Ellingham Diagram MCQ - Practice Questions & Answers

Solve by difficulty

EASY

In the context of the Hall - Heroult process for the extraction $Al$ , which of the following statements is false ?

CO and CO₂ are produced in this process

Al₂O₃ is mixed with CaF₂ which lowers the melting point of the mixture and brings conductivity

Al³⁺ is reduced at the cathode to form Al

Na₃AlF₆ serves as the electrolyte

View Solution

Concepts Covered - 0

Ellingham Diagram

Ellingham diagram normally consists of plots of ∆_fGV vs T for the formation of oxides of common metals and reducing agents i.e., for the reaction given below.
$2 \mathrm{xM}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{M}_{\mathrm{x}} \mathrm{O}(\mathrm{s})$
In this reaction, gas is consumed in the formation of oxide hence, molecular randomness decreases in the formation of oxide which leads to a negative value of ∆S as a result sign of T∆S term becomes positive. Subsequently, ∆_fG^o shifts towards higher side despite rising T. The result is a positive slope in the curve for most of the reactions for the formation of M_xO(s).
Each plot is a straight line and slopes upwards except when some change in phase (s→1 or 1→g) takes place. The temperature at which such change occurs is indicated by an increase in the slope on the positive side (e.g., in the Zn, ZnO plot, the melting is indicated by an abrupt change in the curve).
When the temperature is raised, a point is reached in the curve where it crosses ∆_rG^o=0 line. Below this temperature, ∆_rG^o for the formation of oxide is negative so M_xO is stable. Above this point, free energy of formation of oxide is positive. The oxide, MxO will decompose on its own.
Similar diagrams are constructed for sulfides and halides also. From them it becomes clear that why reduction of M_xS is difficult.

Limitations of Ellingham Diagram

It only tells about the thermodynamic feasibility of the reduction of metal oxide by a reducing agent. It does not give any idea about the kinetics of the reaction.
It always considers an equilibrium between the two reactions or consider an equilibrium between reactants and products. But if solids are there then no equilibrium exist.

Reduction of oxide to metal - Smelting

Reduction of oxides of less electropositive metals like Pb, Fe, Zn, Sn and Cu is carried out by heating them with coal or coke in a blast furnace in presence of limited air at a temperature range nearly 200^oC - 1500^oC.

For example:

$\mathrm{ZnO\: +\: C\: \overset{\Delta }{\rightarrow}\: Zn\: +\: CO}$

$\mathrm{PbO}\: +\: \mathrm{C} \stackrel{\Delta}{\longrightarrow} \mathrm{Pb}\: +\: \mathrm{CO}$

$\mathrm{Fe}_{2} \mathrm{O}_{3}+3 \mathrm{C} \longrightarrow 2 \mathrm{Fe}+3 \mathrm{CO}$

In case of tin, the concentrated cassiterite ore is mixed with 1/5th of its weight of powdered anthracite(carbon) and some limestone for hearing in a reverberatory furnace ar 1473-1573K. Here the ore gets reduced to metallic tin and the impurity of silica can be removed as calcium silicate (slag).

$\mathrm{SnO}_{2}+2 \mathrm{C} \longrightarrow \mathrm{Sn}+2 \mathrm{CO}$

$\mathrm{CaCO}_{3}\: \overset{\Delta }{\rightarrow}\mathrm{CaO}+\mathrm{CO}_{2}$

$\mathrm{CaO}+\mathrm{SiO}_{2} \longrightarrow \mathrm{CaSiO}_{3}$

NOTE: Tin obtained from here is 99.5% pure and known as black tin. Here, use of an excess of lime is avoided otherwise calcium stannate will also be formed.

Reduction by a More Electropositive Metal(Thermite Process)

Some meral oxides which are not reduced by carbon, like chromium trioxide (Cr₂O₃), titanium chloride (TiCl₄), manganese oxide Mn₃O₄are reduced by using highly electropositive merals like Na, K, Al, Mg etc. It is also called electrometallurgy.