Answer:
The reducing nature of \[S{{O}_{2}}\]
in the alkaline medium is shown by the following equation.
\[S{{O}_{2}}+2O{{H}^{-}}\,S{{O}_{4}}^{2-}\,+2{{H}^{+}}\,+2{{e}^{-}}\]
The addition of acid \[({{H}^{+}})\]favours
the reverse reaction. Therefore, sulphur dioxide fails to act as reducing agent
in the acidic medium. In the alkaline medium, \[{{H}^{+}}\] ions formed as the
products are removed. This favours the forward reaction. As a result, \[S{{O}_{2}}\]acts
as a reducing agent in the alkaline medium.
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