Question

Consider the following changes : $\text{SnO}_2 \rightarrow \text{SnO}$ $\Delta G_1^o$ and $\text{PbO}_2 \rightarrow \text{PbO}$ $\Delta G_2^o$. Select the correct option

• A. $\Delta G_1^o>0$, $\Delta G_2^o>0$
• B. $\Delta G_1^o>0$, $\Delta G_2^o<0$
• C. $\Delta G_1^o<0$, $\Delta G_2^o<0$
• D. $\Delta G_1^o<0$, $\Delta G_2^o>0$

Hint:

The Inert Pair Effect causes the stability of the lower oxidation state (Group number $-2$) to increase down Groups 13-16. For Group 14, $\text{Sn}^{4+}$ is stable, while $\text{Pb}^{2+}$ is highly stable. Spontaneous reactions have $\Delta G^o<0$.

Answer 1

The Correct Option is B

The given question involves evaluating the changes from $\text{SnO}_2$ to $\text{SnO}$ and from $\text{PbO}_2$ to $\text{PbO}$ in terms of their standard Gibbs free energy change, denoted as $\Delta G^o$. We need to determine which of these changes have positive or negative $\Delta G^o$ values based on thermodynamic stability and reduction potential considerations. 

1. Understanding the Stability:
   • \(\text{SnO}_2\) and \(\text{PbO}_2\) are both oxides. However, their relative stability and the energies involved in converting these into their respective lower oxides differ significantly.
   • Tin(II) oxide (\(\text{SnO}\)) is less stable compared to Tin(IV) oxide (\(\text{SnO}_2\)). Thus, the conversion from \(\text{SnO}_2\) to \(\text{SnO}\) is less favorable thermodynamically, implying that \(\Delta G_1^o\) is positive (\(\Delta G_1^o>0\)).
   • On the other hand, Lead(IV) oxide (\(\text{PbO}_2\)) is less stable compared to Lead(II) oxide (\(\text{PbO}\)). For lead, the reduction from \(\text{PbO}_2\) to \(\text{PbO}\) is favorable, indicating \(\Delta G_2^o\) is negative (\(\Delta G_2^o<0\)).
2. Conclusion:
   • Based on the stability and thermodynamic feasibility of the changes, the correct answer is: $\Delta G_1^o > 0$, $\Delta G_2^o < 0$.