Potassium permanganate (KMnO4) reacts with sulfur dioxide (SO2) and give manganese sulfate, sulfuric acid and potassium sulfate. Manganese atom in KMnO4 is reduced and sulfur atom in SO2 is oxidized. Purple colour of aqueous KMnO4 solution is reduced to colourless or pale pink colour during the reaction.
Written by: Heshan Nipuna, Eng., Chemical & Process Engineering, University of Peradeniya, last modified: 26/09/2021
Manganese in potassium permanganate is reduced from +7 oxidation state to +2 oxidation state (Mn2+). As well, sulfur in sulfur dioxide is oxidized from +4 oxidation state to +6 oxidation state (SO42-).
Reaction medium should be slightly acidified by adding few drops of sulfuric acid.
Dilute aqueous potassium permanganate solution is a purple colour solution. When colourless sulfur dioxide gas is sent to that solution, purple colour of aqueous solution is decreased and become colourless or pale pink colour solution due to formation of Mn2+ cations.
Sulfur dioxide is a toxic gas and acidic gas. Therefore, be careful to prevent any leakages during the treating sulfur dioxide gas in to the potassium permanganate solution.
Because oxidation number of sulfur and manganese atoms are changed, identify what is the difference of oxidation numbers relevant to oxidation and reduction as below.
Oxidation number difference of manganese atom is 5 and sulfur atom is 2. Then, exchange oxidation number difference as bellow.
Count the total number of oxygen atoms in the left side and right side separately. To balance number oxygen atoms, add 2 H2O molecules to the left.
Count the total number of hydrogen atoms in the left side and right side separately. To balance number hydrogen atoms, add 4 H+ ions to the right side.
Now, we have obtained the balanced ionic equation.
Now, we need to balance the ionic equation.