Question

In the reaction MnO4- + 8H+ + 5e- Mn2+ + 4H20, the manganese in 3.95 grams of KMnO4 is reduced to Mn2+. How many coulombs of charge must have been transferred to the permanganate ion

Answer 1

According to Faraday’s law ,W = (EQ) / 96500

Where W = mass of metal deposited = 3.95 g

           E = Equivalent weight of Mn = Molar mass / number of electrons transferred

               = 55 / 5

               = 11

           Q = charge = ?

Plug the values we get , Q = (96500xW ) / E

                                       = (96500x3.95) / 11

                                       = 34652 C

Therefoere the charge must be transferred is 34652 C