Arrange the following in the increasing order of quantity of electric…

The correct answer is Option (4) → C, B, A, D

By Faraday’s first law, the charge needed to reduce one mole of an ion equals the number of electrons (n) taken up per ion multiplied by one faraday:

    \( \displaystyle Q=nF,\quad F=96487\ \mathrm{C\,mol^{-1}} \)

So the order of charge is the order of the number of electrons gained. For each reduction, n is the fall in oxidation state:

• (C) \( \ce{Ag^{+}} \) → Ag: +1 to 0, n = 1, Q = 1 F

• (B) \( \ce{Cu^{2+}} \) → Cu: +2 to 0, n = 2, Q = 2 F

• (A) \( \ce{Al^{3+}} \) → Al: +3 to 0, n = 3, Q = 3 F

• (D) \( \ce{MnO4^{-}} \) → \( \ce{Mn^{2+}} \): the oxidation state of Mn in \( \ce{MnO4^{-}} \) is found from x + 4(−2) = −1, so x = +7. It falls from +7 to +2, n = 5, Q = 5 F

      \( \displaystyle \ce{MnO4^- + 8H^+ + 5e^- -> Mn^2+ + 4H2O} \)

NCERT (2026-27 edition):
Electrolytic Cells and Electrolysis – “… The amount of electricity (or charge) required for oxidation or reduction depends on the stoichiometry of the electrode reaction. For example, in the reaction:

    Ag⁺(aq) + e^– → Ag(s)    (2.30)

One mole of the electron is required for the reduction of one mole of silver ions. … For the electrode reactions:

    Mg²⁺(l) + 2e^– → Mg(s)    (2.31)
    Al³⁺(l) + 3e^– → Al(s)    (2.32)

It is obvious that one mole of Mg²⁺ and Al³⁺ require 2 mol of electrons (2F) and 3 mol of electrons (3F) respectively.”