Question

(i) Name the method used for the refining of zirconium.

(ii) What is the role of CO in the extraction of Iron ?

(iii) Reduction of metal oxide to metal becomes easier if the metal obtained is in liquid state. Why ?

Answer 1

(i) We use the van Arkel method of vapour-phase refining for refining Zirconium.

(ii) The main role of CO is to bring about the reduction of oxides of iron to Fe in the zone of reduction in blast furnace.

$3F{e}_2{O}_3\left(s\right)+CO\left(g\right)\stackrel{573-673K}{\to }2F{e}_3{O}_4\left(s\right)+C{O}_2\left(g\right)$

$F{e}_3{O}_4\left(s\right)+CO\left(g\right)\stackrel{773-873K}{\to }3FeO\left(s\right)+C{O}_2\left(g\right)$

$F{e}_2{O}_3\left(s\right)+CO\left(g\right)\stackrel{773-873K}{\to }2FeO\left(s\right)+C{O}_2\left(g\right)$

$FeO\left(s\right)+CO\left(g\right)\stackrel{1123K}{\to }Fe\left(s\right)+C{O}_2\left(g\right)$

(iii) Entropy of metal in liquid state is more compared to the solid state. Therefore, the $\Delta S$ of the reduction process is more positive when the metal formed is in the liquid state and the metal oxide being reduced is in the solid state. Since, the value of $T\Delta S$ increases and that of $\Delta H$ remains constant, therefore, the value of ${\Delta }_r{G}^{\circ }(\Delta G=\Delta H-T\Delta S)$ becomes more -ve and hence the reduction becomes easier.