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Thermodynamics question

2001 · Q105
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Thermodynamics question

2001 · Q105

NEETChemistryThermodynamicsMCQ+4 / −1
PbO2  →\to→ PbO; Δ\DeltaΔG298 < 0
SnO2 →\to→ SnO;  Δ\DeltaΔG298 > 0

Most probable oxidation state of Pb and Sn will be
  1. A
    Pb4+, Sn4+
  2. B
    Pb4+, Sn2+
  3. C
    Pb2+, Sn2+
  4. D
    Pb2+, Sn4+
View written solutionFree

Correct answer: D

PbO2 → PbO

Pb has +4 oxidation state In PbO2.

Pb has +2 oxidation state In PbO.

Here Δ\Delta ΔG is negative that is why reaction is spontaneous and Pb4+ reduces Pb2+ thus, Pb2+ is most probable oxidation state of Pb.

SnO2 →\to→ SnO;

Sn has +4 oxidation state In SnO2.

Sn has +2 oxidation state In SnO.

Δ\Delta ΔG is positive that is why reaction is not spontaneous thus, +4 oxidation state of Sn is more stable as it does not change to +2 oxidation state spontaneously.

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