Question 1 Report
When Sulphur(IV) oxide (SO2) is passed into a solution of acidified tetraoxomanganate (VII) (KMnO4), it acts as a reducing agent. This reaction involves the reduction of potassium permanganate (KMnO4), which is characterized by a distinctive color change.
The tetraoxomanganate (VII) ion (MnO4-) is purple in color. During the reaction, SO2 gets oxidized while the MnO4- ion is reduced to Mn2+, which is almost colorless or pale pink, depending on the concentration.
Thus, the color of the solution changes from purple to almost colorless as the reaction progresses.
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