To address the problem, we must identify the product resulting from the oxidation of sulphite by acidified KMnO4.
1. Reaction Analysis:
In acidic conditions, potassium permanganate (KMnO4) acts as a potent oxidizing agent. It converts sulphite ions (SO32-) into sulphate ions (SO42-).
2. Sulphite Oxidation Process:
The sulphite ion (SO32-) is oxidized by KMnO4 in an acidic environment. The manganese within KMnO4 is reduced from a +7 oxidation state to a +2 oxidation state, yielding Mn2+.
3. Balanced Chemical Equation:
The balanced redox reaction in an acidic solution is as follows:
2 KMnO4 + 5 SO32- + 4 H2O → 2 Mn2+ + 5 SO42- + 4 OH-
Final Determination:
Acidified KMnO4 converts sulphite (SO32-) to sulphate (SO42-).
Acidified \(KMnO_4\) oxidizes sulphite to:
Complete and balance the following chemical equations: (a) \[ 2MnO_4^-(aq) + 10I^-(aq) + 16H^+(aq) \rightarrow \] (b) \[ Cr_2O_7^{2-}(aq) + 6Fe^{2+}(aq) + 14H^+(aq) \rightarrow \]
Which element is a strong reducing agent in +2 oxidation state and why?