When `0.1 mol MnO_(4)^(2-)` is oxidized the quantity of electricity required to completely oxidize `MnO_(4)^(2-)` to `MnO_(4)^(-)` is

Options

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

Explanation

To solve the problem of calculating the quantity of electricity required to completely oxidize `0.1 mol MnO4^(2-)` to `MnO4^(-)`, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Oxidation States**:
- For `MnO4^(2-)`, let's find the oxidation state of manganese (Mn). The formula for oxidation state is:
\[ x - 8 = -2 \quad \text{(where x is the oxidation state of Mn)} \]
Solving for x gives:
\[ x = +6 \]
- For `MnO4^(-)`, we find the oxidation state similarly:
\[ x - 8 = -1 \]
Solving for x gives:
\[ x = +7 \]

2. **Determine the Change in Oxidation State**:
- The change in oxidation state from `MnO4^(2-)` to `MnO4^(-)` is:
\[ +6 \to +7 \]
- This indicates that one electron is transferred during the oxidation process.

3. **Calculate the Moles of Electrons Required**:
- Since 1 mole of `MnO4^(2-)` requires 1 mole of electrons for oxidation, for `0.1 mol MnO4^(2-)`, the moles of electrons required will be:
\[ 0.1 \, \text{mol MnO4^(2-)} \times 1 \, \text{mol e}^- = 0.1 \, \text{mol e}^- \]

4. **Convert Moles of Electrons to Quantity of Electricity**:
- The quantity of electricity (Q) can be calculated using Faraday's constant, which is approximately `96500 C/mol`. Therefore, the total charge required is:
\[ Q = \text{moles of electrons} \times \text{Faraday's constant} \]
Substituting the values:
\[ Q = 0.1 \, \text{mol} \times 96500 \, \text{C/mol} = 9650 \, \text{C} \]

### Final Answer:
The quantity of electricity required to completely oxidize `0.1 mol MnO4^(2-)` to `MnO4^(-)` is **9650 coulombs**.