Question

PbO₂ \(\rightarrow\) PbO                            ∆G₂₉₈< 0,

SnO₂ \(\rightarrow\) SnO                            ∆G₂₉₈ > 0,
Most probable oxidation state of Pb & Sn will be

• A. Pb⁺⁴, Sn⁺²
• B. Pb⁺⁴, Sn⁺²
• C. Pb⁺², Sn⁺²
• D. Pb⁺², Sn⁺⁴

Answer 1

The Correct Option is D

The correct option is (D): Pb⁺², Sn⁺⁴.
The most probable oxidation states of Pb and Sn are Pb²⁺ and Sn⁴⁺ because PbO₂ \(\rightarrow\) PbO has a negative change in Gibbs free energy (∆G₂₉₈ < 0), indicating a reduction reaction where Pb goes from a higher oxidation state to a lower one. Similarly, SnO₂ \(\rightarrow\) SnO has a positive (∆G₂₉₈ > 0), indicating an oxidation reaction where Sn goes from a lower oxidation state to a higher one.