Question

We have a solution of $IO_3^ - + {H^ + }$ which will show any observable changes with:
A. Chromium sulphate
B. Ammonium sulphide
C. Copper sulphate
D. Potassium iodide

Answer 1

The iodate ion is the presence of an acidic medium acting as an oxidizing agent, therefore it will oxidize the other reactant to form the compound and itself gets reduced during the reaction. The reaction is related to the iodine clock reaction.

Complete step by step answer:
It is given that the solution contains $IO_3^ - + {H^ + }$. In this reaction oxidation of iodide ion takes place which dissolves the iodine.
The reaction is shown below.
$IO_3^ - + 5I + 6{H^ + } \to 3{I_2} + 3{H_2}O$
The iodate ions in the acidic medium act as an oxidizing agent.
This reaction refers to the iodine clock reaction, where a group of reactions is involved where two colorless solutions are mixed to form a solution which remains colorless for some time but then suddenly changes its colour to deep purple blue color.
The delay in time until the blue color appears is inversely related with the reaction rate but the color development has a direct relation with the reaction rate.
Ammonium sulphide reacts with iodate ion and hydrogen ion changes to ammonium sulphate.
The reaction is shown below.
${(N{H_4})_2}S + IO_3^ - + {H^ + } \to {(N{H_4})_2}S{O_4} + {I_2}$
In this reaction, ammonium sulphide reacts with iodate ion and hydrogen ion to form ammonium sulphate and iodine gas.
The ammonium sulphide gets oxidized in presence of iodate ion to form ammonium sulphate.
So, the correct answer is “Option B”.

Note:
In iodine clock reactions, the time for the change of color is controlled by the temperature or the concentration of the reactants used. The reaction rate depends on the concentration of iodate, iodide and hydrogen ions and on the temperature.