KCET · Chemistry · d and f Block Elements
In the reaction between moist \(\mathrm{SO}_2\) and acidified permanganate solution.
- A \(\mathrm{SO}_2\) is oxidised to \(\mathrm{SO}_4^{2-}\).
\(\mathrm{MnO}_4^{-}\)is reduced to \(\mathrm{Mn}^{2+}\). - B \(\mathrm{SO}_2\) is reduced to S .
\(\mathrm{MnO}_4^{-}\)is oxidised to \(\mathrm{MnO}_4\). - C \(\mathrm{SO}_2\) is oxidised to \(\mathrm{SO}_3^{2-}\).
\(\mathrm{MnO}_4^{-}\)is reduced to \(\mathrm{MnO}_2\). - D \(\mathrm{SO}_2\) is reduced to \(\mathrm{H}_2 \mathrm{~S}\).
\(\mathrm{MnO}_4^{-}\)is oxidised to \(\mathrm{MnO}_4\).
Answer & Solution
Correct Answer
(A) \(\mathrm{SO}_2\) is oxidised to \(\mathrm{SO}_4^{2-}\).
\(\mathrm{MnO}_4^{-}\)is reduced to \(\mathrm{Mn}^{2+}\).
Step-by-step Solution
Detailed explanation
(a) The reaction between moist \(\mathrm{SO}_2\) and acidified permanganate solution is as follows

Thus, from the reaction it is clear that \(\mathrm{SO}_2\) is oxidised to \(\mathrm{SO}_4^{2-}\) while \(\mathrm{MnO}_4^{-}\)is reduced to \(\mathrm{Mn}^{2+}\).

Thus, from the reaction it is clear that \(\mathrm{SO}_2\) is oxidised to \(\mathrm{SO}_4^{2-}\) while \(\mathrm{MnO}_4^{-}\)is reduced to \(\mathrm{Mn}^{2+}\).
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