Question
Chemistry Question on Electrochemistry
How much charge is required for the reduction of 1 mol of MnO-4 to Mn2+ ?
A
1 F
B
5 F
C
3 F
D
6 F
Answer
5 F
Explanation
Solution
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).