4a. Use the Nernst equation to calculate the cell voltage (E) for the following

Question

4a. Use the Nernst equation to calculate the cell voltage (E) for the following redox reaction:

i. Fe3+(aq)+Cu(s)Cu2+(aq)+Fe2+(aq), given that [Fe3+]=0.05 and [[Cu2+]=0.125M at 25°C, and

Fe3++eFe2+             E0=0.77 V

Cu2+2eCu                   E0=0.34 V

4b. Use the information provided in question 4a to calculate the change in free energy (G) and change in entropy (S) for the redox reaction:

i. Fe3+(aq)+Cu(s)Cu2+(aq)+Fe2+(aq)

What do the G and S values indicate about the spontaneity of the redox reaction?

Explanation / Answer

2Fe3+(aq)+Cu(s) ----> Cu2+(aq)+2Fe2+(aq),

E0cell = E0cathode - E0anode

        = (0.77)-(0.337)

       = 0.433 v
Ecell = E0cell - (0.0591/n)log([Cu2+(aq)][Fe2+(aq)]^2/[Fe3+]^2)

       = 0.433 - (0.0591/2)log(0.125* 0.05^2/1)

       = 0.54 v

DG0 = - nFE0cell

     = -2*96500*0.433

     = -83.57 kj/mol

as DG0 = -ve, the process is spontaneous.

DG = - nFEcell

    = -2*96500*0.54

   =-104.22 kj/mol

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