What amount of electric charge is required for the reduction of 1 mole of MnO$_4^{2-}$ into Mn$^{2+}$?
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.


Electricity is passed through an acidic solution of Cu$^{2+}$ till all the Cu$^{2+}$ was exhausted, leading to the deposition of 300 mg of Cu metal. However, a current of 600 mA was continued to pass through the same solution for another 28 minutes by keeping the total volume of the solution fixed at 200 mL. The total volume of oxygen evolved at STP during the entire process is ___ mL. (Nearest integer)
Given:
$\mathrm{Cu^{2+} + 2e^- \rightarrow Cu(s)}$
$\mathrm{O_2 + 4H^+ + 4e^- \rightarrow 2H_2O}$
Faraday constant = 96500 C mol$^{-1}$
Molar volume at STP = 22.4 L

