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

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

Which oxidation states are more characteristic for lead and tin ?

$\stackrel{+4}{\mathrm{Pb}}{\mathrm{O}}_2+\mathrm{Pb}\to 2\stackrel{+2}{\mathrm{Pb}}\mathrm{O}$

Free energy of the reaction is negative which implies that reaction is spontaneous. Lead (Pb) is more stable in +2 oxidation state than +4 due to inert pair effect, therefore the formation of PbO from PbO₂ is spontaneous. The characteristic state of Pb is +2.

$\stackrel{+4}{\mathrm{Sn}}{\mathrm{O}}_2+\mathrm{Sn}\to \stackrel{+2}{\mathrm{Sn}}\mathrm{O}$

Free energy of the reaction is positive which implies that reaction is non-spontaneous. Therefore, the characteristic oxidation state of Sn is +4.