Asked by bekah
Construct a galvanic (voltaic) cell from the half reactions shown below.
Au3+ + 3 e- → Au (s) ξo= 1.420 V
Br2 (l) + 2 e-→ 2 Br- (aq) ξo= 1.087 V
Calculate the equilibrium constant (K) for this cell. Assume that the temperature is 298K.
First, consider what happens to the Nernst equation when the system reaches equilibrium:
ξ = ξo - (RT/nF) ln Q
What is ξ when the system has reached equilibrium? Is there any driving force left?
Remember that ∆G = -nFξ.
At equilibrium, ∆G = 0, so ξ = 0, too.
At equilibrium, Q = K, since all reactants and products are present at their equilibrium concentrations.
This simplifies the Nernst equation to:
ξocell = [(RT)/nF] ln K
Now calculate a value for K.
You may need to use scientific notation to record your answer. If so, enter it as: 4E43 which is equivalent to 4x1043
Au3+ + 3 e- → Au (s) ξo= 1.420 V
Br2 (l) + 2 e-→ 2 Br- (aq) ξo= 1.087 V
Calculate the equilibrium constant (K) for this cell. Assume that the temperature is 298K.
First, consider what happens to the Nernst equation when the system reaches equilibrium:
ξ = ξo - (RT/nF) ln Q
What is ξ when the system has reached equilibrium? Is there any driving force left?
Remember that ∆G = -nFξ.
At equilibrium, ∆G = 0, so ξ = 0, too.
At equilibrium, Q = K, since all reactants and products are present at their equilibrium concentrations.
This simplifies the Nernst equation to:
ξocell = [(RT)/nF] ln K
Now calculate a value for K.
You may need to use scientific notation to record your answer. If so, enter it as: 4E43 which is equivalent to 4x1043
Answers
Answered by
Anonymous
Use second equation
1.420- (-1.087)= .333
Use that as cell potential.
R=8.314
T=298
n= 6 mols electrons because you have to multiply the two reactions to get 6 electrons so you can cancel the electrons out
F=96,500
.333= 0.004279lnQ
divide both sides by .00427 to cancel it from the right side
.333/.00427=77.82
77.82=lnQ
enter in calculator 2nd and ln to get e^ (77.82) because you want to get ride of ln from the right side so its pushed to the left side as its inverse of e^ (77.82) and ln cancels out so youre left with Q.
1.420- (-1.087)= .333
Use that as cell potential.
R=8.314
T=298
n= 6 mols electrons because you have to multiply the two reactions to get 6 electrons so you can cancel the electrons out
F=96,500
.333= 0.004279lnQ
divide both sides by .00427 to cancel it from the right side
.333/.00427=77.82
77.82=lnQ
enter in calculator 2nd and ln to get e^ (77.82) because you want to get ride of ln from the right side so its pushed to the left side as its inverse of e^ (77.82) and ln cancels out so youre left with Q.
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