Question

Consider the cell described below at 259 K:

Fe | Fe²⁺ (0.853 M) || Cd²⁺ (0.907 M) | Cd

Given E^o_Cd²⁺→Cd = -0.403 V, E^o_Fe²⁺→Fe = -0.441 V. Calculate the cell potential after the reaction has operated long enough for the Fe²⁺ to have changed by 0.321 mol/L.

Answer 1

The balanced reaction is   Fe + Cd2+   <---> Cd + Fe2+

given [Fe2+] = 0.853 + 0.321 mol/L = 1.144 M     ( Fe2+ is formed , hence change means added )

then [Cd2+] = 0.907-0.321 =0.586 M

Q cell = [Cd2+] /[Fe2+] = ( 0.586/1.144) = 0.512

Eo cell = Eo ( Cd2+/cd) - Eo ( Fe2+/Fe) = -0.403-(-0.441) = 0.038

Ecell = Eo cell - ( RT/nF) ln Q    where n = number of electrons involved per reatcion = 2

       = 0.038 - ( 8.314 x 259 /2 x 96485) ln (0.512)

     = 0.0455 V