Question

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. 
$PbO_2 + Pb \to 2PbO ; \Delta_rG^\circ(1)$ 
$SnO_2 + Sn \to 2SnO ; \Delta_rG^\circ(2)$ 
Identify the correct set from the following

• A. $\Delta_rG^\circ(1)>0 ; \Delta_rG^\circ(2)<0$
• B. $\Delta_rG^\circ(1)<0 ; \Delta_rG^\circ(2)>0$
• C. $\Delta_rG^\circ(1)>0 ; \Delta_rG^\circ(2)>0$
• D. $\Delta_rG^\circ(1)<0 ; \Delta_rG^\circ(2)<0$

Hint:

The stability of the lower oxidation state (+2) increases down group 14 due to the Inert Pair Effect.

Answer 1

The Correct Option is B

Reaction 1: $Pb(+4) + Pb(0) \to Pb(+2)$. Since $Pb^{2+}$ is more stable than $Pb^{4+}$, the reaction proceeds forward spontaneously. Thus $\Delta G^\circ(1)<0$.
Reaction 2: $Sn(+4) + Sn(0) \to Sn(+2)$. Since $Sn^{4+}$ is more stable than $Sn^{2+}$, the reaction is non-spontaneous in the forward direction. Thus $\Delta G^\circ(2)>0$.