Question

A voltaic cell is constructed with the given solutions of CuSO₄ (0.130 M) and FeSO₄ (0.0500 M); 25.0 g strips of Cu and Fe; and a KNO₃ salt bridge. Using a table displaying standard potentials (E^o_red) to answer the following commands.  

(a) Write the equations for anode and the cathode half-reactions and the overall reaction. Write the line diagram (cell notation).  

(b) Calculate E^o for the cell at 25 ^oC.

(c) Calculate E for the cell at 25 ^oC.  

(d) Calculate the equilibrium constant for the above reaction at 25 ^oC.  

Answer 1

Cu2+ + 2 e− ----> Cu(s)    E0 = +0.337 v

Fe2+ + 2 e− ----> Fe(s)    E0 = −0.44 v

at aonde: oxidation , Fe(s) ---> Fe2+(aq) + 2 e−

at cathode: Reduction, Cu2+ + 2 e− ----> Cu(s)

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over all reaction : Fe(s)+ Cu2+(aq) ---> Fe2+(aq) + Cu(s)
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b) E0cell = E0cathode- E0anode

    = 0.337 - - 0.44

    = 0.777 v

c) Ecell = E0cell-0.0591/nlog[Fe2+][Cu2+]

         = 0.777-(0.0591/2)log(0.05/0.13)

     = 0.79 v

d) DG0 = -nFE0cell

       = -2*96500*0.777

       = -149.96 kj/mol

DG0 =- RTlnK

-149960 = -8.314*298lnk

k = 1.935*10^26