In view of the signs of ∆rG0 for the following reactions PbO2+Pb → 2PbO, ∆rG0 < 0 SnO2+Sn → 2SnO, ∆rG0 > 0Which oxidation states are more characteristic for lead and tin ?

Oxidation statePbO2     +     Pb     →    2PbO +4                    0                     +2 Since ∆G0<0, i.e., it is negative. Therefore, the reaction is spontaneous in the forward direction. This suggest that Pb2+ is more stable than Pb4+.Oxidation stateSnO2     +     Sn     →    2SnO +4                    0                     +2 Since ∆G0>0, i.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 Sn4+. These facts are also supported by the inert pair effect down the group.