Q.The choice of a reducing agent in a particular case depends on thermodynamic factor. How far do you agree with this statement? Support your opinion with two examples.

The choice of reducing agent for a given metal oxide is decided by comparing relative oxide stabilities ($\Delta_fG^\circ$, plotted vs T as the Ellingham diagram) — whichever agent's oxide-formation line is lower at the operating temperature is thermodynamically capable of reducing the other oxide. Example 1: Al reduces $Cr_2O_3$ ($\Delta_rG^\circ = -287$ kJ/mol, thermite process). Example 2: at ~1673 K, carbon (not CO) is used to reduce ZnO, since above ~983 K carbon's Ellingham line is lower than CO's. …

Step 1 – State the underlying principle

Yes, this statement is largely correct: the practical choice of which reducing agent to use for a given metal oxide is fundamentally decided by thermodynamics, specifically by comparing the relative thermodynamic stability of the competing oxides — visualised on the Ellingham diagram ($\Delta_fG^\circ$ vs T for different oxide-formation reactions). At a given temperature, the reducing agent whose own oxide-formation line lies BELOW the target oxide's line is thermodynamically capable of reducing it (its own oxide is more stable, "wins" the oxygen).

Step 2 – Example 1: Al reduces $Cr_2O_3$ (aluminothermic/thermite process)

Given $\Delta_fG^\circ(Cr_2O_3) = -540$ kJ/mol and $\Delta_fG^\circ(Al_2O_3) = -827$ kJ/mol:

Since $Al_2O_3$ formation is much more thermodynamically favourable than $Cr_2O_3$ formation, Al can reduce $Cr_2O_3$ — this is exactly the thermodynamic factor deciding the choice of Al as the reducing agent for chromium extraction (and similarly for iron, in the classic thermite reaction).

Step 3 – Example 2: carbon (not CO) reduces ZnO at high temperature

Below about 983 K, CO is the thermodynamically better reducing agent (its oxide-formation line — for $CO \to CO_2$ — lies lower); above 983 K the C,CO line drops lower instead, making carbon the better reducing agent. Since ZnO must be reduced at a high temperature (~1673 K) to proceed at a reasonable rate/scale, we are above this crossover, and thermodynamics dictates that carbon (coke), not CO, is the effective reducing agent used industrially: $ZnO + C \rightarrow Zn + CO$.

Step 4 – Caveat …

Affirm the statement, then justify it using the Ellingham-diagram logic with two worked examples already established in this chapter (Al/$Cr_2O_3$ via $\Delta_fG^\circ$ comparison; C-vs-CO for ZnO via the ~983 K crossov …

• Agreeing without giving concrete numeric/example support — the question explicitly asks for two examples.
• Overstating the agreement as absolute, ignoring that thermodynamic feasibility is necessary but not always sufficient (economic …