Q. 554.5( 2 Votes )

Explain the following :

(a) CO2 is a better reducing agent below 710K whereas CO is a better reducing agent above 710K.

(b) Generally sulphide ores are converted into oxides before reduction.

(c) Silica is added to the sulphide ore of copper in the reverberatory furnace.

(d) Carbon and hydrogen are not used as reducing agents at high temperatures.

(e) Vapour phase refining method is used for the purification of Ti.

Answer :

(a) At temperatures below 710K, ∆G°(C,CO2) <∆G°(C,CO), so CO2 is a better reducing agent than CO but for temperatures above 710K, ∆G°(C,CO) <∆G°(C,CO2) so CO is a better reducing agent. The lower the value of ∆G°it is better because the difference will then be large and negative which makes the reaction more favourable.

(b) Reduction of sulphide ore is not easy, so it is usually converted into oxide and then reduced. Also, in Ellingham diagram of sulphides, ∆fG of MxS is not compensated and reduction becomes tougher.


(c) Silica is added to the sulphide ore of copper to remove iron impurities as slag. Copper is formed as copper matte which contains Cu2S and FeS.


FeO + SiO2 FeSiO3(slag)


(d) Carbon and hydrogen undergo the following reactions with metal oxides


2Al2O3 + 3C Al4C3 + 3O2


2Al2O3 + 6H2 4AlH3 + 3O2


Carbon and hydrogen will not reduce oxides to metals but they will instead form hydrides and carbides.Therefore, they are not used as reducing agents.


(e) In vapour phase refining method, the impure metal is converted to a volatile complex which on decomposition gives the pure metal.


Van Arkel’s process is used to purify metals like Zr and Ti. The impure metal is heated with Iodine and it forms a volatile complex TiI4 which on decomposition gives pure Ti.


Ti(impure) + 2I2TiI4


TiI4Ti(pure) + 2I2 (1800K)


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