Question

Consider a galvanic cell based on the following two half reactions under standard conditions at 298 K: Fe2+ + 2 e → Fe -0.440

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Answer #1

Answer

cell potential = 0.017V

Explanation

Oxidation half reaction

Fe(s) ----------> Fe2+(aq) + 2e-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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