Question

When \[S{O_2}\] is passed in a solution of potassium iodate, the oxidation state of iodine changes from:
A. + 5 to 0
B. + 5 to - 1
C. - 5 to 0
D. - 7 to - 1

Answer 1

Hint: This is an example of a redox reaction. When \[S{O_2}\] is passed in a solution of potassium iodate then Sulphur dioxide acts as a reducing agent and it reduces the iodate salt to iodine. Additional products of this reaction include potassium sulphate and sulphuric acid. Redox reaction is a type of chemical reaction in which the oxidation states of atoms change (this is due to the actual or formal transfer of electrons between chemical species).

Complete step by step answer:
First, we will write the equation for the above reaction which is as follows:
\[5S{O_2} + 2KI{O_3} + 4{H_2}O \to 4{H_2}S{O_4} + {K_2}S{O_4} + {I_2}\]
Sulphur dioxide gas reacts with a solution of potassium iodate (potassium iodate in water) to form sulphuric acid, potassium sulphate, and iodine gas. We can calculate the oxidation state of iodine in reactants and products:
The oxidation state of iodine in reactants:
\[KI{O_3}\]
Let the oxidation state of iodine be x. Potassium has an oxidation state of \[ + 1\] and oxygen has an oxidation state of \[ - 2\]. Hence three atoms of oxygen will have an oxidation state of \[ - 6\]. Since the compound does not have an overall charge hence total charge is taken to be zero.
\[1 + x - 6 = 0\]
\[ \Rightarrow x - 5 = 0\]
\[ \Rightarrow x = + 5\]
Hence the oxidation state of iodine on the reactant side is \[ + 5\].
The oxidation state of iodine in products is zero since iodine is present in a pure state.
In this equation, we can see that the oxidation state of iodine changes from \[ + 5\] in \[KI{O_3}\]to \[0\] in \[{I_2}\].

So, the correct answer is Option A .

Note: We need to always see if the compound has some overall charge or not.If the compound has some overall charge then we will put it on the right-hand side of the equation.The oxidation state of an atom in any pure element is equal to zero.