MnO4−+8H++5e−→Mn2++4H2O Eo =1.51 V
The quantity of electricity required in Faraday to reduce five moles of MnO−4 is ____.
Answer: 25
MnO4−+8H++5e−→Mn2++4H2O
∵ 1 mol MnO4− requires 5 Faraday charge.
∴ 5 mol MnO4− will require 25 Faraday charge.