19) Use the tabulated half-cell potentials below to calculate AG°for the followi
ID: 696423 • Letter: 1
Question
19) Use the tabulated half-cell potentials below to calculate AG°for the following balanced redox reaction 19) The reaction is a 2 electron transfer Pb2+(aq) + Cu(s) Pb(s) +Cu2+(aq) E" =-0.13 V Cu2+(aq) + 2 e-Cu(s) Ee = 034 V Pb2+(aq) +2e Pb(s) A) 91 k B) 46 k D) -0.47 kJ 20) Calculate the standard cell potential for a reaction that has a equilibrium constand (K) of 8.9 x 1016 The reaction occurs at 298 Kelvin and is a 6 electron transfer redox reaction. A) 0.17 V B) 4.4x 1018v ) 1.00 V D) 3.8 x 1014vExplanation / Answer
Answer 19 Pb2+(aq) + Cu(s) -> Pb(s) + Cu2+(aq).
Step 1: Half-reaction
Cu(s) -----------> Cu2+ + 2e- Eoox = -Eored = -0.34 V
Pb2+ + 2e- (aq) ------> Pb(s) Eored = -0.13
Step 2: Find E*cell
Eocell = Eoox + Eored = -0.34 V + (-0.13 V) = -0.47 V
Step 3: Solve for delta G:
Delta Go = -n * F* Eocell = -2 mol * (96485 C / mol )* (-0.47 J/C) = 90695.9J = 9.07 x 104J = 91.00 x 103
1 KJ =1000 J
Delta Go = 91 KJ
Therefore, Correct answer is (A)
Answer 20
G =-R*T* lnKeq
Keq = equilbrium constant,
R = gas constant = 8.314 J/K.mol
Temperature = T = 298 K
G = 8.314 x 298 x ln(8.9*1016) = 96.6932 KJ/mol
G = -n*F*E°cell
G = (-6 e-)*(96.458 KJ/mol e-/V)*E°cell
96.6932 /578.748 = E°cell
E°cell = 0.167 V or 0.17 V
Correct option is (A)
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