Statement-1: The reduction of a metal oxide is easier if the metal formed is in liquid state at the temperature of reduction.
Statement-2: The value of entropy change $Δ S$ of the reduction process is more on +ve side when the metal formed is in liquid state and the metal oxide being reduced is in solid state. Thus. the value of $ΔG$ becomes more on negative side.

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Statement-1 is False, Statement-2 is True.

Statement-1 is True, Statement-2 is True, Statement-2
is NOT a correct explanation for Statement-1.

Statement-1 is True, Statement-2 is True; Statement-2
is a correct explanation for Statement-1.

Statement-1 is True, Statement-2 is False.

answer is A.

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Detailed Solution

If the metal generated is liquid at the temperature of reduction, reducing a metal oxide will be simpler. Because when the metal created is in a liquid state and the metal oxide being reduced is in a solid state, the value of entropy change S of the reduction process is greater on the positive side. As a result, the value of G shifts further toward the negative.

Hence, option A is correct.

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