For questions 24-26 consider the electrochemical cell shown using line notation

Question

For questions 24-26 consider the electrochemical cell shown using line notation and the standard reduction potentials in the table below: Pt | Fe^2+ (0.30 M), Fe^3+ (0.20 M) || MnO^-1_-4 (0.30 M), H^+ (0.10 M), Mn^2+ (0.20 M) | Pt Which of the following balanced chemical equations corresponds to this line notation? Pt + 5Fe^2+ + MnO^-1_4 + H_3 O^+ rightarrow 5Fe^3+Mn^2+ + 12H_2 O 5Fe^3+ + Mn^2+ + 12H_2 O rightarrow 5Fe^2+ + MnO^-1_4 + 8H_3 O^+ Fe^3+ + Mn^2+ + 12H_2 O rightarrow Fe^2+ + MnO^-1_-4 + 8H_3 O^+ Fe^2+ + MnO^-1_4 + 8H_3 O^- rightarrow Fe^3+ + Mn^2+ + 12H_2 O 5fe^2+ + MnO^-1_4 + 8H_3O^+ rightarrow 5Fe^3+ + Mn^2+ + 12H_2 O The 298 K value of E degree_cell and the change in free energy for this cell is best represented by which of the following expressions? E degree_cell = 0.739 V and Delta G degree = -0.739F E degree_cell = 0.739 V and Delta G degree = -1.48F E degree_cell = 0.739 V and Delta G degree = -3.70F E degree_cell = 2.28 V and Delta G degree = +0.739F E degree_cell = 2.28 V and Delta G degree = +3.70F Now use the concentrations given in the line notation to find the 298 K value of E_cell for this electrochemical cell? E_cell = 2.27 V E_cell = 2.20 V E_cell = 0.739 V E_cell = 0.731 V E_cell = 0.657 V what is the value of the equilibrium constant (K) for this cell at 298K? K = e^356 K = e^-143 K = e^143 K = e^71.3 K = e^28.7

Explanation / Answer

24) The cell reaction would be,

e.

25) For the cell

Eo = 1.50-0.771 = 0.739 V

dGo = -nFEo

Answer : c. Eo = 0.739 V, dGo = -3.70F

26) Ecell = Eo - 0.0592/n logK

               = 0.739 - 0.0592/5 [(0.2/0.3)^6/(0.1)^8]

Answer : e. Ecell = 0.657 V

27) Equilibrium constant

nFEo = RTlnKeq

Keq = e^(5 x 96500 x 0.739/8.314 x 298)

Answer : c. e^143

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