(A) Reduction of a metal oxide is easier if the metal formed is in liquid state at the temperature of reduction.
(R) $∆$ G° for the net reaction, reduction of metal oxide with the reductant, is more -ve when the metal formed is in molten state

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Both (A) and (R) are true and (R) is the correct explanation of (A)

Both (A) and (R) are true but (R) is not the correct explanation of (A)

(A) is true but (R) is false

Both (A) and (R) are false

answer is A.

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Detailed Solution

Reduction of a metal oxide is easier if the metal is formed in a liquid state at the temperatures of the reduction because the entropy is higher if the metal is in a liquid state.

Since the value of T△S increases and that of △H remains constant, therefore, the value of △_rG∘(△G=△H−T△S) becomes more −ve and hence the reduction becomes easier.

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