PbO2 (\rightarrow) PbO ∆G298< 0,... | Chemistry | SolveeIt

Question

PbO2 $\rightarrow$ PbO ∆G298< 0,

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

Pb+4, Sn+2

Pb+2, Sn+2

Pb+2, Sn+4

Answer

Pb+2, Sn+4

Explanation

Solution

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.

Chemistry Question on p -Block Elements

Solution