JEE Main 2023 — Hydrogen Question with Solution

The reaction between iodide ions and hydrogen peroxide (${\mathrm{H}}_2{\mathrm{O}}_2$) occurs in the acidic medium and can be represented in the following manner:

$2{\mathrm{I}}^{-} \left(\mathrm{aq}\right) + {\mathrm{H}}_2{\mathrm{O}}_2 \left(\mathrm{l}\right) + 2{\mathrm{H}}^{+} \left(\mathrm{aq}\right) \to {\mathrm{I}}_2 \left(\mathrm{g}\right) + 2{\mathrm{H}}_2\mathrm{O} \left(\mathrm{l}\right)$

In this reaction, hydrogen peroxide oxidises iodide ions (I^-) to molecular iodine. If a calculated amount of sodium thiosulphate is added in the presence of starch solution as an indicator to the above reaction mixture, the liberated iodine reacts with thiosulphate ions as fast as it is formed and is reduced back to iodide ions till all the thiosulphate ions are oxidised to tetrathionate ions.

${\mathrm{I}}_2 \left(\mathrm{g}\right) + 2{\mathrm{S}}_2{\mathrm{O}}_3^{2-} \left(\mathrm{aq}\right) \to {\mathrm{S}}_4{\mathrm{O}}_6^{2-} \left(\mathrm{aq}\right) + 2{\mathrm{I}}^{-} \left(\mathrm{aq}\right)$

After the complete consumption of thiosulphate ions, the concentration of iodine liberated in the reaction of hydrogen peroxide with iodide ions increases rapidly to a point where iodine forms intense blue complex with starch.