Explain the following:
$(a)$ $CO_2$ is a better reducing agent below $710 \ K$ whereas $CO$ is a better reducing agent above $710 \ K$.
$(b)$ Generally,sulphide ores are converted into oxides before reduction.
$(c)$ Silica $(SiO_2)$ 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 for certain metals.
$(e)$ Vapour phase refining method is used for the purification of $Ti$.

(N/A) According to the Ellingham diagram,at temperatures below $710 \ K$,the formation of $CO_2$ is more spontaneous than $CO$ ($\Delta_{f} G^{\ominus}$ is more negative). Above $710 \ K$,the formation of $CO$ becomes more spontaneous than $CO_2$,making $CO$ a better reducing agent.
$(b)$ Sulphide ores are difficult to reduce directly to metal. It is thermodynamically easier to reduce metal oxides than metal sulphides using carbon or other reducing agents.
$(c)$ Copper ore contains iron as an impurity. Silica $(SiO_2)$ is added as a flux to remove iron impurity as iron silicate slag $(FeSiO_3)$. $FeO SiO_2 \rightarrow FeSiO_3$ (slag).
$(d)$ At very high temperatures,carbon and hydrogen react with metals to form carbides and hydrides,respectively,which makes them unsuitable as reducing agents.
$(e)$ $Ti$ is purified by the Van Arkel method. $Ti$ reacts with iodine to form volatile $TiI_4$,which is then thermally decomposed at $1800 \ K$ to obtain pure $Ti$ metal: $Ti 2I_2$ $\rightarrow TiI_4$ $\xrightarrow{1800 \ K} Ti 2I_2$.