Question

Hydrogen peroxide reacts with iodine in basic medium to give:

• A. $I{O}_3^{-}$
• B. $I{O}^{-}$
• C. $I^{-}$
• D. $I{O}_4^{-}$

Answer 1

The correct option is C $I^{-}$

In a basic medium when an lodine solution reacts with Hydrogen peroxide it gives lodide ions along with water and oxygen. Because of its instability, hydrogen peroxide easily decomposes either to water or oxygen molecules. The oxidation state of oxygen in hydrogen peroxide is $-1$. It can be oxidised to oxygen or reduced to water or hydroxide ion. In aqueous solution, hydrogen peroxide differs from pure substance due to the effects of hydrogen bonding between water and hydrogen peroxide molecules.

The conversion of iodine is due to the oxidizing action of the hydrogen peroxide. The reaction takes place like:

$I_2+H_2{O}_2+2O{H}^{-}\to 2I^{-}+2H_{2}O+O_2$

It clearly shows that when iodine is treated with hydrogen peroxide in the basic medium it gives iodide ions whereas when treated in an acidic medium it gives iodate ions.

Therefore for the aforesaid question the product of the reaction is $I^{-}$.