Due to the inert pair effect, the reluctance of valence $ns^2$ electrons to participate in bonding increases down the group. Therefore, in the carbon family (Group 14), the stability of the $+4$ oxidation state decreases and that of the $+2$ oxidation state increases as we move from $\text{Ge}$ to $\text{Pb}$. Consequently, for Tin ($\text{Sn}$), the $+4$ state is more stable than the $+2$ state, making $\text{Sn}^{2+}$ a reducing agent as it readily oxidises to $\text{Sn}^{4+}$. Conversely, for Lead ($\text{Pb}$), the $+2$ state is more stable than the $+4$ state. Thus, $\text{Pb}^{4+}$ acts as a strong oxidising agent because it readily gets reduced to $\text{Pb}^{2+}$.