When 0.1 mole of MnO42- is oxidised, the quantity of electricity required to completely oxidise MnO42- to MnO4-

To determine the quantity of electricity required to completely oxidize 0.1 mole of MnO42- to MnO4-, we need to follow these steps:

Step 1: Write the balanced chemical equation for the oxidation reaction: 2 MnO42- → 2 MnO4- + 2 e-

Step 2: Determine the number of electrons transferred in the reaction. In this case, 2 electrons are transferred.

Step 3: Use Faraday's law to calculate the quantity of electricity required. Faraday's law states that the quantity of electricity (Q) is equal to the product of the number of moles of electrons (n) and the Faraday constant (F): Q = n * F

Step 4: Calculate the number of moles of electrons transferred. Since 2 electrons are transferred per mole of MnO42-, the number of moles of electrons (n) is equal to 2 times the number of moles of MnO42-: n = 2 * 0.1 = 0.2 moles of electrons

Step 5: Determine the value of the Faraday constant. The Faraday constant (F) is equal to the charge of one mole of electrons, which is approximately 96,485 Coulombs per mole.

Step 6: Calculate the quantity of electricity required by substituting the values into the equation: Q = 0.2 * 96,485 = 19,297 Coulombs

Therefore, the quantity of electricity required to completely oxidize 0.1 mole of MnO42- to MnO4- is 19,297 Coulombs.