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Question: Iodate ion, \(IO_3^ - \) oxidises \(SO_3^ {2 - }\) to \(SO_4^{2 - }\) in acidic medium. If 100 mL sa...

Iodate ion, IO3−IO_3^ - oxidises SO32−SO_3^ {2 - } to SO42−SO_4^{2 - } in acidic medium. If 100 mL sample of solution containing 2.24 g of KIO3KI{O_3} reacts with 60 mL of 0.5 M Na2SO3N{a_2}S{O_3} solution, then final oxidation state of iodine is:
A.+5
B.+3
C.+1
D.-1

Explanation

Solution

Number of electrons in a chemical reaction are always constant. Or you can say that electrons are neither created nor destroyed during a chemical reaction. Number of electrons gained by one species will always be equal to the number lost by another.

Complete step by step answer: Now, we will solve the question step by step:
Step 1. Number of electrons lost by Sulphur (sulphur is getting oxidised therefore loss of electrons):
During the given oxidation reaction SO32 - {\text{SO}}_{\text{3}}^{{\text{2 - }}} (Sulphur with oxidation state +4) is getting oxidised to SO42 - {\text{SO}}_{\text{4}}^{{\text{2 - }}} (Sulphur having oxidation state of +6) i.e., Oxidation state of sulphur changes from +4 to +6. Which implies that oxidation of 1 mole sulphite will involve two moles of electrons.
Now, we are provided with 60 mL of 0.5 M Na2SO3{\text{N}}{{\text{a}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{3}}}. Let the number of moles present here be x.
Now, as you know, Molarity is given by n×1000V(mL)\dfrac{{{\text{n}} \times {\text{1000}}}}{{{\text{V(mL)}}}} , where n is number of moles of solute and V is volume. Putting all the given values we have 0.5 = x×100060{\text{0}}{\text{.5 = }}\dfrac{{{\text{x}} \times {\text{1000}}}}{{{\text{60}}}}.
This gives us the value of x =0.003 moles.
Since oxidation of 1 mole of sulphite ion involves 2 moles of electrons, oxidation of 0.003 moles of sulphite ion will involve 0.003×20.003 \times 2 =0.006 moles of electrons.
Step 2. Calculation of final oxidation state of iodine:
As we know electrons are neither created nor destroyed during a chemical reaction, i.e., the number of electrons lost by one species is equal to the number of electrons gained by another.
In this case if Sulphur has lost 0.006 moles of electrons, it means iodine has taken up 0.006 moles of electrons.
Now, we are given 2.14 g of KIO3{\text{KI}}{{\text{O}}_{\text{3}}} which have a molar mass of 214 g. Therefore, the given number of moles of KIO3{\text{KI}}{{\text{O}}_{\text{3}}} is 2.14214\dfrac{{2.14}}{{214}} = 0.001 moles. This implies that 0.001 moles of KIO3{\text{KI}}{{\text{O}}_{\text{3}}} will require 0.006 moles of electrons.
i.e. 1 mole of KIO3{\text{KI}}{{\text{O}}_{\text{3}}} will require 6 moles of electrons.
The oxidation state of iodine in Potassium iodate is +5, since six electrons are now added to it, the final oxidation state will be 6 - 5 = +1.
i.e., the final oxidation state of iodine is +1.

Hence the correct option is C.

Note: Electrons are lost during oxidation and gained during reduction. And hence during an oxidation reaction the oxidation state of that element will increase and will decrease in case of a reduction reaction.
Since we have mentioned about oxidising agent, an oxidation reagent, in this case iodate ion, is a reagent which itself undergoes reduction but oxidises the other element, sulphur in present case.