Calculate E°cell (in volts) for a fuel cell that employs the reaction between methane gas and oxygen to form carbon dioxide and water. (to 4 significant figures)
∆G⁰f for CH4 = -50.5 kJ/mol
∆G⁰f for CO2 = -394.4 kJ/mol
F = 96485/mol e-
I got 0.4455 V when i tried.
0.3
To calculate the standard cell potential (E°cell) for the fuel cell, we can use the equation:
ΔG° = -nF E°cell
Where:
ΔG° is the standard Gibbs free energy change
n is the number of moles of electrons transferred
F is Faraday's constant (96485 C/mol e-)
E°cell is the standard cell potential
The balanced chemical equation for the reaction is:
CH4 + 2O2 -> CO2 + 2H2O
From the balanced equation, we can see that 8 moles of electrons are transferred because each mole of CH4 is oxidized to produce 4 moles of H2O.
ΔG° is calculated using the formula:
ΔG° = ΔG°f (products) - ΔG°f (reactants)
ΔG°f for CH4 = -50.5 kJ/mol
ΔG°f for CO2 = -394.4 kJ/mol
ΔG° = 1*(-394.4 kJ/mol + 2*(0 kJ/mol)) - (-50.5 kJ/mol + 2*(0 kJ/mol))
ΔG° = -394.4 kJ/mol + 50.5 kJ/mol
ΔG° = -343.9 kJ/mol
Plugging the values into the equation ΔG° = -nF E°cell:
-343.9 kJ/mol = -8*(96485 C/mol e-) * E°cell
Solving for E°cell:
E°cell = -343.9 kJ/mol / (-8*(96485 C/mol e-))
E°cell ≈ 0.4454 V (rounded to 4 significant figures)
Therefore, the correct value for E°cell is 0.4454 V.
To calculate the standard cell potential (E°cell) for the given fuel cell reaction, we can use the equation:
ΔG° = -nFE°cell
Where:
ΔG° is the standard Gibbs free energy change
n is the number of electrons transferred in the reaction
F is the Faraday constant (96485 C/mol e-)
E°cell is the standard cell potential
The balanced chemical equation for the reaction is:
CH4 + 2O2 -> CO2 + 2H2O
From the equation, we can see that 8 electrons are transferred, so n = 8.
Now let's calculate the standard Gibbs free energy change (ΔG°) using the standard Gibbs free energy of formation (∆G°f) values:
ΔG° = ∑ΔG°f(products) - ∑ΔG°f(reactants)
For the given reaction, we have:
∆G°f for CH4 = -50.5 kJ/mol
∆G°f for CO2 = -394.4 kJ/mol
Plugging in the values, we get:
ΔG° = [∆G°f(CO2) + 2∆G°f(H2O)] - [∆G°f(CH4) + 2∆G°f(O2)]
= [-394.4 kJ/mol + 2(0 kJ/mol)] - [-50.5 kJ/mol + 2(0 kJ/mol)]
= -394.4 kJ/mol + 50.5 kJ/mol
= -343.9 kJ/mol
Now, substituting the values into the equation ΔG° = -nFE°cell:
-343.9 kJ/mol = -8 * (96485 C/mol e-) * E°cell
Simplifying, we get:
E°cell = (-343.9 kJ/mol) / (-8 * 96485 C/mol e-)
Converting kJ to volts by dividing by 96.485 kJ/V, we get:
E°cell = (-343.9 kJ/mol) / (-8 * 96485 C/mol e-) / 96.485 kJ/V
Calculating, the value of E°cell is approximately 0.445 V (rounded to four significant figures).
So, you were correct in obtaining a cell potential of 0.4455 V when you tried.