Acidified \(KMnO_4\) oxidizes sulphite to:
\(SO_3^{2-} \)
\(SO_4^{2-} \)
\(SO_2(g) \)
\(S_2O_8^{2-} \)
Solution:
The reaction between acidified potassium permanganate (KMnO4) and sulphite (SO32-) occurs as an oxidation-reduction (redox) process. In this reaction, KMnO4 acts as an oxidizing agent, converting the sulphite ion to another form.
The oxidation state of sulphur in SO32- is +4. The goal of the oxidation process is to increase the oxidation state of sulphur. Acidified KMnO4 effectively oxidizes sulphite ions to peroxodisulfate ions (S2O82-). This involves the formation of a compound where two sulphur atoms are bonded via an oxygen-oxygen bond, and each sulphur atom is in the +6 oxidation state.
The balanced redox reaction is:
2KMnO4 + 5SO32- + 6H+ → 2Mn2+ + 5SO42- + 3H2O
Correct answer:
S2O82-
Convert Propanoic acid to Ethane
Which element is a strong reducing agent in +2 oxidation state and why?

A ladder of fixed length \( h \) is to be placed along the wall such that it is free to move along the height of the wall.
Based upon the above information, answer the following questions:
(iii) (b) If the foot of the ladder, whose length is 5 m, is being pulled towards the wall such that the rate of decrease of distance \( y \) is \( 2 \, \text{m/s} \), then at what rate is the height on the wall \( x \) increasing when the foot of the ladder is 3 m away from the wall?