Consider the electrochemical cell shown using line notation and the standard red

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^28.7

Explanation / Answer

Q24

the Pt will not be SHOWN! since those are only indications of Pt

so:

Fe+2 = Fe+3

MnO4- = H+Mn+2

this is not balanced

so

ignore Fe+3 in reactants... that is, b and c

d is not balanced

therefore, choose E

Q25

Etotal:

Ered - Eox

Fe is being oxidized, so we need 0.771

MnO4- reduces so we need 1.51

Ecell = 1.51-0.771 = 0.739 V

so

dG = nF*Ecell = -5*F*0.739

dG = -3.695F

choose C

Q26

Ecell = Eªcell -0.0592/n*log(Q)

Q= [Fe+3]^5 [Mn+2][H3O+] / ([Fe+2]^5 * [MnO4-] )

Q = (0.2^5)(0.2)(0.1^8)/ ((0.3^5)*(0.3)) = 0.000000000877915

Ecell = 0.739 - 0.0592/5*log(0.000000000877915)= 0.8462 V

nearest answer is C

Q27

dG = -RT*ln(K)

dG = -nF*Ecell

K = exp(nF/(RT)*Ecell)

K = exp(5*96500/(8.314*298)*0.73)

K = exp 356

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