Question

Standard Gibbs free energies of formation of oxides per mole of O2 at 1000 K are given below. SiO2: -728 kJ, TiO2: -737 kJ, VO: -712 kJ, MnO: -624 kJ. Regarding thermodynamic feasibility of oxide reduction, which of the following statements is or are CORRECT under standard conditions at 1000 K?

• A. Si can reduce TiO2.
• B. Mn can reduce VO.
• C. Ti can reduce MnO.
• D. V can reduce SiO2.

Hint:

Use Ellingham diagram logic: the oxide with more negative Gibbs free energy is more stable. A metal forms its own oxide by reducing another oxide only if its oxide has more negative Gibbs free energy.

Answer 1

The Correct Option is A, C

Step 1: Basic principle.
A metal can reduce another metal oxide if the Gibbs free energy of formation of its oxide is more negative than that of the other oxide. Step 2: Arrange oxides by stability (more negative means more stable).
TiO2: -737 kJ (most stable) SiO2: -728 kJ VO: -712 kJ MnO: -624 kJ (least stable) Step 3: Check each option.
(A) Si reducing TiO2: TiO2 is more stable than SiO2, so Si can reduce TiO2. Correct.
(B) Mn reducing VO: MnO is less stable than VO, so Mn cannot reduce VO. Wrong.
(C) Ti reducing MnO: TiO2 is more stable than MnO, so Ti can reduce MnO. Correct.
(D) V reducing SiO2: VO is less stable than SiO2, so V cannot reduce SiO2. Wrong. Step 4: Conclusion.
Correct statements are (A) and (C). \[ \boxed{\text{Correct: (A) and (C)}} \]