Question

Explain the following:

Pb⁴⁺ acts as an oxidising agent but Sn²⁺ acts as a reducing agent.

Answer 1

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.

\[\ce{Pb^{4+} + 2e– -> -Pb^{2+}}\]