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

The question concerns the relative stability of oxidation states of lead and tin, and the thermodynamic feasibility of the following reactions:

• \( \mathrm{PbO_2 + Pb \rightarrow 2\,PbO} \)
• \( \mathrm{SnO_2 + Sn \rightarrow 2\,SnO} \) 

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Analysis of Oxidation State Stability

1. Lead:
   • \( \mathrm{Pb^{2+}} \) is more stable than \( \mathrm{Pb^{4+}} \).
   • The conversion of \( \mathrm{Pb^{4+}} \) (in \( \mathrm{PbO_2} \)) to \( \mathrm{Pb^{2+}} \) (in \( \mathrm{PbO} \)) is thermodynamically favorable.
   • Hence, the reaction is spontaneous:

\[ \Delta_r G^\circ (1) < 0 \]

1. Tin:
   • \( \mathrm{Sn^{4+}} \) is more stable than \( \mathrm{Sn^{2+}} \).
   • The conversion of \( \mathrm{Sn^{4+}} \) (in \( \mathrm{SnO_2} \)) to \( \mathrm{Sn^{2+}} \) (in \( \mathrm{SnO} \)) is thermodynamically unfavorable.
   • Hence, the reaction is non-spontaneous:

\[ \Delta_r G^\circ (2) > 0 \]

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Final Conclusion:

\[ \boxed{\Delta_r G^\circ (1) < 0 \;;\; \Delta_r G^\circ (2) > 0} \]