During the electorlysis of cryolite, aluminium and fluorine are formed in ….molar ratio

Options

A. `1:2`
B. `2:3`
C. `1:1`
D. `1:3`

Explanation

To determine the molar ratio of aluminum and fluorine produced during the electrolysis of cryolite (Na3AlF6), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Composition of Cryolite**:
- The chemical formula of cryolite is Na3AlF6.

2. **Decomposition of Cryolite**:
- When cryolite undergoes electrolysis, it decomposes into sodium fluoride (NaF) and aluminum fluoride (AlF3):
\[ \text{Na}_3\text{AlF}_6 \rightarrow 3\text{NaF} + \text{AlF}_3 \]

3. **Electrolysis Reactions**:
- During electrolysis, aluminum ions (from AlF3) are reduced at the cathode, while fluoride ions (from NaF) are oxidized at the anode.
- The reduction reaction at the cathode is:
\[ \text{Al}^{3+} + 3e^- \rightarrow \text{Al} \]
- The oxidation reaction at the anode is:
\[ 2\text{F}^- \rightarrow \text{F}_2 + 2e^- \]

4. **Determine the Number of Electrons Exchanged**:
- For each mole of aluminum produced, 3 moles of electrons are required.
- For every 1 mole of F2 produced, 2 moles of electrons are released.

5. **Calculate the Molar Ratio**:
- From the reduction of aluminum, we see that 3 moles of electrons yield 1 mole of aluminum.
- From the oxidation of fluoride, 2 moles of electrons yield 1 mole of fluorine gas (F2).
- Therefore, for every 3 moles of aluminum produced, 2 moles of fluorine gas are produced.
- This gives us a molar ratio of:
\[ \text{Al} : \text{F}_2 = 3 : 2 \]

6. **Final Molar Ratio**:
- The molar ratio of aluminum to fluorine produced during the electrolysis of cryolite is:
\[ 3 : 2 \]

### Final Answer:
The molar ratio of aluminum to fluorine formed during the electrolysis of cryolite is **3:2**.

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