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-
Complete and balance the following chemical equations: (a) \[ 2MnO_4^-(aq) + 10I^-(aq) + 16H^+(aq) \rightarrow \] (b) \[ Cr_2O_7^{2-}(aq) + 6Fe^{2+}(aq) + 14H^+(aq) \rightarrow \]