Answer
cell potential = 0.017V
Explanation
Oxidation half reaction
Fe(s) ----------> Fe2+(aq) + 2e- E°red = -0.440V
Reduction half reaction
Cd2+(aq) + 2e- ----------> Cd(s) E°red = -0.403V
Overall reaction
Fe(s) + Cd2+(aq) ---------> Fe2+(aq) + Cd(s)(aq)
E° = E°red, cathode - E°red,anode
E° = - 0.403V - (- 0.440V)
E°= -0.403V + 0.440V
E° = 0.037V
number of electron transfer , n = 2
Reaction quotient ,Q = [Fe2+]/[Cd2+]
When cathaode solution concentration decreased by 0.657M ande solution concentration increases by 0.657M
Q = ( 1.000M + 0.657M)/( 1.000M - 0.657M)
Q = 4.8309
Nernst equation at 298K is
Ecell = E° - (0.0592V/n)logQ
Ecell = 0.037V - ( 0.0592V/2)log4.8309
Ecell = 0.037V - 0.020
Ecell = 0.017V
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