It is noticed that $$Pb^{2+}$$ is more stable than $$Pb^{4+}$$ but $$Sn^{2+}$$ is less stable than $$Sn^{4+}$$
Observe the following reactions.

$$ PbO_2+Pb\rightarrow 2PbO;\triangle_rG^{o}(1) $$
$$ SnO_2+Sn\rightarrow 2SnO;\triangle_rG^{o}(2) $$

Identify the correct set from the following

For any reaction under standard conditions, the criterion for spontaneity is given by the sign of the standard Gibbs free energy change:

$$\Delta_{r}G^{o} \; < \; 0\quad\text{(reaction is spontaneous)}$$
$$\Delta_{r}G^{o} \; > \; 0\quad\text{(reaction is non-spontaneous)}$$

Case 1: Reaction $$PbO_{2} + Pb \;\longrightarrow\; 2\,PbO$$

Oxidation states involved are $$Pb(+4)$$ in $$PbO_{2}$$, $$Pb(0)$$, and $$Pb(+2)$$ in $$PbO$$. It is given that $$Pb^{2+}$$ is more stable than both $$Pb^{4+}$$ and $$Pb^{0}$$. Therefore the reaction that converts $$Pb(+4)$$ and $$Pb(0)$$ into the more stable $$Pb(+2)$$ will proceed spontaneously.

Hence for reaction (1): $$\Delta_{r}G^{o}(1) \; < \; 0$$.

Case 2: Reaction $$SnO_{2} + Sn \;\longrightarrow\; 2\,SnO$$

Oxidation states involved are $$Sn(+4)$$ in $$SnO_{2}$$, $$Sn(0)$$, and $$Sn(+2)$$ in $$SnO$$. It is given that $$Sn^{4+}$$ is more stable than both $$Sn^{2+}$$ and $$Sn^{0}$$. Thus the reverse reaction

$$2\,SnO \;\longrightarrow\; SnO_{2} + Sn$$

is spontaneous, implying the forward reaction is non-spontaneous.

Therefore for reaction (2): $$\Delta_{r}G^{o}(2) \; > \; 0$$.

Thus the correct signs are $$\Delta_{r}G^{o}(1)\; < \;0$$ and $$\Delta_{r}G^{o}(2)\; > \;0$$. This corresponds to Option B.