Consider the electrochemical cell shown using line notation and the standard red
ID: 1068979 • Letter: C
Question
Consider the electrochemical cell shown using line notation and the standard reduction potentials in the table below: Pt|Fe^2+(0.30M), Fe^3+(0.20M) || MnO_4^-1 (0.30M), H^+(0.10M), Mn^2+(0.20M)|Pt Which of the following balanced chemical equations corresponds to this line notation? Pt + 5Fe^2+ + MnO_4^-1 + H_3O rightarrow 5Fe^3+ + Mn^2+ + 12H_2O 5Fe^3+ + Mn^2+ + 12H_2O rightarrow 6Fe^+ + MnO_4^-3 + 8H_3O Fe^3+ + Mn^2+ 12H_2O rightarrow Fe^2+ + MnO_4^-1 + 8H_3O^+ Fe^2 + MnO_4^-1 + 8H_3O_2 rightarrow Fe^3+ Mn^2+ + 12H_2O 5Fe^2+ + MnO_4^-1 + 8H_3O^+ + Mn^2+ + 12H_2O The 298 K value of K degree _cell is ____________ 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 = -0.739 F E degree _cell = 0.739 V and delta G degree = -1.48 F E degree _cell = 0.739 V and delta G degree = -3.70 F E degree _cell = 2.28 V and delta G degree = +0.739 F E degree _cell = 2.28 V and delta G degree = +3.70 F Now use the concentrations given in the line notation to find the 298 K value of E_cell for this eletrochemical cell? E_cell = 2.27 V E_cell = 2.20 V E_cell = 0.739 V E_cell =02.731 V E_cell = 0.657 V What is the value of the equilibrium constant (K) for this cell at 298 K? K = e^356 K = e^143 K = e^143 K = e^71.3 K = e^28.7Explanation / Answer
(24) (d)
Because for every 1 mol MnO4- i(it gains 5 electrons)t need 5 mol Fe2+(it loses only one electron)
(25) (c)
E0cell = E0cathode - E0anode = E0R - E0L = 1.51 - 0.771 = 0.739 V
deltaG0 = - n F E0cell = - 5 * F * 0.739 = - 3.70 F
(26) (d)
Ecell = E0cell - (0.05916/n)Log[Fe3+]5[Mn2+]/[Fe2+]5[MnO4-][H+]8
Ecell = 0.739 - (0.05916 / 5) Log(0.20)5(0.20)/(0.30)5(0.30)(0.1)8
Ecell = + 7.31
(27) (c)
lnK = n F E0cell / R T = 5 * 96500 * 0.739 / (8.314*298)
K = e^143