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  • Explain the following: (i) CO_{2} is a better reducing agent below 710K whereas CO is a better reducing agent above 710K. (ii) Generally sulphide ores are converted into oxides before reduction.

Explain the following:

(i) CO_{2} is a better reducing agent below 710K whereas CO is a better reducing agent above 710K.
(ii) Generally sulphide ores are converted into oxides before reduction.
(iii) Silica is added to the sulphide ore of copper in the reverberatory furnace.
(iv) Carbon and hydrogen are not used as reducing agents at high temperatures.
(v) Vapour phase refining method is used for the purification of Ti.

Answers (1)

(i) As per the Ellingham diagram, below 710 K

\Delta G_{C,CO_{2}}^{0}<\Delta G_{C,CO}^{0}

Hence, CO_{2} is better reducing agent compared to CO.

At temperature above 710 K, \Delta G_{C,CO_{2}}^{0}<\Delta G_{C,CO}^{0}

Hence CO is better reducing agent.

(ii) Sulphide ores cannot be reduced easily but oxide ores can be easily reduced compared to them, hence sulphide ores are generally converted into oxides before reduction 

(iii) In addition to copper sulphide, copper pyrites contain iron sulphide.  In reverberatory furnace copper ore is roasted to give oxides.

FeO is removed by adding silica from the matte containing copper sulphide and some iron sulphide.  

                                     2FeS+3O_{2}\rightarrow 2FeO+2SO_{2}

                                      FeO+SiO_{2}\rightarrow FeSiO_{3}

(iv) Carbon and hydrogen react with metals at high temperature to form carbides and hydrides respectively hence they are not used as reducing agents.

(v) Ti reacts with iodine to form TiI_{4} which is volatile and decomposes to give Ti at high temperature to give extra pure titanium. That is why vapour phase refining method is used for purification of Ti

                                            Ti + 2I_{2}\overset{530 K}{\rightarrow}TiI_{4}\overset{1800 K}{\rightarrow}Ti+2I_{2}

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