Explain the following:

A $C{O}_2$ 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 $Ti$.

$\text{Comparing reducing property of}CO\text{and}C{O}_2\text{below}710K.$

From the graph of Gibbs energy $\left(\Delta G^0\right)$ vs temperature $\left(T\right)$, given in the figure, we clearly observe:

At temperature below $710K$, the $C$, $C{O}_2$ line comes below the $C,CO$ line.

$\Delta G^0(C,C{O}_2)<\Delta G^0(C,CO)$

So, in this range $C{O}_2$ is a better reducing agent.


$\text{Comparing reducing property of}CO\text{and}C{O}_2\text{above}710K.$

From the graph of Gibbs energy $\left(\Delta G^0\right)$ vs temperature $\left(T\right)$, given in the figure, we clearly observe:

At temperature above $710K$, the $C,CO$ lines come below the $C,C{O}_2$ line.

$\Delta G^0(C,CO)<\Delta G^0(C,C{O}_2)$

So, in this range $CO$ is a better reducing agent.


In the graph of $\Delta G^0$ vs T for the formation of oxides.

The $C{u}_{2}O$ line is almost at the top.

So, it is quite easy to reduce oxide ores of copper.


Therefore, sulphide ores are converted to oxide before reduction.

The sulphide ores are roasted/smelted to give oxides:

$2C{u}_{2}S+3O_2\to 2C{u}_{2}O+2S{O}_2$

The oxide can then be easily reduced to metallic copper using coke:

$C{u}_{2}O+C\to 2Cu+C$


The sulphide ores of copper also contain iron sulphides as impurities.

Silica is added to the sulphide ore of copper in the reverberatory furnace to remove the iron oxide. In the furnace, iron oxide ‘slags off’ as iron silicate.

$FeO+Si{O}_2\to FeSi{O}_3$ (Slag)


Carbon and hydrogen are not used as reducing agents at high temperature because at high temperature carbon and hydrogen react with metals and form carbides and hydrides respectively.

$CaO+3C\to Ca{C}_2+CO$ (high temperature)

$CaO+H_2\to Ca{H}_2+CO$ (high temperature)


Vapour phase refining method is used for the purification of $Ti$.

In this process, the crude metal is heated in an evacuated vessel with iodine. The metal iodide being more covalent, volatilizes:

$Ti+2I_2\to Ti{I}_4$

The metal iodide is decomposed on a tungsten filament, when it is electrically heated to about $1800K$.

The pure metal then deposits on the filament.

$Ti{I}_4\to Ti+2I_2$