Question

Consider the following Galvanic Cell : Al (s) | Al ⁺³ (0.010M) || Fe ⁺²(0.10M) | Fe(s) where,

Al⁺³ + 3e > Al(s) E = -1.662 V

Fe⁺² + 2e > Fe (s) E -.440 V

A. What is potential of the above voltaic cell at 25C ?
B.) Calculate Delta G of the above cell at 25C ?

Answer 1

overall reaction of the cell

{AlAl3+ + 3e }*2 oxidation-anode

2Al2Al3+ + 6e.............(1)

{Fe2+ + 2eFe}*3 reduction -cathode

3Fe2+ + 6e3Fe.................(2)

adding equations (1) and (2),

2Al +3Fe2+2Al3+ + 3Fe

Eo=Ecathode-Eanode=-0.440-(-1.662)=1.222V

n=6 electron exchange in redox reaction,

part A) E=Eo-0.0592/n log [Al3+]^2/[Fe2+]^3

   =1.222V-0.0592/6 V log (0.010)^2/(0.10)^3

   =1.222V-0.009867 V log (0.0001/0.001)

    =1.222V+0.00987V

    =1.2319V

part B) G0=-nFEo

                   =-6 * 96458C/mol * 1.2319 J/C

                   =-712959.66J

                    =-712.96KJ/mol