Consider the following reaction
MnO₄ - + 8H + + 5e - → Mn⁺² + 4H₂O,Eᵒ = 1.51V.
The quantity of electricity required in Faraday to reduce five moles of MnO₄ - is ___________. (Integer answer)

Solution

Chemistry ma’am · 3 steps

Read the electrons straight off the half-equation: 5 per MnO₄⁻, then scale up to 5 moles.

1. Electrons per ion. The half-equation shows that one MnO₄⁻ takes up 5 electrons. Manganese goes from +7 to +2, a drop of 5, which confirms it.
2. Electrons to faradays. One faraday is the charge on one mole of electrons. So reducing 1 mol of MnO₄⁻ needs 5 F.

   Q = 5 mol × 5 F mol⁻¹ = 25 F

3. So. Reducing 5 moles of MnO₄⁻ needs 25 faradays. The value E° = 1.51 V is not needed here.

The common slip: 5. Taking 5 F as the total for the whole question gives 5, not 25. That is the charge for one mole of MnO₄⁻, so it still has to be multiplied by 5 moles.

The idea to keep. Faradays needed = moles of substance × electrons per formula unit. Take the electron count from the balanced half-equation or from the change in oxidation state.

Answer from the NTA final answer key. Solution by PrepWiser.