In view of the signs of ${∆}_{\mathrm{r}}{\mathrm{G}}^0$ for the following reactions 

$$\begin{gathered} {\mathrm{PbO}}_2+\mathrm{Pb} \to 2\mathrm{PbO}, {∆}_{\mathrm{r}}{\mathrm{G}}^0 < 0 \\ {\mathrm{SnO}}_2+\mathrm{Sn} \to 2\mathrm{SnO}, {∆}_{\mathrm{r}}{\mathrm{G}}^0 > 0 \end{gathered}$$

Which oxidation states are more characteristic for lead and tin ?

Solution:

Oxidation state

$$\begin{gathered} {\mathrm{PbO}}_2 + \mathrm{Pb} \stackrel{ }{\to } 2\mathrm{PbO} \\ +4 0 +2 \end{gathered}$$

Since $∆{\mathrm{G}}^0<0, \mathrm{i}.\mathrm{e}.,$ it is negative. Therefore, the reaction is spontaneous in the forward direction. This suggest that ${\mathrm{Pb}}^{2+}$ is more stable than ${\mathrm{Pb}}^{4+}.$

Oxidation state

$$\begin{gathered} {\mathrm{SnO}}_2 + \mathrm{Sn} \stackrel{ }{\to } 2\mathrm{SnO} \\ +4 0 +2 \end{gathered}$$

Since $∆{\mathrm{G}}^0>0, \mathrm{i}.\mathrm{e}.,$ it is positive, therefore, the reaction is non-spontaneous in the forward direction. But it will be spontaneous in the backward direction. This suggests that ${\mathrm{Sn}}^{4+}$. These facts are also supported by the inert pair effect down the group.