How much charge is required for the following reductions: (i) 1 mol of Al^(3+) to Al (ii) 1 mol of Cu^(2+) to Cu (iii) 1 mol of MnO_(4)^(-) to Mn^(2+).

How much charge is required for the following reductions: (i) 1 mol o

1.	How much charge is required for the following reductions: (i) 1 mol of Al^(3+) to Al (ii) 1 mol of Cu^(2+) to Cu (iii) 1 mol of MnO_(4)^(-) to Mn^(2+).
Answer» Solution :(i) Required electric charge for reduction of 1 mol `Al^(3+)` to Al : Reduction on Al electrode : `underset (1" mol")(Al^(3+))+underset(3" mol")(3e^(-)) to underset(1" mol")(Al)` According to this reaction, 1 mol of `Al^(3+)` to get reduced into Al it requires 3 mol of electrons. So, electric charge of 1 mol of electron, 1F=96500 coulomb So, electric charge of 3 mol of electron, 3F=3(96500)=289500 coulomb (II) Required electric charge for reduction of 1 mol `Cu^(2+) ` to Cu : Reduction on Al electrode : `underset(1" mol")(Cu^(2+)) + underset(2" mol")(2e^(-)) to underset(1" mol")(Cu)` According to this reaction, 1 mol of `Cu^(2+)` to get reduced into Cu it requires 2 mol of electrons So, 2 mol electron=2 feraday electric charge is required. `=2xx96500` `=193000` coulomb electric charge (iii) Electric charge required for reduction of 1 mol `MnO_(4)^(-)` to `Mn^(2+)` : The reaction occurred on electrode when 1 mol of `MnO_(4)^(-)` is reduced to `Mn^(2+)` `underset(1MOL)(MnO_(4)^(-)) +8H^(+) + underset(5mol)(5E^(-)) to Mn^(2+)+4H_(2)O` According to this reaction, for reduction of 1 mol `MnO_(4)^(-)` it requires 5 mol electron and So, electric charge of 5 mol of electron =5Ferady `(5xx96500)` =193000 Coulomb So, 5 Ferady=193000 Coulomb electric charge is required.