To reduce MnO$_4^{2-}$ to Mn$^{2+}$, we need to consider the change in oxidation state of Mn. In MnO$_4^{2-}$, Mn is in the +2 oxidation state, and in Mn$^{2+}$, Mn is also in the +2 state. Thus, no change in oxidation state occurs. However, the reduction process involves the addition of 2 electrons per ion, and for 1 mole of MnO$_4^{2-}$, 4 moles of electrons are required. This corresponds to a charge of 4 Faradays.