Question

Pt/Ce3+ (0.01M),Ce4+ (0.1M),H2SO4 (1M)//Fe2+ (0.01M),Fe3+ (0.1M),HC1(1M)/Pt ,What would the composition of the system be at the end of a galvanic discharge to an equilibrium condition? What would the cell potential be? What would the potential of each electrode be vs. NHE? Vs. SCE? Take equal volumes on both sides.

Answer 1

Pt/Ce3+(0.01 M), Ce4+(0.1 M), H2SO4(1 M)//Fe2+(0.01 M), Fe3+(0.1 M), HCl(1 M)/Pt

Fe3+ + e = Fe2+ Eo = 0.77V (w.r.t the NHE)

Ce4+ + e = Ce3+ Eo = 1.61V

Adjusting for conditions using the Nerst equation: Ec = 0.77 + 0.0257 ln [Fe3+]/[Fe2+] = 0.829V

Ea = 1.61 + 0.0257 ln [Ce4+] [Ce3+] = 1.669V

Erxn = Ec − Ea = −0.84V δG > 0;reaction is not spontaneous.

For the SCE the potential of the electrode is to be added to these values of the cathode and the anode emf's.