Sulphur dioxide is a stronger reducing agent in alkaline medium than in acidic medium-explain.

Explanation

In acidic and alkaline media, the oxidation half-reactions of `SO_(2)` are as follows:
Acidic medium: `SO_(2)+2H_(2)O-2ehArrH_(2)SO_(4)+2H^(+)` …[1]
Alkaline medium: `SO_(2)+4OH^(-)-2ehArrSO_(4)^(2-)+2H_(2)O` ….[2]
In equation [1], `H^(+)` ions are on the right side of the equilibrium. Therefore, in the acidic medium, the reverse reaction becomes favourable. Thus, the tendency of `SO_(2)` to be oxidised or its reducing property decreases.
On the other hand, in equation [2], `OH^(-)` ions are on the left side of the equilibrium. Hence, in alkaline medium, the forward reaction becomes favourable. Thus the tendency of `SO_(2)` to be oxidised or its reducing property increases. Therefore, `SO_(2)` behaves as a stronger reducing agent in alkaline medium than in acidic medium.