Consider the cell described below at 255 K: Sn | Sn2+ (0.969 M) || Pb2+ (0.941 M

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. 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. 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.

Explanation / Answer

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,

Ecell = Ecello - (0.05916 V/n)log ([Sn2+]/[Pb2+])

Where, n is the number of electrons.

Ecello = Eo (cathode) - Eo (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

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

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