In Hall's process , in the production of `Al,` carbon is used as the anode material. The reactions are

Options

A. Anode `: Al^(3+)+3e^(-) rarr Al`
Cathode `:C(s)+2O^(2-) rarr CO_(2)(s)+4e^(-)`
B. Anode`:C(s)+2O^(2-) rarr CO_(2)(g)+4e^(-)`
Cathode `:Al^(3+)+3e^(-) rarrAl`
C. Anode `: Alrarr Ae^(3+)+e^(-)`
Cathode `:CO_(2)+4e^(-) rarr C(s)+2O^(2-)`
D. None of these

Explanation

To solve the question regarding the reactions occurring in Hall's process for the production of aluminum, we will break down the steps involved in the electrochemical reactions at both the anode and cathode.

### Step-by-Step Solution:

1. **Identify the Anode Reaction**:
- In Hall's process, carbon is used as the anode material. At the anode, oxidation occurs.
- The oxidation reaction can be represented as:
\[ C_{(s)} + O_2 \rightarrow CO_2_{(g)} + 4e^- \]
- Here, solid carbon (C) reacts with oxygen (O2) to produce carbon dioxide (CO2) gas and releases 4 electrons.

2. **Identify the Cathode Reaction**:
- At the cathode, the reduction of aluminum ions takes place.
- The reduction reaction can be represented as:
\[ Al^{3+} + 3e^- \rightarrow Al_{(l)} \]
- In this reaction, aluminum ions (Al3+) gain 3 electrons to form liquid aluminum (Al).

3. **Overall Reaction**:
- The overall process combines both the anode and cathode reactions. However, since the number of electrons lost at the anode (4) does not match the number of electrons gained at the cathode (3), we need to balance the reactions for a complete overall reaction.
- To balance the electrons, we can multiply the cathode reaction by 4 and the anode reaction by 3:
\[ 3C_{(s)} + 3O_2 \rightarrow 3CO_2_{(g)} + 12e^- \]
\[ 4Al^{3+} + 12e^- \rightarrow 4Al_{(l)} \]
- The balanced overall reaction would then be:
\[ 3C_{(s)} + 4Al^{3+} + 3O_2 \rightarrow 3CO_2_{(g)} + 4Al_{(l)} \]

### Final Answer:
- At the anode: Oxidation of carbon occurs, producing CO2 and releasing electrons.
- At the cathode: Reduction of aluminum ions occurs, producing liquid aluminum.