Question:

How does the acidified \( KMnO_4 \) solution react with ferrous sulphate, \( SO_2 \), oxalic acid, and sodium thiosulphate?

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Acidified \( KMnO_4 \) acts as an oxidizing agent, converting reducing agents into their oxidized forms while itself being reduced to \( Mn^{2+} \).
Updated On: Mar 5, 2025
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Solution and Explanation

Step 1: The acidified \( KMnO_4 \) acts as a strong oxidizing agent and reacts with various reducing agents. Step 2: Reactions: (i) With ferrous sulphate: \[ 2 KMnO_4 + 10 FeSO_4 + 8 H_2SO_4 \rightarrow 5 Fe_2(SO_4)_3 + 2 MnSO_4 + K_2SO_4 + 8 H_2O \] (ii) With \( SO_2 \): \[ 2 KMnO_4 + 5 SO_2 + 2 H_2O \rightarrow 2 MnSO_4 + K_2SO_4 + 2 H_2SO_4 \] (iii) With oxalic acid: \[ 2 KMnO_4 + 5 (COOH)_2 + 3 H_2SO_4 \rightarrow 2 MnSO_4 + 10 CO_2 + 8 H_2O + K_2SO_4 \] (iv) With sodium thiosulphate: \[ 2 KMnO_4 + 5 Na_2S_2O_3 + 6 H_2SO_4 \rightarrow 2 MnSO_4 + 5 Na_2SO_4 + 3 S + K_2SO_4 + 3 H_2O \]
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