In acidic solution, potassium permanganate ( $KMn O_{4}$ ) oxidises oxalic acid ( $H_{2} C_{2} O_{4}$ ) according to the equation $2Mn O_{4}^{-} + 5 C_{2} O_{4}^{2 -} + 16 H^{+} \to 2M n^{2 +} + 10C O_{2} + 8 H_{2} O$ . Which of the following species acts as the oxidising agent?

medium MCQ 2 marks Class 10 · Science · Chemical Reactions and Equations

Answer

Mn O_{4}^{-}

The correct answer is $Mn O_{4}^{-}$ — Hence $Mn O_{4}^{-}$ is the oxidising agent.

Step-by-step solution

Identify oxidation numbers: Mn in $Mn O_{4}^{-}$ is +7; in $M n^{2 +}$ it is +2 (reduction).
C in $C_{2} O_{4}^{2 -}$ is +3; in $C O_{2}$ it becomes +4 (oxidation).
The species whose oxidation number decreases (from +7 to +2) gains electrons – it is reduced and therefore acts as the oxidising agent.
Hence $Mn O_{4}^{-}$ is the oxidising agent.

Common mistake: Assuming oxalic acid, being an acid, is the oxidising agent.

Concept tested: Identifying oxidising agent