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Question
Explain the following:
Pb4+ acts as an oxidising agent but Sn2+ acts as a reducing agent.
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Solution
Inert pair effect is less prominent in Sn than in Pb. Therefore, +2 oxidation of Sn is less stable than its +4 oxidation state. In other words, Sn2+ can easily lose two electrons to form Sn4+ and hence Sn2+ acts as a reducing agent.
\[\ce{Sn^{2+} -> Sn^{4+} + 2e}\]
In contrast, the inert pair effect is more prominent in Pb than in Sn. Therefore, +2 oxidation state of Pb is more stable than its +4 oxidation state. In other words, Pb4+ can easily lose two electrons to form Pb2+ and hence Pb4+ acts as an oxidising agent.
\[\ce{Pb^{4+} + 2e– -> -Pb^{2+}}\]
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