How many coulombs of electricity is required to reduce 1 mole of `Cr_2O_7^(2-)` in acidic medium?

Options

A. 4 x 96500 C
B. 6 x 96500 C
C. 2 x 96500 C
D. 1 x 96500 C

Explanation

To determine how many coulombs of electricity are required to reduce 1 mole of `Cr_2O_7^(2-)` (dichromate ion) in acidic medium, we can follow these steps:

### Step-by-Step Solution:

1. **Write the Reduction Half-Reaction**:
The first step is to write the balanced reduction half-reaction for the dichromate ion in acidic medium. The balanced equation is:
\[ Cr_2O_7^{2-} + 14 H^+ + 6 e^- \rightarrow 2 Cr^{3+} + 7 H_2O \]
This equation shows that 1 mole of dichromate ion requires 6 moles of electrons for its reduction.

2. **Determine the Number of Electrons**:
From the balanced equation, we see that to reduce 1 mole of `Cr_2O_7^{2-}`, 6 moles of electrons (6 electrons) are needed.

3. **Use Faraday's Constant**:
The charge of one mole of electrons is given by Faraday's constant, which is approximately 96500 coulombs. Therefore, the total charge (in coulombs) required to transfer 6 moles of electrons is:
\[ \text{Total Charge} = \text{Number of moles of electrons} \times \text{Faraday's constant} \]
\[ \text{Total Charge} = 6 \times 96500 \, \text{C} \]

4. **Calculate the Total Charge**:
Now we can calculate the total charge:
\[ \text{Total Charge} = 6 \times 96500 = 579000 \, \text{C} \]

5. **Final Answer**:
Therefore, the amount of electricity required to reduce 1 mole of `Cr_2O_7^{2-}` in acidic medium is:
\[ 579000 \, \text{C} \]

### Summary:
The total amount of electricity required to reduce 1 mole of `Cr_2O_7^{2-}` in acidic medium is **579000 coulombs**.