Question

What is the total number of moles (n) of electrons exchanged between the oxidizing agent and the reducing agent in the overall redox equation: 5 ag+(aq) + mn2+(aq) + 4h2o(l) → 5ag(s) + mno4–(aq) + 8h+(aq)?

Answer 1

The equation is as below,

       5 Ag⁺ (aq) + Mn²⁺ (aq) + 4H₂O (l) → 5Ag (s) + MnO4⁻(aq) + 8H⁺ (aq)

Reduction:
                According to eq. Silver is reduced as,

                                5 Ag⁺  + 5 e⁻    →    5 Ag

One electron is gained per Ag ion, so 5 Ag ions will gain 5 electrons.

Oxidation:
               According to eq. Manganese is Oxidized as, 

                                        Mn²⁺    →    MnO4⁻

Oxidation number of Mn in MnO₄⁻ is +7, so Mn²⁺ is loosing 5 electrons, Hence,

                                        Mn²⁺    →    MnO4⁻  + 5 e⁻

Results:
             Hence, 5 moles of electrons are exchanged between oxidizing and reducing agents.