Consider the electrochemical cell shown using line notation and the standard red
ID: 1068998 • Letter: C
Question
Consider the electrochemical cell shown using line notation and the standard reduction potentials in the table below: Pt l Fe^2+ (0.30M), Fe^3+ (0.20M) Il 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_30^+ rightarrow 5Fe^3+ + Mn^2+ + 12H_20 5Fe^3+ + Mn^2+ + 12H_20 rightarrow 5Fe^2+ + MnO_4^-1 + 8H_30+ Fe^3+ + Mn^2 + 12H_20 rightarrow Fe^2+ + MnO_4^-1 + 8H_30^+ Fe^2+ + MnO_^-1 + 8H_30^+ rightarrow Fe^3+ + Mn^2+ + 12H_20 5Fe^2+ + MnO_4^-1 + 8H_3O^+ rightarrow 5Fe^3+ + Mn^2+ + 12H^2O The 298 K value of E degree_cell is ______ and the change free energy for this cell is best represented by which of the following expressions? E degree_cell = 0.739 V and deltaG 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^3.56 K = e^-143 K = e^143 K = e^71.3 K = e^28.7Explanation / Answer
24)
from half reaction given,
answer:e
25) E0cell = E0cathode - E0anode
cathode : Mno4-/Mn2+ electrode
anode : Fe2+/Fe3+ electrode
E0cell = 1.51 - 0.771
= 0.739 v
DG0 = -nFE0cell
n = no of electrons transfered = 5
F = faraday constant = 96500 C
= -5*F*0.739
= -3.7F
answer: C
26) from answer , in q.no : 24
Ecell = E0cell - 0.0591/nlog([Mn2+][Fe3+]^5/[H+]^8[MnO4-][Fe3+]^5)
= 0.739 -(0.0591/5)log(0.2*0.2^5/(0.1^8*0.3*0.3^5))
= 0.657 V
answer: e
27) DG0 = -RTlnk
-3.7*96500 = -8.314*298lnk
lnk = 144
k = e^144
answer: c