Question

Consider the cell described below at 255 K:

Sn | Sn2+ (0.969 M) || Pb2+ (0.941 M) | Pb

Given EoPb2+→Pb = -0.131 V, EoSn2+→Sn = -0.143 V. Calculate the cell potential after the reaction has operated long enough for the Sn2+ to have changed by 0.323 mol/L.

Answer 1

In shorthand notation of a cell, the left side to the double vertical line (II) indicates the anode and right side indicates the cathod reaction.

Oxidation occurs at anode: Sn(s) ------> Sn²⁺ + 2e^-

Reduction occurs at cathode: Pb²⁺ + 2 e^- -------> Pb(s)

Overall reaction: Sn(s) + Pb²⁺ (aq) ---> Sn²⁺ (aq) + Pb(s)

Nernst equation for the cell potential is,

E_cell = E_cell^o - (0.05916 V/n)log ([Sn²⁺]/[Pb²⁺])

Where, n is the number of electrons.

E_cell^o = E^o (cathode) - E^o (anode) = (- 0.131) - (- 0.143) = 0.012 V

Change in [Sn²⁺] = Initial [Sn²⁺] - Final [Sn²⁺] = 0.969 M - 0.323 M = 0.646 M

Since one mole Pb²⁺ reacts to give one Sn²⁺ ,

Final [Pb²⁺] = 0.941 M + 0.646 M = 1.587 M

E_cell = 0.012 V - (0.05916 V/2) log (0.323/1.587) = 0.032 V