+2e−.....................(i)2{I^ - }(aq) \rightleftharpoons {I_2} + 2{e^ - }.....................(i)
12H+(aq)+2IO3−+10e−→6I2+6H2O..............(ii)12{H^ + }(aq) + 2I{O_3}^ - + 10{e^ - } \to 6{I_2} + 6{H_2}O..............(ii)
On solving the equation (i) and (ii):
12H++2IO3−+10I−→6I2+6H2O12{H^ + } + 2I{O_3}^ - + 10{I^ - } \to 6{I_2} + 6{H_2}O
Further simplifying the equation we have:
6H++IO3−+5I−→3I2+3H2O6{H^ + } + I{O_3}^ - + 5{I^ - } \to 3{I_2} + 3{H_2}O
Thus option C. is the correct answer.

Note:
Combustion is an important process where the concept of Redox reaction is applied; it is a fast form of the reaction and can be regarded as the redox reaction since it is an explosive and fast process. Thus the redox reactions have very wide applications.