KMnO_4 oxidises I^{-} in acidic and neutral/f | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) In acidic medium,$KMnO_4$ acts as a strong oxidizing agent and oxidizes iodide ions $(I^{-})$ to iodine $(I_2)$: $2 MnO_4^{-} + 10 I^{-} + 16 H^{+

Vedclass · Accepted Answer

$I_2$ and $IO_3^-$

Vedclass · Answer

(A) In acidic medium,$KMnO_4$ acts as a strong oxidizing agent and oxidizes iodide ions $(I^{-})$ to iodine $(I_2)$: $2 MnO_4^{-} + 10 I^{-} + 16 H^{+} \rightarrow 2 Mn^{2+} + 5 I_2 + 8 H_2 O$ In neutral or faintly alkaline medium,$KMnO_4$ oxidizes iodide ions $(I^{-})$ to iodate ions $(IO_3^{-})$: $2 MnO_4^{-} + I^{-} + H_2 O \rightarrow 2 MnO_2 + IO_3^{-} + 2 OH^{-}$ Therefore,the products are $I_2$ and $IO_3^{-}$ respectively.

$KMnO_4$ oxidises $I^{-}$ in acidic and neutral/faintly alkaline solution,respectively to

Similar Questions

What change does not occur when $K_2Cr_2O_7$ reacts with $H_2O_2$ solution in an acidic medium?

In acidic medium, one mole each of $MnO_4^-$ and $Cr_2O_7^{2-}$ is reduced by $x$ and $y$ moles of ferrous ions $(Fe^{2+})$ respectively. The sum of $x$ and $y$ is:

When mercury $(II)$ chloride is treated with excess of stannous chloride,the products obtained are

In the following reaction,$\text{SO}_2 + 2\text{H}_2\text{S} \to 3\text{S} + 2\text{H}_2\text{O}$,which choice has a value twice that of the equivalent mass of the oxidising agent?

When mercury $(II)$ chloride is treated with excess of $SnCl_2$,the products obtained are

Vedclass Test Series

Exam Paper Generator

Online Exam Module