Question:

$KMnO _{4}$ oxidises $S _{2} O _{3}^{2-}$ to $SO _{4}^{2-}$ in medium $x$ and $NO^{-}_2$ to $NO^{-}_3$ in medium $y, x$ and $y$ are respectively

Updated On: Apr 4, 2024
  • acidic, basic
  • acidic, acidic
  • acidic, neutral
  • neutral, acidic
Hide Solution
collegedunia
Verified By Collegedunia

The Correct Option is D

Solution and Explanation

In neutral or faintly alkaline solution. $8 MnO _{4}^{-}+3 S _{2} O _{3}^{-2}+ H _{2} O \longrightarrow 8 MnO _{2} +6 SO _{4}^{-2}+2 OH ^{-}$ In acidic medium, $2 MnO _{4}^{-}+5 NO _{2}^{-}+6 H ^{+} \longrightarrow 2 Mn ^{+2}+5 NO _{3}^{-}+3 H _{2} O$
Was this answer helpful?
0
0

Concepts Used:

Balancing a redox reaction

The method that is based on the difference in oxidation number of oxidizing agent and the reducing agent is known as oxidation number. The half-reaction method entirely depends on the division of the redox reactions into oxidation half and reduction half. It entirely depends on the individual which method to choose and use.

Oxidation Number Method:

Like various other reactions, it is very important to write the correct compositions and formulas. A very important thing to keep in mind while writing oxidation-reduction reactions is to write the compositions and formulas of the substances and the products present in the chemical reaction in a very correct manner.

Half-Reaction Method:

In this procedure, we decouple the equation into two halves. After that, we balance both the parts of the reaction separately. Finally, we add them together to get a balanced equation.