How many coulombs of electricity are required for the process :

(iii) reduction of 1 mol of `F_2` to `2F^(-)` ?

Explanation

To determine how many coulombs of electricity are required for the reduction of 1 mole of \( F_2 \) to \( 2F^- \), we can follow these steps:

### Step 1: Write the Reduction Reaction
The reduction of fluorine gas (\( F_2 \)) to fluoride ions (\( F^- \)) can be represented by the following half-reaction:
\[ F_2 + 2e^- \rightarrow 2F^- \]
This equation indicates that 1 mole of \( F_2 \) gains 2 moles of electrons to form 2 moles of \( F^- \).

### Step 2: Determine Moles of Electrons
From the balanced equation, we see that 1 mole of \( F_2 \) requires 2 moles of electrons for the reduction process.

### Step 3: Calculate the Total Charge in Coulombs
We know that 1 mole of electrons corresponds to approximately 96500 coulombs (Faraday's constant). Therefore, for 2 moles of electrons:
\[ \text{Charge (C)} = \text{moles of electrons} \times \text{Faraday's constant} \]
\[ \text{Charge (C)} = 2 \, \text{moles} \times 96500 \, \text{C/mole} \]
\[ \text{Charge (C)} = 193000 \, \text{C} \]

### Conclusion
Thus, the total charge required for the reduction of 1 mole of \( F_2 \) to \( 2F^- \) is **193000 coulombs**.

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