In order to completely oxidize `0.1 mol` of `MnO_(4)^(2-)` to permanganate ion. The quantity of electricity required is

Options

A. `96500C`
B. `2xx96500C`
C. `9650C`
D. `96.50C`

Explanation

To determine the quantity of electricity required to completely oxidize \(0.1 \, \text{mol}\) of \( \text{MnO}_4^{2-} \) to permanganate ion \( \text{MnO}_4^{-} \), we can follow these steps:

### Step 1: Identify the oxidation states
In \( \text{MnO}_4^{2-} \), the oxidation state of manganese (Mn) is +6. In the permanganate ion \( \text{MnO}_4^{-} \), the oxidation state of manganese is +7.

### Step 2: Determine the change in oxidation state
The oxidation process involves the conversion of manganese from +6 in manganate to +7 in permanganate. This means that each \( \text{MnO}_4^{2-} \) ion loses 1 electron during the oxidation process.

### Step 3: Calculate the number of moles of electrons required
Since \(0.1 \, \text{mol}\) of \( \text{MnO}_4^{2-} \) is being oxidized, the number of moles of electrons required will be equal to the number of moles of manganate ions:
\[ \text{Moles of electrons} = 0.1 \, \text{mol} \]

### Step 4: Use Faraday's constant to find the total charge
Faraday's constant (\(F\)) is approximately \(96500 \, \text{C/mol}\). The total quantity of electricity (\(Q\)) required can be calculated using the formula:
\[ Q = \text{moles of electrons} \times F \]
Substituting the values:
\[ Q = 0.1 \, \text{mol} \times 96500 \, \text{C/mol} = 9650 \, \text{C} \]

### Conclusion
The quantity of electricity required to completely oxidize \(0.1 \, \text{mol}\) of \( \text{MnO}_4^{2-} \) to permanganate ion is \(9650 \, \text{C}\).

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