Chemistry · Redox Equilibrium · NEET
Look at the oxidation number of sulphur. In SO2, S is +4. Sulphur's full range is -2 (lowest, as in H2S) to +6 (highest, as in SO3 or H2SO4). Since +4 is in the MIDDLE, sulphur still has room to move both ways. If S loses 2 electrons it becomes +6 (so SO2 is oxidised, meaning SO2 was the reducing agent). If S gains 4 electrons it becomes 0 (elemental S, so SO2 is reduced, meaning SO2 was the oxidising agent). An element locked at its highest state, like S in SO3 (+6), can only go down, so it can only oxidise others.
Most of the time SO2 acts as a REDUCING AGENT. This is what NEET tests again and again: SO2 turns acidified KMnO4 (purple) colourless, and turns acidified K2Cr2O7 (orange) green. In both, S goes +4 to +6 and SO2 is oxidised, so it reduced the other reagent. SO2 acts as an oxidiser only against strong reducing agents (for example with H2S, where S in SO2 goes +4 to 0).
With hydrogen sulphide: SO2 + 2H2S gives 3S + 2H2O. Here the S of SO2 goes from +4 down to 0, so SO2 is reduced. Since SO2 got reduced, it made H2S lose electrons, so SO2 is the oxidising agent here. This is the standard example NCERT expects when asked to prove SO2 can also oxidise.
In SO3 and H2SO4 sulphur is already at +6, its highest possible oxidation state. From +6 sulphur has no higher state to move to, so it cannot lose more electrons and cannot be oxidised. That means it cannot act as a reducing agent. It can only gain electrons (go down), so at +6 sulphur is only an oxidising agent. This is exactly why SO2 (+4, middle) is special while SO3 (+6, top) is not.
Both are strong oxidising agents, so they force SO2 to act as a reducer. With acidified KMnO4: the purple colour disappears (Mn +7 goes to +2). With acidified K2Cr2O7: the orange colour turns green (Cr +6 goes to +3, forming green Cr2(SO4)3). In both, sulphur goes +4 to +6. NEET often asks which gas decolourises KMnO4 - the answer is SO2 because it is a good reducing agent.
Which one of the following statements is correct when SO2 is passed through acidified K2Cr2O7 solution?
Name the gas that can readily decolourise acidified KMnO4 solution.
Try the real previous-year questions from this chapter — each with the answer and a full solution.
+4. Each oxygen is -2, so two oxygens give -4; the molecule is neutral, so S must be +4. This middle value is the reason SO2 can act both ways.
Only SO2. Its sulphur is +4 and can rise to +6, so it can be oxidised (reduce others). In SO3 sulphur is already +6 (maximum), so it cannot be oxidised and cannot act as a reducer.
SO2 bleaches by REDUCTION (it removes oxygen / adds hydrogen to the coloured substance). This bleaching is temporary because air slowly re-oxidises the material. This again shows SO2's usual role as a reducing agent.
H2O2 (hydrogen peroxide). In H2O2 oxygen is -1, a middle value between -2 and 0, so it too can go both up and down. NCERT groups SO2 and H2O2 together for this exact reason.
