When 0.1 mole `Cr_(2)O_(7)^(-2)` is oxidised then quantity of elecricity required to completely oxidise `Cr_(2)O_(7_^(-2))` to `Cr^(+3)` is :-

Options

A. 9650 C
B. 96500 C
C. 57900 C
D. 54900 C

Explanation

To solve the problem of calculating the quantity of electricity required to completely oxidize 0.1 mole of \( \text{Cr}_2\text{O}_7^{2-} \) to \( \text{Cr}^{3+} \), we can follow these steps:

### Step 1: Write the half-reaction for the oxidation of \( \text{Cr}_2\text{O}_7^{2-} \)
The half-reaction for the oxidation of dichromate ion \( \text{Cr}_2\text{O}_7^{2-} \) to chromium ion \( \text{Cr}^{3+} \) can be written as:
\[ \text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{e}^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O} \]
This shows that 1 mole of \( \text{Cr}_2\text{O}_7^{2-} \) requires 6 moles of electrons for complete oxidation.

### Step 2: Calculate the number of moles of electrons required for 0.1 mole of \( \text{Cr}_2\text{O}_7^{2-} \)
Since 1 mole of \( \text{Cr}_2\text{O}_7^{2-} \) requires 6 moles of electrons, for 0.1 mole of \( \text{Cr}_2\text{O}_7^{2-} \):
\[ \text{Moles of electrons} = 0.1 \text{ moles of } \text{Cr}_2\text{O}_7^{2-} \times 6 = 0.6 \text{ moles of electrons} \]

### Step 3: Calculate the total charge using Faraday's constant
The total charge (Q) can be calculated using the formula:
\[ Q = n \times F \]
where \( n \) is the number of moles of electrons and \( F \) is Faraday's constant, approximately \( 96500 \, \text{C/mol} \).

Substituting the values:
\[ Q = 0.6 \text{ moles} \times 96500 \, \text{C/mol} = 57900 \, \text{C} \]

### Step 4: Conclusion
The quantity of electricity required to completely oxidize 0.1 mole of \( \text{Cr}_2\text{O}_7^{2-} \) to \( \text{Cr}^{3+} \) is approximately \( 57900 \, \text{C} \).

---