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Why is the reduction of a metal oxide easier if the metal formed is in liquid state at the temperature of reduction?

Why is the reduction of a metal oxide easier if the metal formed is in liquid state at the temperature of reduction?

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4 Answers

Aarti Gupta
askIITians Faculty 300 Points
9 years ago
The entropy of a metal is higher in its liquid state than in its solid state. Therefore,
entropy change, ∆S of the reduction process is more on the positive side when the metal formed in liquid state and metal oxide being reduced is in the solid state. As a result, the value of ∆G becomes more on negative side and therefore, reduction becomes easier.
Kratika sharma
19 Points
6 years ago
The entropy of metal is more in its liquid state as compared to solid state . As a result the delta G in more negative , ans the reaction is more feasible
Moyukh Chakrabarti
24 Points
6 years ago
The entropy of a metal is higher in its liquid state than in its solid state. Therefore,entropy change, ∆S of the reduction process is more on the positive side when the metal formed in liquid state and metal oxide being reduced is in the solid state. As a result, the value of ∆G becomes more on negative side and therefore, reduction becomes easier.
The entropy of metal is more in its liquid state as compared to solid state . As a result the delta G in more negative , ans the reaction is more feasible
Kushagra Madhukar
askIITians Faculty 628 Points
3 years ago
Dear student,
Please find the answer to your question.
 
The entropy of a metal is higher in its liquid state than in its solid state. Therefore, entropy change, ∆S of the reduction process is more on the positive side when the metal formed in liquid state and metal oxide being reduced is in the solid state. As a result, the value of ∆G becomes more on negative side and therefore, reduction becomes easier.
 
Thanks and regards,
Kushagra

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