Question

B) Pt(s) ΙSn2 (0.0055 M), Sn4 (0.12 M) ΙΙFe2 (0.0020 M), Fe3 (0.15 M) ΙPt(s)

Ecell=? V

Answer 1

Pt(s) ΙSn2 (0.0055 M), Sn4 (0.12 M) ΙΙFe2 (0.0020 M), Fe3 (0.15 M) ΙPt(s)

              anode                                     cathode

So,

Sn⁺² + 2Fe³⁺ ----------> 2Fe⁺² (aq) + Sn⁺⁴

At anode :   Sn²⁺ --------------> Sn⁴⁺ + 2e^-      E^o = - 0.15 V

At cathode : Fe³⁺ +e- -------------> Fe²⁺          E^o = + 0.77 V

E^o_cell = E^o_reduction of reaction at cathode+E^o_oxidation of reaction at anode

        = -0.15 V + 0.77 V

       = + 0.62 V

E^o_cell = + 0.62 V

no of electrons transferred n = 2

Then,

E_cell = E^o_cell - RT/nF In {[Fe²⁺]² [Sn⁴⁺] / [Sn²⁺] [Fe³⁺]²}

E_cell = E^o_cell - [0.059/n] Iog {[Fe²⁺]² [Sn⁴⁺] / [Sn²⁺] [Fe³⁺]²} where n = no of electrons transferred

       = 0.62 -[0.059/2] Iog {[0.0002]² [0.12] / [0.0055] [0.15]²}

       = + 0.75 V

Therefore,

E_cell = + 0.75 V