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 \( \text{Cr}_2\text{O}_7^{2-} \) in acidic medium, we can follow these steps:

### Step 1: Identify the reduction half-reaction
The reduction of dichromate ion \( \text{Cr}_2\text{O}_7^{2-} \) in acidic medium can be represented by the following half-reaction:

\[ \text{Cr}_2\text{O}_7^{2-} + 14 \text{H}^+ + 6 \text{e}^- \rightarrow 2 \text{Cr}^{3+} + 7 \text{H}_2\text{O} \]

### Step 2: Determine the number of electrons transferred
From the half-reaction, we can see that 6 electrons (\( 6 \text{e}^- \)) are required to reduce 1 mole of \( \text{Cr}_2\text{O}_7^{2-} \).

### Step 3: Calculate the total charge in coulombs
To find the total charge (in coulombs) required to transfer 6 moles of electrons, we use Faraday's constant, which is approximately \( 96485 \, \text{C/mol} \). The total charge \( Q \) can be calculated using the formula:

\[ Q = n \times F \]

where:
- \( n \) = number of moles of electrons (6 moles)
- \( F \) = Faraday's constant (\( 96485 \, \text{C/mol} \))

Substituting the values:

\[ Q = 6 \, \text{mol} \times 96485 \, \text{C/mol} = 578910 \, \text{C} \]

### Step 4: Conclusion
Thus, the total charge required to reduce 1 mole of \( \text{Cr}_2\text{O}_7^{2-} \) in acidic medium is \( 578910 \, \text{C} \).