When 0.1 mol MnO2−4 is oxidised the quantity of electricity required to completely MnO2−4 to MnO4− is:

To determine the quantity of electricity required to completely oxidize 0.1 mol of MnO2−4 to MnO4−, we can use Faraday's law of electrolysis.

1. First, we need to determine the balanced chemical equation for the oxidation reaction. The oxidation of MnO2−4 to MnO4− can be represented as follows: MnO2−4 → MnO4− + 2e−

2. From the balanced equation, we can see that for every 1 mole of MnO2−4 oxidized, 2 moles of electrons (e−) are involved in the reaction.

3. Next, we need to calculate the total charge (in coulombs) required to transfer 2 moles of electrons. The charge carried by one mole of electrons is given by the Faraday constant, which is approximately 96,485 C/mol.

   Charge = (2 moles of electrons) × (96,485 C/mol) = 192,970 C

4. Finally, we can calculate the quantity of electricity required to completely oxidize 0.1 mol of MnO2−4 to MnO4− by multiplying the charge by the number of moles of MnO2−4.

   Quantity of electricity = (0.1 mol) × (192,970 C/mol) = 19,297 C

Therefore, the quantity of electricity required to completely oxidize 0.1 mol of MnO2−4 to MnO4− is 19,297 coulombs.