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).
Battery or cell converts chemical energy of the redox reaction to electrical energy. In fuel cells (a galvanic cell), the chemical energy of combustion of fuels like H2, ethanol, etc., is directly converted to electrical energy. In a fuel cell, H2 and O2 react to produce electricity, where H2 gas is oxidized at the anode and oxygen is reduced at the cathode, and the reactions involved are:
Anode reaction: H2 + 2OH- → 2H2O + 2e-
Cathode reaction: O2 + 2H2O + 4e- → 4OH-
67.2 L of H2 at STP reacts in 15 minutes.
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.