Answer:
(i) Both tin and lead show two oxidation states of +2 and +4 due to inert pair effect. But the inert pair effect is more prominent in case of Pb than in Sn. In other words, +2 oxidation state of Sn is less stable than its +4oxidation state. Therefore, Sn \[(II)\] acts as a reducing agent and gets converted into more stable Sn (IV) by losing two electrons. For example, it reduces \[F{{e}^{3+}}\] to \[F{{e}^{2+}}\] ions.
\[2\,F{{e}^{3+}}+S{{n}^{2+}}\xrightarrow{\,}\,2\,F{{e}^{2+}}+S{{n}^{4+}}\]
In contrast, +2 oxidation state of Pb is more stable than its +4 oxidation state. In other words, Pb \[(II)\] does not lose electrons easily and hence does not act as a reducing agent.
(ii) Being a reducing agent, tin \[(II)\] chloride reduces mercuric chloride \[(HgC{{l}_{2}})\] to mercurous chloride \[(H{{g}_{2}}C{{l}_{2}})\]and ferric salts to ferrous salts.
\[SnC{{l}_{2}}+2\,HgC{{l}_{2}}\xrightarrow{\,}\,SnC{{l}_{4}}+H{{g}_{2}}C{{l}_{2}}\]
\[SnC{{l}_{2}}+2\,FeC{{l}_{3}}\xrightarrow{\,}\,SnC{{l}_{4}}+2\,FeC{{l}_{2}}\]
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