Question

In the metallurgy of aluminium:
(A)- $A{{l}^{3+}}$ is oxidized to Al(s)
(B)- graphite anode is oxidized to carbon monoxide and carbon dioxide
(C)- oxidation state of oxygen changes in the reaction at anode
(D)- oxidation state of oxygen changes in the overall reaction involved in the reaction

Answer 1

Hint: The process of extraction of metals by electrolysis of their salts is called electrometallurgy. Aluminium metal is extracted from electrolytic reduction of alumina ($A{{l}_{2}}{{O}_{3}}$). Loss of electrons by a specie results in an increase in its oxidation state and vice-versa


Complete step by step solution:
In the process of extraction of aluminum, purified$A{{l}_{2}}{{O}_{3}}$ is mixed with cryolite ($N{{a}_{3}}Al{{F}_{6}}$) and fluorspar ($Ca{{F}_{2}}$) to form a molten mass which is covered with a layer of powdered coke.
The molten matrix is carried out in an electrolytic cell made of iron maintained around 1173 K. This process of extraction of Al is known as Hall and Heroult’s process.
Graphite is used as anode which is useful in the reduction of alumina oxide to the metal and steel is used as cathode.
The reactions taking place at anode and cathode during the electrolytic reduction of alumina are given below:
At cathode: \[2A{{l}_{2}}{{O}_{3}}\to 4Al+3{{O}_{2}}\] or $A{{l}^{3+}}(melt)+3{{e}^{-}}\to Al$
Oxidation state of $A{{l}^{3+}}$ is decreasing from +3 to zero in the elemental state as it gains 3 electrons at cathode. We can see that $A{{l}^{3+}}$ is not oxidized but reduced to Al metal. Therefore, option A is not the correct answer.
At anode: \[\begin{align}
  & C(s)+{{O}^{2-}}(melt)\to CO(g)+2{{e}^{-}} \\
 & C(s)+2{{O}^{2-}}(melt)\to C{{O}_{2}}(g)+4{{e}^{-}} \\
\end{align}\]
It is clear that graphite (carbon electrode) reacts with the oxygen released at anode to form carbon dioxide ($C{{O}_{2}}$) and carbon monoxide (CO).
The overall reaction may be written as
     \[2A{{l}_{2}}{{O}_{3}}+3C\to 4Al+3C{{O}_{2}}\]
Oxidation state of oxygen in CO and $C{{O}_{2}}$ is -2. Since the oxidation state of oxygen does not undergo any change during the reaction at anode or in the overall reaction, neither option (C) or (D) can be correct.

Hence, the correct option is (B).

Note: Points to be noted here are that oxidation (loss of electrons) occurs at anode and reduction (gain of electrons) occurs at cathode. Oxidation state of oxygen is generally taken as -2 in all of its compounds except for its elemental state, peroxides and superoxide where it is 0, -1 and $-\dfrac{1}{2}$, respectively.