How much charge is required for the reduction of 1 mol of MnO-4 to Mn2+ ?
1 F
5 F
3 F
6 F
The reduction of MnO₄⁻ to Mn²⁺ involves the following balanced equation:
MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O
From the balanced equation, we can see that 5 electrons (5e⁻) are required for the reduction of 1 mole of MnO₄⁻ to Mn²⁺.
The unit of charge, Faraday (F), represents 1 mole of electrons, which is equivalent to 96,485.3329 Coulombs.
Therefore, the charge required for the reduction of 1 mole of MnO₄⁻ to Mn²⁺ is:
5 × 96,485.3329 C = 482,426.6645 C
Rounded to the nearest whole number, the charge required is approximately 482,427 C.
Therefore, the correct answer is 482,427 C, which is approximately 5 F (Faraday).
An electrochemical cell is a device that is used to create electrical energy through the chemical reactions which are involved in it. The electrical energy supplied to electrochemical cells is used to smooth the chemical reactions. In the electrochemical cell, the involved devices have the ability to convert the chemical energy to electrical energy or vice-versa.