Q.What happens when sulphur dioxide gas is passed through lime water first slowly and then in excess?

SO₂ can either be oxidised further (acting as a reducing agent, e.g. with Cl₂) or itself oxidise a stronger reductant like H₂S (acting as an oxidising agent) — one reaction of each type illustrates the dual behaviour.

In $SO_2$, sulphur is in the +4 oxidation state, which is intermediate between S's lowest (−2) and highest (+6) common states. This lets $SO_2$ act as either a reducing agent (S going up to +6) or an oxidising agent (S coming down to 0), depending on what it reacts with.

As a reducing agent (S: +4 → +6), e.g. reaction with chlorine water:

$SO_2 + Cl_2 + 2H_2O \rightarrow H_2SO_4 + 2HCl$

As an oxidising agent (S: +4 → 0), e.g. reaction with hydrogen sulphide:

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SO2 passed slowly into lime water gives a white CaSO3 precipitate; passing excess SO2 redissolves it as soluble Ca(HSO3)2.

When SO2 gas is passed slowly (in limited amount) through lime water (a dilute solution of Ca(OH)2), it reacts to form insoluble calcium sulphite, seen as a white turbidity/precipitate:

Ca(OH)2 + SO2 -> CaSO3 (white ppt) + H2O

When SO2 is passed in excess, the excess SO2 (in presence of water) reacts further with the CaSO3 precipitate to form the soluble calcium hydrogen sulphite (calcium bisulphite):

CaSO3 + SO2 + H2O -> Ca(HSO3)2 (soluble)

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A reducing gas that decolourises acidified $KMnO_4$ and is oxidised by $H_2O_2$ to a sulphate (confirmed by an acid-insoluble white $BaSO_4$ precipitate) is sulphur dioxide.

$SO_2$ is a reducing gas: it decolourises acidified $KMnO_4$ solution as it is oxidised (with $Mn^{7+}$ reduced to $Mn^{2+}$, discharging the purple colour):

$5SO_2 + 2KMnO_4 + 2H_2O \rightarrow K_2SO_4 + 2MnSO_4 + 2H_2SO_4$

Boiling the resulting solution with $H_2O_2$ fully oxidises the sulphur to sulphate:

$SO_2 + H_2O_2 \rightarrow H_2SO_4$

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