Gibbs energy vs T plot for the formation of oxides is given below.
For the given diagram, the correct statement is -

Solution

Chemistry ma’am · 3 steps

At 600°C, a reducer works only if its line lies below the metal oxide's line.

1. The rule. Each line is Δ G° for forming an oxide, per mole of O₂. Subtract one line from the other to get Δ G° for the reduction. It is negative only when the reducer's line lies below the metal oxide's line at that temperature.

   Option	Reducer line at 600°C	Metal oxide line at 600°C	Reduction possible?
   (a) C reduces FeO	C + O₂ → CO₂: about -340	FeO: about -250	Yes, C line is below
   (b) C reduces ZnO	C + O₂ → CO₂: about -340	ZnO: about -515	No, C line is above
   (c) CO reduces ZnO	2CO + O₂ → 2CO₂: about -260	ZnO: about -515	No, CO line is above
   (d) CO cannot reduce FeO	2CO + O₂ → 2CO₂: about -260	FeO: about -250	CO can reduce it, so the statement is false

2. Check option (a). Subtract the FeO line from the carbon line to get 2FeO + C → 2Fe + CO₂. Its Δ G° is negative, so the reaction is spontaneous.

   Δ G° ≈ (-340) - (-250) = -90 kJ mol⁻¹

3. So. At 600°C, carbon can reduce FeO, so (a) is the correct statement.

The idea to keep. On an Ellingham diagram, the element whose line lies lower at a given temperature can reduce the oxide of any line above it. Always read both lines at the same temperature.

Answer from the NTA final answer key. Solution by PrepWiser.