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Which of the following observations based on the Ellingham diagram \mathrm{\left ( \Delta G^{0} < 0\right )} is not correct?

Option: 1

A metal can reduce the oxide of another metal that lies above it in the Ellingham diagram.


Option: 2

\mathrm{CO} is more effective than \mathrm{C} as a reducing agent below \mathrm{170^{0}C.}.


Option: 3

\mathrm{\Delta G^{0}} of metal oxide is higher than that of \mathrm{CO}, hence oxidation of metal sulfides to oxides is not favorable.


Option: 4

There is a need for the conversion of metal sulfide to metal oxide before reduction, which can be explained thermodynamically.


Answers (1)

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The Ellingham diagram is a graphical representation of the thermodynamic stability of metal oxides as a function of temperature. It helps to predict the feasibility of a particular redox reaction based on the standard Gibbs free energy change\mathrm{\left ( \Delta G^{0} \right )} values.

A metal can reduce the oxide of another metal that lies above it in the Ellingham diagram: This statement is correct. If a metal lies above another metal in the Ellingham diagram, it means that the standard Gibbs free energy change for the formation of its oxide is more negative, indicating greater thermodynamic stability. Therefore, the metal oxide can be reduced by the metal below it.

\mathrm{CO} is more effective than \mathrm{C} as a reducing agent below \mathrm{170^{0}C} : This statement is also correct. At lower temperatures, \mathrm{CO} is a stronger reducing agent compared to \mathrm{C} (carbon). This is because the formation of \mathrm{\mathrm{CO}_2} (carbon dioxide) from \mathrm{CO} is thermodynamically favored at lower temperatures.

\mathrm{\Delta G^{0}} of metal oxide is higher than that of \mathrm{CO} , hence oxidation of metal sulfides to oxides is not favorable: This statement is incorrect. If the \mathrm{\Delta G^{0}} of metal oxide is higher than that of \mathrm{CO}, it means that the formation of metal oxide from the metal and oxygen is less thermodynamically favorable than the formation of \mathrm{CO}_2 from \mathrm{CO} and oxygen. Therefore, the oxidation of metal sulfides to oxides would be favorable in the presence of \mathrm{CO} .

There is a need for the conversion of metal sulfide to metal oxide before reduction, which can be explained thermodynamically: This statement is correct. The thermodynamic stability of metal sulfides is generally lower than that of metal oxides. Therefore, to facilitate the reduction of metal sulfides, it is often necessary to convert them into metal oxides first, which can be achieved through oxidation. This conversion can be explained by the thermodynamic considerations based on the Ellingham diagram.

Therefore, the incorrect observation is 3. \mathrm{\Delta G^{0}} of metal oxide is higher than that of \mathrm{CO} , hence

oxidation of metal sulfides to oxides is not favourable.

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Gaurav

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