In view of the signs of Delta_{r} G^{circ} fo | vedclass

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(A) For the reaction $PbO_2 + Pb \longrightarrow 2 PbO,$ the value of $\Delta_{r} G^{\circ} < 0$ indicates that the reaction is spontaneous,meaning $P

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For lead $+2,$ for tin $+4$

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(A) For the reaction $PbO_2 + Pb \longrightarrow 2 PbO,$ the value of $\Delta_{r} G^{\circ} For the reaction $SnO_2 + Sn \longrightarrow 2 SnO,$ the value of $\Delta_{r} G^{\circ} > 0$ indicates that the reaction is non-spontaneous,meaning $SnO_2$ (where $Sn$ is in $+4$ state) is more stable than $SnO$ (where $Sn$ is in $+2$ state). Thus,$+4$ is the more stable and characteristic oxidation state for tin.Therefore,the characteristic oxidation states are $+2$ for lead and $+4$ for tin.

In view of the signs of $\Delta_{r} G^{\circ}$ for the following reactions:
$PbO_2 + Pb \longrightarrow 2 PbO, \Delta_{r} G^{\circ} < 0$
$SnO_2 + Sn \longrightarrow 2 SnO, \Delta_{r} G^{\circ} > 0$
Which oxidation states are more characteristic for lead and tin?

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In view of the signs of $\Delta_{r} G^{\circ}$ for the following reactions:$PbO_2 + Pb \longrightarrow 2 PbO, \Delta_{r} G^{\circ} < 0$$SnO_2 + Sn \longrightarrow 2 SnO, \Delta_{r} G^{\circ} > 0$Which oxidation states are more characteristic for lead and tin?